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Topics: - sections 7.5 (additional techniques of integration) and 7.6

(applications of integration)* Read these sections and study solved examples in your

textbook!

Work On:- Practice problems from the textbook and assignments

from the coursepack as assigned on the course web page (under the link “SCHEDULE + HOMEWORK”)

The Product Rule and Integration by Parts

The product rule for derivatives leads to a technique of integration that breaks a complicated integral into simpler parts.

Integration by Parts Formula:

€

udv = uv − vdu∫∫hopefully this is a simpler Integral to evaluate

given integral that we cannot solve

The Product Rule and Integration by Parts

Deriving the Formula

Start by writing out the Product Rule:

Solve for

€

d

dx[u(x) ⋅v(x)] =

du

dx⋅v(x) + u(x) ⋅

dv

dx

€

u(x) ⋅dv

dx:

€

u(x) ⋅dv

dx=d

dx[u(x) ⋅v(x)]−

du

dx⋅v(x)

The Product Rule and Integration by Parts

Deriving the Formula

Start by writing out the Product Rule:

Solve for

€

d

dx[u(x) ⋅v(x)] =

du

dx⋅v(x) + u(x) ⋅

dv

dx

€

u(x) ⋅dv

dx:

€

u(x) ⋅dv

dx=d

dx[u(x) ⋅v(x)]−

du

dx⋅v(x)

The Product Rule and Integration by Parts

Deriving the Formula

Integrate both sides with respect to x:

€

u∫ (x)dv

dxdx = ∫ d

dx[u(x) ⋅v(x)] dx − v(x)∫ du

dxdx

The Product Rule and Integration by Parts

Deriving the Formula

Simplify:

€

u∫ (x)dv

dxdx = ∫ d

dx[u(x) ⋅v(x)] dx − v(x)∫ du

dxdx

€

⇒ u∫ (x)dv = u(x) ⋅v(x) − v(x)∫ du

Integration by Parts

Template:

Choose:

Compute:

€

udv = uv − vdu∫∫

€

u =

€

dv = easy to integrate partpart which gets simplerafter differentiation

€

du =

€

v =

Integration by Parts

Example:Integrate each using integration by parts.

(a)(b)

(c)(d)

€

arcsin xdx∫€

x cos4xdx∫

€

x 2ex2dx∫

€

ln x dx1

2

∫

Strategy for IntegrationMethod Applies when…Basic antiderivative …the integrand is recognized as the reversal of a differentiation formula, such as

Guess-and-check …the integrand differs from a basic antiderivative in that “x” is replaced by “ax+b”, for example

Substitution …both a function and its derivative (up to a constant) appear in the integrand, such as

Integration by parts …the integrand is the product of a power of x and one of sin x, cos x, and ex, such as

…the integrand contains a single function whose derivative we know, such as

Strategy for Integration

What if the integrand does not have a formula for its antiderivative?

Example:

impossible to integrate

€

e−x 2

dx0

1

∫

Approximating Functions with Polynomials

Recall:The quadratic approximation toaround the base point x=0 is

€

T2(x) =1− x 2 .

€

f (x) = e−x 2

€

T2(x) =1− x 2

base point

€

f (x) = e−x 2

Integration Using Taylor Polynomials

We approximate the function with an appropriate Taylor polynomial and then integrate this Taylor polynomial instead!

Example:

easy to integrateimpossible to integrate

€

e−x 2

dx0

1

∫ ≈ (1− x 2) dx0

1

∫for x-values near 0

Integration Using Taylor Polynomials

We can obtain a better approximation by using a higher degree Taylor polynomial to represent the integrand.

Example:

€

e−x 2

dx0

1

∫

≈ (1− x 2 +1

2x 4 −

1

6x 6) dx

0

1

∫

= 0.74286

The Definite Integral – Area Between Curves

The area between the curves and and between and is

Recall:

€

A = f (x) − g(x) dxa

b

∫

€

f (x) − g(x) =f (x) − g(x) when f (x) ≥ g(x)

g(x) − f (x) when f (x) ≤ g(x)

⎧ ⎨ ⎩

€

y = f (x)

€

y = g(x)

€

x = a

€

x = b

Area Between Curves

€

f (x) − g(x) dxa

b

∫

€

= f (x) − g(x)( )dxa

c

∫ + g(x) − f (x)( )dxc

d

∫ + f (x) − g(x)( )dxd

b

∫

Area between f and g on [a,b] =

The Definite Integral – Area Between Curves

Examples: Sketch the region enclosed by the given curves and then find the area of the region.

(a)

(b)€

y = x 2 − 2x, y = x + 4

€

y = x , y =1

x, x =

1

2, x = 2

estimate of the surface area of lake ontario

aerial distance hamilton - kingston approx. 290 km

aerial distance hamilton - kingston = 290km

represents approximately 38km

aerial distance hamilton - kingston = 290km

represents approximately 38km

aerial distance hamilton - kingston = 290km

represents approximately 38km

38

41.0

area = 38*41=1558 km^2

38

55.3

area = 38*55.3=2101.4 km^2

38

58.2

area = 38*58.2=2211.6 km^2

38

63.0

area = 38*63.0=2394.0 km^2

63.0

total area = 38*41.0+ 38*55.3+ 38*58.2+ 38*64.9+ 38*82.0+ 38*70.6+ 38*97.3+ 38*63.0= 20,227.4 km^2

38

41.0

55.358.2

64.9 82.070.6

97.3

represents approximately 17km

area = 17.2*17.2=295.8 km^2

area = 17.2*43.9=764.8 km^2

area = 17.2*55.3=951.7 km^2

area = 17.2*61.0=1050.1 km^2

total area = 19,233.5

average of the two estimates = (20,227.4 + 19,233.5)/2= 19,730.5 km^2

our estimate = 19,730.5 km^2

New York Times Almanac … 19,500 km^2

NOAA (National Oceanographic and Atmospheric Administration, U.S.) … 19,009 km^2

EPA (Environmental Protection Agency) … 18,960 km^2

Britannica Online … 19,011km^2

The Definite Integral - Average Value

The average value of a function f on the interval from a to b is

For a positive function,

€

f =1

b− af (x)dx

a

b

∫

average height = area width

€

f (x)

€

f

The Definite Integral - Average Value

€

f (x)dxa

b

∫ = (b− a) f

€

f (x)

€

f

€

area = base × height

= (b− a) × f

€

area = f (x)dxa

b

∫

€

Average Value

Example: Find the average value of the function on the interval

€

f (x) = x 2

€

[0, 2].

Application

Example:

Several very skinny 2.0-m-long snakes are collected in the Amazon. Each snake has a density of

where is measured in grams per centimeter and is measured in centimeters from the tip of the tail.

Find the average density of the snake.

€

ρ(x) =1+ 2 ×10−8 x 2(300 − x)

€

ρ

€

x

Application

€

x

€

ρ(x)

€

ρ(x)

€

x

€

x

Application

(a) Find the total mass of each snake.

(b) Find the average density of each snake.

estimate of the volume of a heart chamber from echocardiogram

echocardiogram (ECHO, cardiac ultrasound) is a sonogram of the heart

ECHO uses standard ultrasound techniques to image two-dimensional slices of the heart

latest ultrasound systems now employ 3D real-time imaging

ECHO uses standard ultrasound techniques to image two-dimensional slices of the heart

latest ultrasound systems now employ 3D real-time imaging

uses of ECHO

(a) creating 2D picture of the cardiovascular system (shape of the heart)

(b) assessment of quality of cardiac tissue (damage, thickening of walls within the heart)

(c) estimate of the velocity of blood

uses of ECHO

(d) investigate features of blood flow

• functioning of cardiac valves• detection of abnormal communication between the left and right side of the heart• leaking of blood through the valves• strength at which blood is pumped out of heart (cardiac output)

5.02 cm

5.02 cm

5.02/15= 0.335 cm

5.02 cm

5.02/15= 0.335 cm

2.00 cm

2.50 cm

5.02 cm

5.02/15= 0.335 cm

2.00 cm

2.50 cm

5.02 cm

5.02/15= 0.335 cm

2.00 cm

2.50 cm

0.335 cm

1.00 cm

€

volume =

π (1)20.335 =

1.052 cm3

0.335 cm

1.25 cm

€

volume =

π (1.25)20.335 =

1.644 cm3

0.335 cm

1.50 cm

€

volume =

π (1.50)20.335 =

2.368 cm3

0.335 cm

1.61 cm

€

volume =

π (1.61)20.335 =

2.728 cm3

0.335 cm

1.07 cm

€

volume =

π (1.07)20.335 =

1.205 cm3

€

volume of heart chamber =

1.052 +1.644 + 2.368 + 2.728 + ...+1.205

= 50.342 cm3

all diameters, from bottom to top:all in cm

2.00, 2.50, 3.00, 3.21, 3.43, 3.57,3.93, 4.07, 4.29, 4.29, 4.29, 4.14,4.07, 3.43, 2.14

height: 5.02/15 cm

Approximating Volumes

€

Δx =b − a

n

€

Vn = A(x1)Δx + A(x2)Δx +L + A(xn )Δx

€

= A(x i)Δxi=1

n

∑Riemann Sum

€

So, the volume V of the solid S ≈Vn .

€

A(x i) = area of base

Integrals and Volumes

Definition:Denote by A(x) the area of the cross-section of S by the plane perpendicular to the x-axis that passes through x. Assume that A(x) is continuous on [a,b].

Then the volume V of S is given by

provided that the limit exists.€

V = limn→ ∞Vn = lim

n→ ∞A(x i

i=1

n

∑ )Δx = A(x)dxa

b

∫

Volumes of Solids of Revolution

Examples:Find the volume of the solid obtained by rotating the region R enclosed (bounded) by the given curves about the given axis.

(a)

(b)

€

y =1

x, y = 0, x =1, and x = 2 about the x - axis

€

y = 8 − x, y = 3, x = 2, and x = 5 about the y - axis