Chapter 5 .3 Riemann Sums and Definite Integrals

5.3 The Area Problem• We want to find the area of the region S

(shown on the next slide) that lies under the curve y = f(x) from a to b, where f is a continuous function with f(x) ≥ 0.

• But what do we mean by “area”?• For regions with straight sides, this

question is easy to answer:

Area Problem (cont’d)

• For (see the figures on the next slide)…– a rectangle, the area is defined as length

times width;– a triangle, the area is half the base times

the height;– a polygon, the area is found by dividing the

polygon into triangles and adding the areas.

• But what if the region has curved sides?

time

velocity

After 4 seconds, the object has gone 12 feet.

Consider an object moving at a constant rate of 3 ft/sec.

Since rate . time = distance:

If we draw a graph of the velocity, the distance that the object travels is equal to the area under the line.

ft3 4 sec 12 ft

sec

3t d

b

af x dx

IntegrationSymbol

lower limit of integration

upper limit of integration

integrandvariable of integration

(dummy variable)

It is called a dummy variable because the answer does not

depend on the variable chosen.

If f(x) is continuous and non-negative on [a, b], then the definite integral represents the area of the

region under the curve and above the x-axis between the vertical lines x = a and x = b

0

1

limn

k kP

k

f c x

b

af x dx

F b F a

Area

Area from x=0to x=1

Example: 2y x

Find the area under the curve from x=1 to x=2.

2 2

1x dx

23

1

1

3x

31 12 1

3 3

8 1

3 3

7

3

Area from x=0to x=2

Area under the curve from x=1 to x=2.

Example:

Find the area between the

x-axis and the curve

from to.

cosy x

0x 3

2x

2

3

2

3

2 2

02

cos cos x dx x dx

/ 2 3 / 2

0 / 2sin sinx x

3sin sin 0 sin sin

2 2 2

1 0 1 1 3

pos.

neg.

Note 2: is a number; it does not depend on x. In fact, we have

b

adxxf )(

b

a

b

a

b

adrrfdttfdxxf )()()(

Note 3: The sum is usually called a Riemann sum.

n

iii xxf

1

)(

Note 4: Geometric interpretations

For the special case where f(x)>0 ,

= the area under the graph of f from a to b. b

a

dxxf )(

In general, a definite integral can be interpreted as a difference of areas:

21)( AAdxxfb

a

x

y

o a b+ +

-

Solution We compute the integral as the difference of the areas of the two triangles:

5.1)1( 21

3

0

AAdxx

-11 3o

x

y

y=x-1 A1

A2

the rules for working with integrals, the most important of which are:

2. 0a

af x dx If the upper and lower limits are equal,

then the integral is zero.

1. b a

a bf x dx f x dx Reversing the limits changes

the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

1.

0a

af x dx If the upper and lower limits are equal,

then the integral is zero.2.

b a

a bf x dx f x dx Reversing the limits changes

the sign.

b b

a ak f x dx k f x dx 3. Constant multiples can be

moved outside.

b b b

a a af x g x dx f x dx g x dx 4.

Integrals can be added and subtracted.

b b b

a a af x g x dx f x dx g x dx 4.

Integrals can be added and subtracted.

5. b c c

a b af x dx f x dx f x dx

Intervals can be added(or subtracted).

a b c

y f x

5.7 The Fundamental Theorem

If you were being sent to a desert island and could take only one equation with you,

x

a

df t dt f x

dx

might well be your choice.

The Fundamental Theorem of Calculus, Part 1

If f is continuous on , then the function ,a b

x

aF x f t dt

has a derivative at every point in , and ,a b

x

a

dF df t dt f x

dx dx

x

a

df t dt f x

dx

First Fundamental Theorem:

1 .Derivative of an integral.

a

xdf t dt

xf x

d

2 .Derivative matches upper limit of integration.

First Fundamental Theorem:

1 .Derivative of an integral.

a

xdf t dt f x

dx

1 .Derivative of an integral.

2 .Derivative matches upper limit of integration.

3 .Lower limit of integration is a constant.

First Fundamental Theorem:

x

a

df t dt f x

dx

1 .Derivative of an integral.

2 .Derivative matches upper limit of integration.

3 .Lower limit of integration is a constant.

New variable.

First Fundamental Theorem:

cos xd

t dtdx cos x 1. Derivative of an integral.

2. Derivative matches upper limit of integration.

3. Lower limit of integration is a constant.

sinxd

tdx

sin sind

xdx

0

sind

xdx

cos x

The long way:First Fundamental Theorem:

20

1

1+t

xddt

dx 2

1

1 x

1. Derivative of an integral.

2. Derivative matches upper limit of integration.

3. Lower limit of integration is a constant.

2

0cos

xdt dt

dx

2 2cosd

x xdx

2cos 2x x

22 cosx x

The upper limit of integration does not match the derivative, but we could use the chain rule.

53 sin

x

dt t dt

dx The lower limit of integration is not a constant, but the upper limit is.

53 sin

xdt t dt

dx

3 sinx x

We can change the sign of the integral and reverse the limits.

The Fundamental Theorem of Calculus, Part 2

If f is continuous at every point of , and if F is

any antiderivative of f on , then

,a b

b

af x dx F b F a

,a b

(Also called the Integral Evaluation Theorem)

We already know this!To evaluate an integral, take the anti-derivatives and subtract.

Example 10:

24

0tan sec x x dx

The technique is a little different for definite integrals.

Let tanu x2sec du x dx

0 tan 0 0u

tan 14 4

u

1

0 u du

We can find new limits,

and then we don’t have to

substitute back.

new limit

new limit

12

0

1

2u

1

2We could have substituted back and used the

original limits.

Example 8:

24

0tan sec x x dx

Let tanu x

2sec du x dx4

0 u du

Wrong!The limits don’t match!

42

0

1tan

2x

2

21 1tan tan 0

2 4 2

2 21 11 0

2 2

u du21

2u

1

2

Using the original limits:

Leave the limits out until you substitute

back.

This is usually

more work than finding

new limits

Example 9 as a definite integral:

1 2 3

13 x 1 x dx

3Let 1u x 23 du x dx

11 3 22

1( 1) 3xx dx

133 2

1

2( 1)

3x

33 3/ 2 3/ 22(1 1) ( 1 1)

3

22 2

3 4 2

3

Rewrite in form of

∫undu

No constants needed- just integrate using

the power rule.

Substitution with definite integrals

7

0

4 3xdx4 3

3

3

u x

du dx

dudx

71 3 3 3 3

2 2 2 2 2

0

1 1 2 2 2 234(4 3 ) (25 4 )

3 3 3 9 9 9u du u C x

2525 1 3 3 3

2 2 2 2

4 4

1 2 2 234(25 4 )

3 9 9 9u du u

Using a change in limits