Announcements
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
given integral that
we cannot solve
hopefully this is a simpler
Integral to evaluate
The Product Rule and
Integration by Parts
Deriving the Formula
Start by writing out the Product Rule:
d
du
dv
[u(x)  v(x)]   v(x)  u(x) 
dx
dx
dx
dv
Solve for u(x)  :
dx

dv d
du
u(x)   [u(x)  v(x)]   v(x)
dx dx
dx
The Product Rule and
Integration by Parts
Deriving the Formula
Start by writing out the Product Rule:
d
du
dv
[u(x)  v(x)]   v(x)  u(x) 
dx
dx
dx

Solve for

u(x) 
dv
:
dx
dv d
du
u(x)   [u(x)  v(x)]   v(x)
dx dx
dx
The Product Rule and
Integration by Parts
Deriving the Formula
Integrate both sides with respect to x:
dv
 u(x) dx dx 

d
du
[u(x)  v(x)] dx   v(x) dx
dx
dx
The Product Rule and
Integration by Parts
Deriving the Formula
Simplify:
dv
 u(x) dx dx 


d
du
[u(x)  v(x)] dx   v(x) dx
dx
dx
 u(x)dv  u(x) v(x)   v(x)du
Integration by Parts
 udv  uv   vdu
Template:
Choose:

u
part which gets simpler
after differentiation
Compute:
du 

dv 
v
easy to integrate part
Integration by Parts
Example:
Integrate each using integration by parts.
(a)
(c)
(b)  x e dx
 x cos 4 xdx
2
 ln x dx
1
2

x
2
(d)  arcsin xdx
Strategy for Integration
Method
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
1
e
0
x 2
dx
Approximating Functions with
Polynomials
Recall:
The quadratic approximation to f (x)  ex
around the base point x=0 is T2 (x)  1 x 2 .

base point

x 2
f (x)  e
T2 (x)  1 x 2

2
Integration Using Taylor Polynomials
We approximate the function with an
appropriate Taylor polynomial and then
integrate this Taylor polynomial instead!
Example:
impossible to integrate
1
e
0
easy to integrate
x
2
dx 
1
 (1 x ) dx
2
0
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:
1
e
x 2
dx
0
1
1 4 1 6
  (1 x  x  x ) dx
2
6
0
2
 0.74286

The Definite Integral –
Area Between Curves
The area between the curves y  f (x) and
y  g(x) and between x  a and x  b is
A

b

a
  g(x) dx
f (x)

Recall:
f (x)  g(x) when f (x)  g(x)

f (x)  g(x)  
g(x)  f (x) when
f (x)  g(x)
Area Between Curves
b
Area between f and g on [a,b] =

f (x)  g(x) dx
a

c
d
b
a
c
d
  f (x)  g(x)dx   g(x)  f (x)dx    f (x)  g(x)dx

The Definite Integral –
Area Between Curves
Examples:
Sketch the region enclosed by the given curves
and then find the area of the region.
(a) y  x 2  2x, y  x  4
1
1
(b) y  x, y  , x  , x  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
41.0
38
area = 38*41=1558 km^2
55.3
38
area = 38*55.3=2101.4 km^2
58.2
38
area = 38*58.2=2211.6 km^2
63.0
38
area = 38*63.0=2394.0 km^2
55.3
58.2
64.9
82.0
97.3
70.6
41.0
38
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
63.0
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
1
f 
ba
f (x)

b
 f (x) dx
f
a

For a positive function,
area
average height =
width
The Definite Integral - Average Value
f (x)

area 
b

area  base  height
f (x)dx
 (b  a)  f
a
f



b
 f (x) dx  (b  a) f
a

Average Value
Example:
Find the average value of the function
f (x)  x 2 on the interval [0, 2].

Application
Example:
Several very skinny 2.0-m-long snakes are collected
in the Amazon. Each snake has a density of
(x)  1 2 108 x 2 (300  x)
where  is measured in grams per centimeter and x
is measured in centimeters from the tip of the tail.

Find the average density of the snake.


Application
(x)
(x)


x
xx



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 realtime imaging
ECHO uses standard ultrasound techniques to
image two-dimensional slices of the heart
latest ultrasound systems now employ 3D realtime 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
2.50 cm
2.00 cm
5.02 cm
5.02/15
= 0.335 cm
2.50 cm
2.00 cm
5.02 cm
5.02/15
= 0.335 cm
2.50 cm
2.00 cm
1.00 cm
0.335 cm
volume 
 (1) 2 0.335 
1.052 cm 3

1.25 cm
0.335 cm
volume 
 (1.25) 2 0.335 
1.644 cm 3

1.50 cm
0.335 cm
volume 
 (1.50) 2 0.335 
2.368 cm 3

1.61 cm
0.335 cm
volume 
 (1.61) 2 0.335 
2.728 cm 3

1.07 cm
0.335 cm
volume 
 (1.07) 2 0.335 
1.205 cm 3

volume of heart chamber 
1.052  1.644  2.368  2.728  ... 1.205
 50.342 cm 3
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
A(xi )  area of base
x 


Vn  A(x1)x  A(x2 )x L  A(xn )x
n
  A(x i )x
i1

Riemann Sum
So, the volume V of the solid S  Vn .
ba
n
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
n
V  lim Vn  lim  A(x i )x 
n
n
i1
provided that the limit exists.
b
 A(x) dx
a
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.
1
(a) y  , y  0, x  1, and x  2 about the x - axis
x
(b) y  8  x, y  3, x  2, and x  5 about the y - axis