Chapter 17 IB Maths HL Cambridge Textbook

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17 Basic integration and its applications 569

17

As in many areas of mathematics as soon as we learn a newprocess we must then learn how to undo it However it turnsout that undoing the process of differentiation opens up thepossibility of answering a seemingly unconnected problemwhat is the area under a curve

17A Reversing differentiationWe saw in the last chapter how differentiation gives us thegradient of a curve or the rate of change of one quantity withanother What then if we already know the function describinga curversquos gradient or the expression for a rate of change andwish to 1047297nd the original function Our only way of proceedingis to lsquoundorsquo the differentiation that has already taken place andthis process of reverse differentiation is known as integration

Basic

integrationand itsapplications

Introductory problem

Te amount of charge stored in a capacitor is given by thearea under the graph of current (I ) against time (t ) Whenit contains alternating current the relationship between I and t is given by When it contains direct currentthe relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whetheralternating or direct current is used

In this chapter youwill learn

to reverse the processbullof differentiation (this

process is calledintegration)

to find the equationbullof a curve given itsderivative and a pointon the curve

to integrate sinbull x cos x and tan x

to integratebull ex and 1

x to find the areabullbetween a curve andthe x - or y -axis

to find the areabullenclosed between twocurves

copy Cambridge University Press 2012

Not for printing sharing or distribution



570 Topic 6 Calculus

Let us look at two particular cases to get a feel for this process

Each time we will be given and need to answer the question

lsquoWhat was differentiated to give thisrsquo

If then the original function y must have contained

x 2 as we know that differentiation decreases the power by 1Differentiating x 2 gives exactly 2x so we have found that ifd y

x x = then 2

If12 then the original function y must have contained x

3

2

Differentiating32 will give

12 and we do not

want the3

However if we multiply the3

2 by then when we

differentiate the coeffi cient cancels to 1 so if x 1

2

then2 3

2

Writing out lsquoifd

d

y

x

1

2 then y x 32 is descriptive but rather

laborious and so the notation used for integration is

1 32

Here the dx simply states that the integration is taking place

with respect to the variable x in exactly the same way that in y

it states that the differentiation is taking place with respect to x

We could equally well write for example int 1 3

2

The integration symbolcomes from the oldEnglish way ofwriting the letter lsquoSrsquoOriginally it stood for theword lsquoSumrsquo (or rather m)

As you will see in latersections the integral doesindeed represent a sum ofinfinitesimally small quantities

Exercise 17A1 Find a possible expression for y in terms of x

(a) (i) 2 (ii) 4

(b) (i) = minus1

2 (ii)

x x = minus

5

You may have heardof the term lsquodifferentialequationrsquo These arethe simplest types ofdifferential equation






(c) (i)d

d

y

x x

1 (ii)

d

d

y

x x

12

(d) (i)x

10 4 (ii) 1 2

17B Constant of integration

We have seen how to integrate some functions of the form x n byreversing the process of differentiation but the process as carriedout above was not complete

Let us consider again the 1047297rst example where we stated that

int 2 2

Were there any other possible answers here

We could have given int 2 or

int = minus3

2x x x

Both of these are just as valid as our original answer we know

that when we differentiate the constant ( +1 or minus ) we just

get 0 We could therefore have given any constant withoutfurther information we cannot know what this constant on the

original function was before it was differentiatedHence our complete answers to the integrals considered inSection 17A are

int =2

x c1 32

3

where the c is an unknown constant of integration

We will see later that given further information we can 1047297nd

this constant

Exercise 17B

1 Give three possible functions which when differentiatedwith respect to x give the following

(a) 3

(b) 0

We will see how to 1047297nd the constant

of integration in

Section 17F






2 Find the integrals

(a) (i) 7 (ii) int 1

x x

(b) (i)1

2 (ii)

3

17C Rules of integration

o 1047297nd integrals so far we have used the idea of reversingdifferentiation for each speci1047297c function Let us now thinkabout applying the reverse process to the general rule ofdifferentiation

We know that for n n nminus1 or in words

o differentiate x n multiply by the old power then decrease thepower by 1

We can express the reverse of this process as follows

o integrate x n increase the power by 1 then divide by thenew power

Using integral notation

KEY POINT 171

Te general rule for integrating x n for any rational powern ne minus1 is

x c+int 1

1

Note the condition n ne minus1 which ensures that we are notdividing by zero

It is worth remembering the formula below for integrating aconstant = +k x c which is a special case of the rule inKey point 171

1

In Key point 163 we saw that if we differentiate kf )x we getkf prime( )x

we can reverse this logic to show that

KEY POINT 172

o integrate multiples of functions

We will see how

to integrate x ndash1 inSection 17D

T he + c is a par t o f

t he ans wer and you

mus t wri te i t e ver y

time

e x a m h i n t

T his ru le on l y wor ks

i f k is a cons tan t

e x a m h i n t






Since we can differentiate term by term (also in Key point 164)then we can also split up integrals of sums

KEY POINT 173

For the sum of integrals

int int +

By combining Key points 172 and 173 with k = minus1 we canalso show that the integral of a difference is the difference of theintegrals of the separate parts

Tese ideas are demonstrated in the following examples

Be warned You

canno t in tegra te

produc ts or quo tien ts

b y in tegra ting eac h

par t separa te l y

e x a m h i n t

Worked example 171

Find (a) int x x (b) int minus

( )+4

minus x d

Add one to the power anddivide by this new power

(a) 6xint minus

minus c x = +xminus

Tidy up

= minus

22

x

= minus +minus2

x

Go through term by termadding one to the powerof x and dividing by this

new powerRemember the rule forintegrating a constant

(b) 3 3 8

4

1x x+x minus +x

minus +minus minus+ +

int c

Tidy up

=minus

+minus

5 1xminus c

= + +

minus 1

3x c

Just as for differentiation it may be necessary to change termsinto the form n before integrating






Exercise 17C

1 Find the following integrals

(a) (i) int 9x x (ii) int 12x x

(b) (i) int (ii) int (c) (i) 9 (ii) int

1dx

(d) (i) (ii)

(e) (i) int x x (ii) int 33 x x

(f) (i)2

(ii)3


(a) (i) int 3 (ii) int 7 z

(b) (i) int q (ii) int r r

In t he in tegra l do no t

forge t t he d x or t he

equi va len t We wi l l

ma ke more use o f i t

la ter T he func tion

you are in tegra ting

is ac tua l l y being

mu l tip lied b y d x

so you cou ld

wri te ques tion

1( f )(ii ) as int2 x

e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d

Write the cube root as a powerand use rules of exponents

(a)

int xx

= int x x

Dividing by10

3

7

31+ is the

same as multiplying by3

10

Expand the brackets first thenuse rules of exponents

(b) int +

x

x minus

Dividing by a fraction isthe same as multiplying by

its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57

(d) (i) int 2h (ii) int

d p

p4

3 Find the following integrals (a) (i) int +x minus x 2 d (ii) int x minus x + d

(b) (i) int +1 1

4d (ii) int times minus times

1 15

v d

(c) (i) int x x x d (ii) int 3

(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]

17D Integrating x ndash1 and ex

When integrating int +n

1

1 we were careful to exclude

the case n = minus1

In Key point 168 we saw that ( )n Reversing this gives

KEY POINT 174

int minusx c=x

In Key point 167 we saw that x e( We can use this to

integrate the exponential function

KEY POINT 175

e

We will modify

this rule in Section17H






Exercise 17D 1 Find the following integrals

(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x

(c) (i) int minus x 2 1

d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2


(a) (i) int ex (ii) int ex x

(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +

17E Integrating trigonometric functions

We can expand the set of functions that we can integrate bycontinuing to refer back to work covered in chapter 16

We saw in Key point 166 that (s x which means

that

Similarly as ( then int si minus

KEY POINT 176

Te integrals of trigonometric functions

int s minus s

=

We do not have a function whose derivative is ta and so

have no way (yet) of 1047297nding ta We will meet a method that

enables us to establish this in chapter 19 but for completeness

the result is given here

KEY POINT 176a

tan

See Exercise 19B

for establishing this

result

T he in tegra l o f

tan x is no t gi ven

in t he Formu la

boo k le t and is wor t h

remem bering

e x a m h i n t






Exercise 17E


(a) (i) int s n x cminus os x d (ii) x s n x d

(b) (i) int t n (ii)

sin

2 3

(c) (i)7

(ii)

(d) (i) int minus x s n x d (ii) cos minus x sminus ncos x

2 Find int s n

cos

x cos

x x d [5 marks]

3 Find int cos x sminus n [5 marks]

17F Finding the equation of a curve

We have seen how we can integrate the function y

to 1047297nd the

equation of the original curve except for the unknown constant

of integration Tis is because the gradientd y

x determines the

shape of the curve but not exactly where it is However if weare also given the coordinates of a point on the curve we can

then determine the constant and hence specify the originalfunction precisely

If we again consider x 2 which we met at the start of this

chapter we know that the original function must have equation y = x 2 + c for some constant value c

If we are also told that the curve passes through the point(1 minus1) we can 1047297nd c and specify which of the family of curvesour function must be

(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked

example 162 where given the gradient

we could draw many

different curves bychanging the starting

point






Te above example illustrates the general procedure for 1047297ndingthe equation of a curve from its gradient function

KEY POINT 177

o 1047297nd the equation for y given the gradientx

and onepoint ( p q) on the curve

1 Integrate remembering +c

2 Find the constant of integration by substitutingx = p y = q

Exercise 17F

1 Find the equation of the original curve if

(a) (i) and the curve passes through (ndash2 7)

(ii) x = 2 and the curve passes through (0 5)

(b) (i)d

d

y

x x

1 and the curve passes through (4 8)

(ii)1


(c) (i) y

= + and the curve passes through (1 1)

(ii)d y

x = e and the curve passes through (ln5 0)

Worked example 173

Te gradient of a curve is given by = + and the curve passes through the point

(1 ndash4) Find the equation of the curve

To find y from ddy x

we need tointegrate

Donrsquot forget + c

= + 5xminus d

= +minus +

The coordinates of the givenpoint must satisfy this

equation so we can find c

When x = = minus so

minus ) + ) += 53 2

c

rArr minus minus + == 4 + 6c rArr

there4 = minusx minus x5+ 6






(d) (i)+

and the curve passes through (e e)

(ii)1

and the curve passes through (e2 5)

(e) (i)d y

x x = x and the curve passesthrough (π 1)

(ii) x 3ta and the curve passes through (0 4)

2 Te derivative of the curve y f )x is1

(a) Find an expression for all possible functions f(x)

(b) If the curve passes through the point (2 7) 1047297nd theequation of the curve [5 marks]

3 Te gradient of a curve is found to be minus2

(a) Find the x -coordinate of the maximum point justifyingthat it is a maximum

(b) Given that the curve passes through the point (0 2)show that the y -coordinate of the maximum point

is minus7 [5 marks]

4 Te gradient of the normal to a curve at any point isequal to the x -coordinate at that point If the curve passesthrough the point (e2 3) 1047297nd the equation of the curvein the form ( ) where is a rationalfunction x gt 0 [6 marks]

17G Definite integration

Until now we have been carrying out a process known as

inde1047297nite integration inde1047297nite in the sense that we have anunknown constant each time for example

1

However there is also a process called de1047297nite integration which yields a numerical answer without the involvementof the constant of integration for example

bb

3 3 3

3 3 minus

int

Here a and b are known as the limits of integration a is the

lower limit and b the upper limit






Te square bracket notation means that the integration hastaken place but the limits have not yet been applied o do thiswe simply evaluate the integrated expression at the upper limitand subtract the integrated expression evaluated at the lowerlimit

You may be wondering where the constant of integration hasgone We could write it in as before but we quickly realise thatthis is unnecessary as it will just cancel out at the upper andlower limit each time

b c

bb

3

3

1

1

1

Te value of x is a dummy variable it does not come intothe answer But both a and b can vary and affect the resultChanging x to a different variable does not change the answerFor example

u a x bb

31 1

int

Worked example 174

Find the exact value of1

1+

Integrate and write in squarebrackets x

+ ]xint

Evaluate the integratedexpression at the upper

and lower limits andsubtract the lower from

the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you

kno w ho w to e va lua te

de fini te in tegra ls on your

ca lcu la tor as e xp lained

on Ca lcu la tor s ki l ls s hee t

1 0 on t he C D - R OM

I t can sa ve you time

and you can e va lua te

in tegra ls you donrsquo t kno w

ho w to do a lge braica l l y

E ven w hen you are

as ked to find t he e xac t

va lue o f t he in tegra l you

can c hec k your ans wer

on t he ca lcu la tor

e x a m h i n t






Exercise 17G

1 Evaluate the following de1047297nite integrals giving exact answers

(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint

2 Evaluate correct to three signi1047297cant 1047297gures

(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne

1int 13 Find the exact value of the integral e x int

π [5 marks]

4 Show that the value of the integral12

x k

k

is independentof k [4 marks]

5 If x ) 73

evaluate 13

[4 marks]

6 Solve the equation 1 [5 marks]

y

x

y = f (x)

a b

A

17H Geometrical significance of definiteintegration

Now we have a method that gives a numerical value for anintegral the natural question to ask is what does this numbermean

On Fill-in proof sheet 20 on the CD-ROM Te fundamentaltheorem of calculus we show that as long as f )x is positivethe de1047297nite integral of f )x between the limits a and b is thearea enclosed between the curve the x -axis and the lines x = a and x = b

KEY POINT 178

rea )x b






If you are sketching the graph on the calculator you can get it toshade and evaluate the required area see Calculator skillssheet 10 on the CD-ROM You need to show the sketch as a partof your working if it is not already shown in the question

Worked example 175

Find the exact area enclosed between the x -axis the curve y = sin x and the lines x = 0 and x = π

Sketch the graph and identifythe area required = in

3

Integrate and write in squarebrackets

A = ]= minus

0

3ππ

Evaluate the integratedexpression at the upper andlower limit and subtract the

lower from the upper

= minus )cos cminus minus os

= minus minus ( )minus =






The Ancient Greekshad developed ideasof limiting processessimilar to those used

in calculus but it took nearly2000 years for these ideas

to be formalised This wasdone almost simultaneouslyby Isaac Newton andGottried Leibniz in the 17thCentury Is this a coincidenceor is it often the case that along period of slow progressis often necessary to get tothe stage of majorbreakthroughsSupplementary sheet 10

looks at some other peoplewho can claim to haveinvented calculus

In the 17th Century integration was defined as thearea under a curve The area was broken downinto small rectangles each with a height f (x ) and a

width of a small bit of x called ∆x The total areawas approximately the sum of all of these rectangles

x a

x

f x x sum )∆

Isaac Newton one of the pioneers of calculus was also abig fan of writing in English rather than Greek So sigmabecame the English letter lsquoSrsquo and delta became the Englishletter d so that when the limit is taken as the width of therectangles become vanishingly small then the expressionbecomes

x a

)x dint This illustrates another very important interpretation of

integration ndash the infinite sum of infinitesimally small parts

Worked example 176

Find the area A in this graphy

x

y = x(xminus1)(2minusx)

1 2A

0

Write down the integral to beevaluated then use calculator

x x xminusint 1

x = minus y D

The area must be positive there4 =

When the curve is entirely below the x -axis the integral will givea negative value Te modulus of this value is the area






Unfortunately the relationship between integrals and areas isnot so simple when there are parts of the curve above and belowthe axis Tose bits above the axis contribute positively to thearea but bits below the axis contribute negatively to the areaWe must separate out the sections above the axis and belowthe axis

Worked example 177

(a) Find x x 1

4

x int 1 d

(b) Find the area enclosed between the x -axis the curve y minus x + and the lines= 1 and 4

Apply standard integration (a) x x

4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4

The value found above canrsquot bethe correct area for (b)

Sketch the curve to see exactlywhich area we are being asked

to find

(b) y

2 minus + 3

1 3






Te fact that the integral was zero in Worked example 177 part(a) means that the area above the axis is exactly cancelled by thearea below the axis

Tis example warns us that when asked to 1047297nd an area we mustalways sketch the graph and identify exactly where each partof the area is If we are evaluating the area on the calculator wecan use the modulus function to ensure that the entire graph isabove the x -axis Using the function from Worked example 177

1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued

The area is made up of twoparts so evaluate each of

them separately

x

1

33

xminusint

( ) minus = minus

there4 rea below the axis is

x x3

4

2xminus int

minus ( ) =

4

there4 rea above the axis is

Total area = + =3 3

Transformationsof graphs using the

modulus function

were covered inchapter 7






KEY POINT 179

Te area bounded by the curve y f )x the x-axis and the

lines x = a and x = b is given byb

x int

When working without a calculator if the curve crosses

the x -axis between a and b we need to split the area intoseveral parts and 1047297nd each one separately

Te interpretation of integrals as areas causes one inconsistency

with our previous work Consider the integral1

2

1

x dminus

int

Graphically we can see that this area should exist

x

y

minus1minus2

However if we do the integration we 1047297nd that

12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n

Tis is the correct answer (which we could have found usingthe symmetry of the curve) but it goes through a stage wherewe had to take logarithms of negative numbers and this issomething we are not allowed to do We avoid this by rede1047297ning

the integral ofx

as






KEY POINT 174 AGAIN

minusint =

With this de1047297nition we can integrate y =1

over negative

numbers and the integral above becomes

12

1

2

1

x

e oreminus

minusint

n

Notice that the answer is negative since the required area is below

the x -axis We can still not integrate1

with a negative lower and

positive upper limit since the graph has an asymptote at x = 0

You may rightly be alittle uncomfortablewith inserting amodulus function lsquojust

because it worksrsquo Inmathematics do the ends

justify the means

Exercise 17H

1 Find the shaded areas

(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1






(c) (i)y = x3 minus x

y

xminus1 2

(ii)y = x2minus3x

y

x

5

2 Te area enclosed by the x -axis the curve andthe line x = k is 18 Find the value of k [6 marks]

3 (a) Find3

int (b) Find the area between the curve minus2 and the

x -axis between x = 0 and x = 3 [5 marks]

4 Between x = 0 and x = 3 the area of the graph y x minus2 below the x -axis equals the area above thex -axis Find the value of k [6 marks]

5 Find the area enclosed by the curve 1x minus minus 0

and the x -axis [7 marks]

lsquo Find t he area

enc losedrsquo means firs t

find a c losed region

bounded b y t he

cur ves men tioned

t hen find i ts area

A s ke tc h is a ver y

use fu l too l

e x a m h i n t

17I The area between a curve andthe y -axis

Consider the diagram alongside How can we 1047297nd the shadedarea A

One possible strategy is to construct a box around the graph todivide up the regions of interest You can integrate to 1047297nd thearea labelled A1 and then by adding and subtracting the areas ofthe blue and red rectangles shown calculate A

y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1






Happily there is a quicker way we can treat x as a functionof y effectively re1047298ecting the whole diagram in the line y = x and then use the same method as in the previous section

KEY POINT 1710

Te area bounded by the curve y the y -axis and the

lines y = c and y = d is given by y d

) dint where is

the expression for x in terms of y

You may have realised that this is related to inverse functions from Section 5E

x

y

x = f (y)

cd

A

Worked example 178

Te curve shown has equation y 1minus Find the shaded area

y

x

y = 2radic x minus 1

10

Express x in terms of y x = 2

rArr = + y






Exercise 17I

continued

Find the limits on the y -axisIt may help to label them on the

graph

When x = minusWhen x = 1 = =minus 6

y

y = 2radic minus

1

10

Write down the integral andevaluate using calculator

Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6






2 Te diagram shows the curve If the shadedarea is 504 1047297nd the value of a [6 marks]

y

x

y = radic

x

a

2a

0

3 Find the exact value of the area enclosed by the graphof + the line y = 2 and the y -axis [6 marks]

4 Te diagram shows the graph of

Te shaded area is 39 units Find the value of a [7 marks]y

x

y =radic x

4 a

5 Te diagram shows the graph of 2 where a isin infin Te area of the pink region is equal to the area of the blue

region Give two equations for a in terms of b and hencegive a in exact form and determine the size of theblue area [8 marks]y

x

y = x2

b

1 a

17J The area between two curves

So far we have only looked at areas bounded by a curve and oneof the coordinate axes but we can also 1047297nd areas bounded bytwo curvesTe area A in the diagram can be found by taking the areabounded by f )x and the x -axis and subtracting the areabounded by and the x -axis that is

A

b

minus

y

x

y = f (x)

y = g(x)

a b

A






We can do the subtraction before integrating so that we onlyhave to integrate one expression instead of two Tis gives analternative formula for the area

KEY POINT 1711

Te area A between two curves f (x ) and g (x ) is

Ab

(

where a and b are the x-coordinates of the intersectionpoints of the two curves

Worked example 179

Find the area A enclosed between + and minus2 +

First find the x -coordinates ofintersection

For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0

Make a rough sketch to see therelative positions of the two curves

y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x

Subtract the lower curve from thehigher before integrating

= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8






Subtracting the two equations before integrating is particularlyuseful when one of the curves is partly below the x-axis If x is always above then the expression we are integrating f g )x minus )x is always positive so we do not have to worryabout the signs of f )x and g )x themselves

Worked example 1710

Find the area bounded by the curves y x minus and y x 2

Sketch the graph to see therelative position of two

curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A

Find the intersection points ndash

use calculatorintersections x = minus2 658

Write down the integralrepresenting the area

rea =minus

minus818

1 658

xminus

= minus

1 6 8

x

Evaluate the integral usingcalculator = 21 6 (3SF)






Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)

y =minusx2minus4x+ 12

y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x

2 Find the area enclosed between the graphs of y x +x minus and + [6 marks]

3 Find the area enclosed by the curve 2

the y-axisand the line [6 marks]

4 Find the area between the curves1

and in theregion lt lt π [6 marks]

5 Show that the area of the shaded region alongside is2

[6 marks]

y = x2

y

xminus1 2






6 Te diagram alongside shows the graphs of and Find the shaded area [6 marks]

7 Find the total area enclosed between the graphs of

)minus 2 and minus2 7 1+ [6 marks]

8 Te area enclosed between the curve y x 2 and the line

y mx is 10 Find the value of m if m 0 [7 marks]

9 Show that the shaded area in the diagram below is [8 marks]

y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary

bull Integration is the reverse process of differentiation

bull Any integral without limits (inde1047297nite) will generate a constant of integration

bull For all rational n ne minus1 int +x c

1

1

bull If minus we get the natural logarithm function int minus =

bull Te integral of the exponential function is int e c

bull Te integrals of the trigonometric functions are

int si minus=

int tan =

bull Te de1047297nite integral has limits f( ) x b

dint is found by evaluating the integrated expression

at b and then subtracting the integrated expression evaluated at a






bull Te area between the curve y = f (x ) the x -axis and lines and x is given by

Ab

If the curve goes below the x-axis the value of this integral will be negative

bull On the calculator we can use the modulus function to ensure we are always integrating apositive function

bull Te area between the curve the y -axis and lines y c and y = d is given by A g y

bull Te area between two curves is given by

Ab

(

where x a and x b are the intersection points

Introductory problem revisited

Te amount of charge stored in a capacitor is given by the area under the graph ofcurrent (I ) against time (t ) When there is alternating current the relationship betweenI and t is given by t When it contains direct current the relationship between I and t is given by I = k What value of k means that the amount of charge stored in thecapacitor from t = 0 to t = π is the same whether alternating or direct current is used

Te area under the curve of I against t is given by sin cosint 0

ππ For a rectangle of

width π to have the same area the height must beπ

t

l

I = sin t

π

t

l

k

π

You can look at integration as a quite sophisticated way of finding an average value ofa function






Short questions

1 If ) = f prime x s n and

1047297nd x [4 marks]

2 Calculate the area enclosed by the curves and

[6 marks]

[copy IB Organization 2003]

3 Find the area enclosed between the graph of y minus2 2 and the x -axisgiving your answer in terms of k [6 marks]

4 Te diagram shows the graph of n for 1

y

xa b

0

Te red area is three times larger than the blue area Find the value of n [6 marks]

5 Find the inde1047297nite integral

int [5 marks]

6 (a) Solve the equation

a

int

(b) For this value of a 1047297nd the total area enclosed between the x -axis andthe curve y x minus3 for le a [6 marks]

Mixed examination practice 17





7 Find the area enclosed between the graphs of andfor lt lt π [3 marks]

8 (a) Te function )x has a stationary point at 1( ) and ) f primeprime +

What kind of stationary point is ( )1 [5 marks]

(b) Find f )x

Long questions

1 (a) Find the coordinates of the points of intersection of the graphsminus 2 and minus2 2

(b) Find the area enclosed between these two graphs

(c) Show that the fraction of this area above the axis is independentof a and state the value that this fraction takes [10 marks]

2 (a) Use the identity 1= to show that cos x arcs n x 2

(b) Te diagram below shows part of the curve

y

x

a

P

y = sinx

Write down the x -coordinate of the point P in terms of a

(c) Find the red shaded area in terms of a writing your answer in a formwithout trigonometric functions

(d) By considering the blue shaded area 1047297nd arcsin x x int for 1

[12 marks]




Let us look at two particular cases to get a feel for this process

Each time we will be given and need to answer the question

lsquoWhat was differentiated to give thisrsquo

If then the original function y must have contained

x 2 as we know that differentiation decreases the power by 1Differentiating x 2 gives exactly 2x so we have found that ifd y

x x = then 2

If12 then the original function y must have contained x

3

2

Differentiating32 will give

12 and we do not

want the3

However if we multiply the3

2 by then when we

differentiate the coeffi cient cancels to 1 so if x 1

2

then2 3

2

Writing out lsquoifd

d

y

x

1

2 then y x 32 is descriptive but rather

laborious and so the notation used for integration is

1 32

Here the dx simply states that the integration is taking place

with respect to the variable x in exactly the same way that in y

it states that the differentiation is taking place with respect to x

We could equally well write for example int 1 3

2

The integration symbolcomes from the oldEnglish way ofwriting the letter lsquoSrsquoOriginally it stood for theword lsquoSumrsquo (or rather m)

As you will see in latersections the integral doesindeed represent a sum ofinfinitesimally small quantities

Exercise 17A1 Find a possible expression for y in terms of x

(a) (i) 2 (ii) 4

(b) (i) = minus1

2 (ii)

x x = minus

5

You may have heardof the term lsquodifferentialequationrsquo These arethe simplest types ofdifferential equation






(c) (i)d

d

y

x x

1 (ii)

d

d

y

x x

12

(d) (i)x

10 4 (ii) 1 2




int 2 2



int = minus3

2x x x





int =2

x c1 32

3



this constant

Exercise 17B


(a) 3

(b) 0


of integration in

Section 17F







(a) (i) 7 (ii) int 1

x x

(b) (i)1

2 (ii)

3








KEY POINT 171


x c+int 1

1



1



KEY POINT 172


We will see how





time

e x a m h i n t



e x a m h i n t







KEY POINT 173


int int +



Be warned You

canno t in tegra te



par t separa te l y

e x a m h i n t

Worked example 171


( )+4

minus x d


(a) 6xint minus

minus c x = +xminus

Tidy up

= minus

22

x

= minus +minus2

x



(b) 3 3 8

4

1x x+x minus +x


int c

Tidy up

=minus

+minus

5 1xminus c

= + +

minus 1

3x c







Exercise 17C




1dx

(d) (i) (ii)


(f) (i)2

(ii)3












so you cou ld

wri te ques tion


e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d


(a)

int xx

= int x x

Dividing by10

3

7

31+ is the


10


(b) int +

x

x minus


its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




(c) (i)d

d

y

x x

1 (ii)

d

d

y

x x

12

(d) (i)x

10 4 (ii) 1 2




int 2 2



int = minus3

2x x x





int =2

x c1 32

3



this constant

Exercise 17B


(a) 3

(b) 0


of integration in

Section 17F







(a) (i) 7 (ii) int 1

x x

(b) (i)1

2 (ii)

3








KEY POINT 171


x c+int 1

1



1



KEY POINT 172


We will see how





time

e x a m h i n t



e x a m h i n t







KEY POINT 173


int int +



Be warned You

canno t in tegra te



par t separa te l y

e x a m h i n t

Worked example 171


( )+4

minus x d


(a) 6xint minus

minus c x = +xminus

Tidy up

= minus

22

x

= minus +minus2

x



(b) 3 3 8

4

1x x+x minus +x


int c

Tidy up

=minus

+minus

5 1xminus c

= + +

minus 1

3x c







Exercise 17C




1dx

(d) (i) (ii)


(f) (i)2

(ii)3












so you cou ld

wri te ques tion


e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d


(a)

int xx

= int x x

Dividing by10

3

7

31+ is the


10


(b) int +

x

x minus


its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





(a) (i) 7 (ii) int 1

x x

(b) (i)1

2 (ii)

3








KEY POINT 171


x c+int 1

1



1



KEY POINT 172


We will see how





time

e x a m h i n t



e x a m h i n t







KEY POINT 173


int int +



Be warned You

canno t in tegra te



par t separa te l y

e x a m h i n t

Worked example 171


( )+4

minus x d


(a) 6xint minus

minus c x = +xminus

Tidy up

= minus

22

x

= minus +minus2

x



(b) 3 3 8

4

1x x+x minus +x


int c

Tidy up

=minus

+minus

5 1xminus c

= + +

minus 1

3x c







Exercise 17C




1dx

(d) (i) (ii)


(f) (i)2

(ii)3












so you cou ld

wri te ques tion


e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d


(a)

int xx

= int x x

Dividing by10

3

7

31+ is the


10


(b) int +

x

x minus


its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





KEY POINT 173


int int +



Be warned You

canno t in tegra te



par t separa te l y

e x a m h i n t

Worked example 171


( )+4

minus x d


(a) 6xint minus

minus c x = +xminus

Tidy up

= minus

22

x

= minus +minus2

x



(b) 3 3 8

4

1x x+x minus +x


int c

Tidy up

=minus

+minus

5 1xminus c

= + +

minus 1

3x c







Exercise 17C




1dx

(d) (i) (ii)


(f) (i)2

(ii)3












so you cou ld

wri te ques tion


e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d


(a)

int xx

= int x x

Dividing by10

3

7

31+ is the


10


(b) int +

x

x minus


its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Exercise 17C




1dx

(d) (i) (ii)


(f) (i)2

(ii)3












so you cou ld

wri te ques tion


e x a m h i n t

Worked example 172

Find (a) 3

(b) int

( )minusx

2

d


(a)

int xx

= int x x

Dividing by10

3

7

31+ is the


10


(b) int +

x

x minus


its reciprocal

times10

10

3x c = +10

= minusminus xx 2

= + +5

95 3

minus times c

= ++5 3 1

minus c






(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




(c) (i) 13

(ii) int 57


d p

p4


(b) (i) int +1 1


1 15

v d


(d) (i) int ( 3 (ii) int )2

4 Find int 1

x [4 marks]



1


the case n = minus1


KEY POINT 174

int minusx c=x



KEY POINT 175

e

We will modify








(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





(a) (i) (ii)

(b) (i) int 1

x (ii) int

1

x


d (ii) int + x 3

d

(d) (i) int +

2x (ii) int

x x minusx 2



(b) (i)x

(ii)7 x

(c) (i) int + x

(ii) int +




that


KEY POINT 176


int s minus s

=





KEY POINT 176a

tan

See Exercise 19B


result

T he in tegra l o f


in t he Formu la


remem bering

e x a m h i n t






Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Exercise 17E




sin

2 3

(c) (i)7

(ii)


2 Find int s n

cos

x cos

x x d [5 marks]




to 1047297nd the



x determines the






(1minus1)

c = minus6

c = minus4

c = minus2

c = 4

c = 2

c = 0

y

x0

Look back to Worked


we could draw many


point







KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





KEY POINT 177





Exercise 17F




(b) (i)d

d

y

x x


(ii)1


(c) (i) y


(ii)d y


Worked example 173




we need tointegrate


= + 5xminus d

= +minus +



When x = = minus so

minus ) + ) += 53 2

c








(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




(d) (i)+


(ii)1


(e) (i)d y









is minus7 [5 marks]





1


bb

3 3 3

3 3 minus

int










b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]






b c

bb

3

3

1

1

1


u a x bb

31 1

int

Worked example 174


1+


+ ]xint



the upper

=(In (e) + 4 (e)) minus (In (1) + 4 (1))

= + minusminus =

Ma ke sure you










E ven w hen you are





e x a m h i n t






Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Exercise 17G


(a) (i) x x 2

(ii)1

4

1

(b) (i)2π

int (ii) sπ

π2

(c) (i) ex 1

int (ii) 31

1

eminusint


(a) (i)3

1 4

int (ii)1

int

(b) (i) e1

int (ii) ne


π [5 marks]


x k

k


5 If x ) 73

evaluate 13

[4 marks]


y

x

y = f (x)

a b

A




KEY POINT 178

rea )x b







Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





Worked example 175



3


A = ]= minus

0

3ππ
















x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]










x a

x

f x x sum )∆


x a



Worked example 176


x


1 2A

0


x x xminusint 1

x = minus y D









Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





Worked example 177

(a) Find x x 1

4

x int 1 d



4

xminus1

= ( ( ) ( ( )minus + minus +

= minus =3

4



to find

(b) y

2 minus + 3

1 3








1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]






1

4

int 1y

x

y = |x2 minus 4x + 3|

1 3 4

continued


them separately

x

1

33

xminusint

( ) minus = minus


x x3

4

2xminus int

minus ( ) =

4


Total area = + =3 3


modulus function







KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




KEY POINT 179



x int





2

1

x dminus

int


x

y

minus1minus2


12

1

2

1

]minus

minus

minus

minus

minus n minus

minusn

1

= ln

1

minus n


the integral ofx

as






KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




KEY POINT 174 AGAIN

minusint =


over negative


12

1

2

1

x

e oreminus

minusint

n







justify the means

Exercise 17H


(a) (i)y =x2

y

x1 2

(ii)y = 1

x2

y

x2 4

(b) (i) y =x2

minus4x+ 3

y

x1 2

(ii)y =x2

minus4

y

x1







y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





y

xminus1 2

(ii)y = x2minus3x

y

x

5


3 (a) Find3









bounded b y t he

cur ves men tioned



use fu l too l

e x a m h i n t




y

x

y = f (x)

c

d

A

y

x

y = f (x)

c

d A1







KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





KEY POINT 1710



) dint where is



x

y

x = f (y)

cd

A

Worked example 178


y

x


10


rArr = + y






Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Exercise 17I

continued


graph


y

y = 2radic minus

1

10


Area fro GDC= +2

1 2=d


(a) (i) y

x

y = x2

1

6

(ii) y

x

y = x3

1

3

(b) (i) y

xy = 1

x2

2

1

(ii) y

x

y =radic x

1

5

(c) (i)y

x

y = ln x

1e

e

(ii) y

x

y = 3x

1 6







y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





y

x

y = radic

x

a

2a

0




x

y =radic x

4 a



x

y = x2

b

1 a



A

b

minus

y

x

y = f (x)

y = g(x)

a b

A







KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





KEY POINT 1711


Ab

(


Worked example 179



For intersection

x x x

rArr =

rArr ) =rArr =

minus

minus

+x

4minusx 0


y

x1 4

A

2x + 1y =

2 minus 3x + 5y = x


= x

minusx x

4

minus

minus minusx

2

1

4

2

minus =

8







Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





Worked example 1710



curves

Using GDC

= ex minus 5

y = 3 2

y

minus2 2

A




rea =minus

minus818

1 658

xminus

= minus

1 6 8

x







Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Exercise 17J


(a) (i)

y = 2x+ 1

y = (xminus1)2

y

x

(ii) y =x+ 1

y = 4xminusx2minus1

y

x

(b) (i)


y

x

y =x2 + 2x+ 12

(ii)

y =x2minus2x+ 9

y

x

y = 4xminusx2 + 5

(c) (i)y =x2 minusx

y = 2xminusx2

y

x

(ii)y =x2minus7x+ 7

y = 3minusxminusx2

y

x







[6 marks]

y = x2

y

xminus1 2












y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]










y = 2 minus xy2 = x

y

x

y = cos x y = sin x

y

x

Summary




1

1




int si minus=

int tan =










Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]





Ab





Ab

(







t

l

I = sin t

π

t

l

k

π







Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]




Short questions


1047297nd x [4 marks]


[6 marks]




y

xa b

0



int [5 marks]


a

int










(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]






(b) Find f )x

Long questions






y

x

a

P

y = sinx




[12 marks]

Date post:	20-Feb-2018
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Chapter 17 IB Maths HL Cambridge Textbook

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