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∫∫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
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
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
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
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)
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
∫