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Chapter 7: Techniques of Integration
Section 7.1: Integration by Parts
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Find the indefinite integral using integration by parts technique
Find the definite integral using integration by parts technique
Vocabulary:
Integration by Parts – an integration technique that sometimes helps to simplify hard integrals
Key Concept:
Integration by Parts
• Similar to the Product Rule in Differentiation
∫u dv = uv – ∫v du
Used to make integrals simplier
Strategies to use integration by parts
1) Use derivative to drive a polynomial function to zero
2) Reduce polynomials to get a u-substitution
3) Use derivative to get the original integral
And the simplify using addition/subtraction
See pg. 475 for the basic integration rules that we should be familiar with so far….
Integration by Parts
d
uv   u v  uv .
The product rule states
dx
d
uv   u v  uv
So,
OR
dx

Therefore,


uv    u v   uv 
 uv  uv   uv .
Generally, we start with
 udv  uv   vdu . This is the rule called Integration by Parts.
Chapter 7: Techniques of Integration
Examples:
 x cos xdx
we need to write this in the form u dv
If we let u = x and dv = cos x dx,
then du = dx and v = sin x
So
 x cos xdx  x sin x   sin xdx
 x sin x  cos x  C
You can check your answers by taking the derivative!
Generally, we want to choose u so that taking its derivative makes a simpler function.
Examples:
 xe dx
x
 x ln xdx
 x sin 3xdx
See example 2 on page 477. This shows
 ln xdx - u  ln x, dv  dx
Chapter 7: Techniques of Integration
Repeated Integration by Parts
Perform integration by parts until the integral you began with appears on the right or you get to a point that you can use a
basic integration rule. Then add or subtract accordingly and then multiply or divide.
Example:
e
x
sin xdx
More Examples:
e
x
x
cos 2 xdx
2
sin xdx
Definite Integration by parts…
 f xg xdx  f xg x   g x f xdx
combine this formula with the fundamental theorem of calculus, assume f  and g  are continuous, and you get

b
b
a
example:

b
f x g x dx  f x g x a  g x  f x dx
a


2

4
x csc 2 xdx
Chapter 7: Techniques of Integration
Tabular view of repeated Integration by Parts
If you have a polynomial as one of the two factors in a integration by parts problem, then the following is a short cut to
solving the problem.
Steps:
1.
2.
3.
4.
5.
6.
7.
Draw a 3 column table and label the first column “Dif” and the third column “Int”
Put p(x) (the polynomial) in the first column and differentiate it until you obtain 0
Put f(x) (the other function) in the third column and integrate repeatedly until you reach the 0 in the first column
Draw an arrow from the Dif column to the row below it in the Int column
Label the arrows, starting with “+” and alternating with “-“
From each arrow form the product of the expressions at the tail and the tip of the arrow and multiply that expression
by “+”1 or “-“1 based on the sign on the arrow
Add the results together to obtain the value of the integral
Example:
Find the integral of ∫ (2x³ - 7x² + 3x – 4) ex dx
Dif
2x³ - 7x² + 3x – 4
6x² - 14x + 3
12x - 14
12
0
+
+
+
Int
ex
ex
ex
ex
ex
∫ (2x³ - 7x² + 3x – 4) ex dx = (2x³ - 7x² + 3x – 4) ex – (6x² - 14x + 3) ex + (12x - 14) ex – (12) ex
= ex [(2x³ - 7x² + 3x – 4) – (6x² - 14x + 3) + (12x - 14) – (12)]
= ex [(2x³ - 13x² + 29x – 33)
Note that lots of work in setting up ‘u=’ and ‘dv=’ over and over again has been saved making this a great time saver as well
as something that is less prone to making sign errors.
Try it on the following problem:
Find the integral of ∫ (6x³ + 3x² - 5x – 7) ex dx
Dif
Int
∫ (6x³ + 3x² - 5x – 7) ex dx =
Concept Summary:
Integration by parts is analogous to the product rule of differentiation
The goal of integration by parts is either to get an integral that is simpler or one that repeats
Homework: pg 480 – 482: Day1: 3, 4, 7, 9, 36;
Read: Section 7.2
Day 2: 1, 14, 19, 51
Chapter 7: Techniques of Integration
Section 7.2: Trigonometric Integrals
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Solve integrals involving trigonometric functions
Vocabulary: None new
Key Concepts:
Strategies for Hard Trig Integrals
Substitution
sinn x or cosn x,
n is odd
Keep one sin x or cos x or for dx
Convert remainder
sinn x or cosn x,
n is even
Use half angle formulas:
sin² x = ½(1 – cos 2x)
cos² x = ½(1 + cos 2x)
sinm x • cosn x,
n or m is odd
From odd power,
keep one sin x or cos x, for dx
Use identities to substitute
sin² x + cos² x = 1
sinm x • cosn x,
n & m are even
Use half angle identities
sin² x = ½(1 – cos 2x)
cos² x = ½(1 + cos 2x)
tann x or cotn x
From power pull out tan2 x or cot2 x
and substitute using identities
cot2 x = csc2 x - 1 or
tan2 x = sec2 x – 1
tanm x• secn x or
cotm x • cscn x ,
where n is even
Pull out sec2 x or csc2 x for dx
sec² θ – 1 = tan² θ
Type I: sinn x or cosn x, n is odd


Keep one sin x or cos x or for dx
Convert remainder with sin² x + cos² x = 1
Examples:
∫ cos
5
x dx
Trig
Identity
Expression
∫ sin
3
x dx
sin² x + cos² x = 1
Chapter 7: Techniques of Integration
Type II: sinn x or cosn x, n is even

Use half angle formulas: sin² x = ½(1 - cos 2x)
Examples:
∫ sin² x dx
Type III: sinm x • cosn x, n or m is odd


From odd power, keep one sin x or cos x, for dx
Use identities to substitute
Example:
∫ sin
3
x cos4 x dx
Type IV: : sinm x • cosn x, n and m are even.

Use half angle identities
Example:
∫ sin² x cos² x dx
cos² x = ½(1 + cos 2x)
∫ cos
4
x dx
Chapter 7: Techniques of Integration
Type V: tann x or cotn x

From power pull out tan2 x or cot2 x and substitute cot2 x = csc2 x - 1 or tan2 x = sec2 x – 1
∫ cot
4
Examples:
∫ tan
5
x dx
x dx
Type VI: tanm x• secn x or cotm x • cscn x , where n is even

Pull out sec2 x or csc2 x for dx
Example:

tan
3
2
x sec 4 xdx
Chapter 7: Techniques of Integration
Trigonometric Reduction Formulas
Expression
∫ sinn x dx
∫ cosn x dx
∫ tann x dx
∫ secn x dx
Reduction Formula
1
n–1
= - --- sinn-1 x cos x + ------- ∫ sinn-2 x dx
n
n
1
n–1
= --- cosn-1 x sin x + ------- ∫ cosn-2 x dx
n
n
1
= ------- tann-1 x - ∫ tann-2 x dx
n-1
1
n-2
= ------- secn-2 x tan x + ------- ∫ secn-2 x dx
n-1
n-1
Remember the following integrals: (when n=1 in the above)
∫ tan x dx = ln |sec x| + C
∫ sec x dx = ln |sec x + tan x| + C
Examples:
∫ sin² x dx
∫ tan
5
(check the double angle formula!)
x dx
(check your work from previous page!)
Homework – Problems: pg 488-489,
Read: Section 7.3
Day 1: 1, 2, 5, 9, 10
Day 2: 3, 7, 11, 14, 17
Chapter 7: Techniques of Integration
Section 7.3: Trigonometric Substitution
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Use trigonometric identities to simply certain “hard” integrals
Vocabulary: None
Key Concept:
Table of Trigonometric Substitutions
Example 1:  4 - x²
Example 2:  x² (4-x2)-3/2
Example 3:  1/(x² + 9)
Expression
Substitution
Trig Identity
a² - x²
x = a sin θ
-π/2 ≤ θ ≤ π/2
1 - sin² θ = cos² θ
a² + x²
x = a tan θ
-π/2 ≤ θ ≤ π/2
1 + tan² θ = sec² θ
x² - a²
x = a sec θ
0 ≤ θ ≤ π/2 or π ≤ θ ≤ 3π/2
sec²θ – 1 = tan² θ
Chapter 7: Techniques of Integration
Example 4:  4 + x²
Example 5:  1/(x²9 + 9x²)
Example 6:  (x² - 16)/x
Application Problem: Find the area under the curve y = 16 - 4x² between x = 0 and x = 2.
Homework – Problems: pg 494 – 495 Day 1:
Day 2:
Read: Section 7.4
1, 5, 9, 17
6, 13, 22, 33
Chapter 7: Techniques of Integration
Section 7.4: Integration of Rational Functions by Partial Fractions
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Solve integral problems using the technique of partial fractions
Vocabulary: None new
Key Concept:
Type I – Improper: (degree of numerator ≥ degree of denominator) Start with long division
Examples:

x
 x  1 dx
x3
dx
x 1
Chapter 7: Techniques of Integration
Type II – Proper: decompose into partial Partial
fractions
Example:
3x  1
Example:
x
2
Fractions Example
3x – 1
-------------- dx
∫ dx
 x  6 x² - x – 6
x² - x – 6 = (x – 3) (x – 2)
I) factor denominator
3x – 1
A
B
------------- = --------- + ---------x² – x – 6
(x – 3) (x – 2)
II) rewrite fraction
3x – 1 = A(x + 2) + B(x – 3)
III) multiply through by common denominator
IV) solve one factor and substitute
when x = -2 then B = 7/5
V) repeat with remaining factor(s
VI) substitute A and B and integrate
when x = 3 then A = 8/5
8
1
7
1
-- -------- dx + -- --------- dx
5 (x-3)
5 (x + 2)
∫
∫
= (8/5) ln|x-3| + (7/5) ln|x+2| + C
x

2
x7
dx
 x6
5 x 2  20 x  6
dx
x3  2x 2  x
(If a factor is repeated, i.e.
1
A
B
, we must consider all possibilities and thus write
)

2
( x  1)
x  1 ( x  1) 2
Chapter 7: Techniques of Integration
Type III – Variations of Arctan:
Examples:
 2 x
2
2

dx
1
x
 arctan  C
2
a
a
a
dx
 x  1
dx
9
Type IV – Variations of Arcsin:
Examples:
x
dx
9  x2

dx
a x
2
2
 arcsin

2
4

dx
x  4x  5

dx
2
x
a
dx
16  1  x 
Homework – Problems: pg 504-505, Day 1: 1, 2, 3, 7
Day 2: 4, 10, 19, 40
Read: Section 7.5
2
 4x  x 2
Chapter 7: Techniques of Integration
Section 7.5: Strategy for Integration
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Nothing at this time
Vocabulary:
none new
Key Concept:
Not Covered at this Time
Strategy for Integration
•
•
•
•
Know basic integration table (pg 506)
Simplify the Integrand if Possible
Look for an Obvious Substitution
Classify the Integrand According to its Form
–
–
–
–
Trigonometric Functions
Rational Functions
Integration by Parts
Radicals
• Try Again
– Basically Only Two Methods of Integration
• Substitution
• Parts
– Try Algebraic Manipulation
Homework – Problems: none
Read: section 7.6
Chapter 7: Techniques of Integration
Section 7.6: Integration Using Tables and Computer Algebra Systems
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Solve problems involving density
Vocabulary:
None
Key Concept:
Not Covered at this Time
Homework – Problems: none
Read: read 7.7
Chapter 7: Techniques of Integration
Section 7.7: Approximate Integration
SOL: APC.16: The student will compute an approximate value for a definite integral. This will include numerical
calculations using Riemann Sums and the Trapezoidal Rule.
Objectives: Students will be able to:
Approximate integrals using Riemann Sums and the Trapezoidal Rule
Vocabulary:
None
Key Concept:
Some elementary functions do not possess antiderivatives that are elementary functions, i.e.
3
x 1- x ,
cos x
2
, sin x .
x
Approximation techniques must be used. We have done approximations using rectangles, primarily with left endpoints, right
endpoints and midpoints. Another technique for approximations is the trapezoidal rule.
What is the area of the first trapezoid?
1
ba
A  h  P0  x   P1  x   , where h 
n
2
Now add each trapezoid together to get: A 
n
1
1
 2 h  P  P   2 h  P  2P  2P  ...  2P
i
i 1
i 1
Examples:

Use the trapezoidal rule with n = 4 to approximate

Use the trapezoidal rule with n = 5 to approximate
e
0
1
0
sin xdx .
x2
dx .
1
2
3
n 1
 Pn 
Chapter 7: Techniques of Integration
Pond Problem: Use the trapezoidal rule to estimate the surface area of the pond. Suppose measurements are taken every 20
feet.
50
30
45
40
45
40
The trapezoidal rule averages the results of the left endpoint and right endpoint rules.
If f is an increasing function:
 f  x  dx ≤ right end
right end ≤  f  x  dx ≤ left end
left end ≤
b
a
b
If f is a decreasing function:
a
If the graph of f is concave down, the trapezoidal rule underestimates the area; if it is concave up, it overestimates the area.
How far off is our estimate? What is the error?
Error Approximation & the Trapezoidal Rule:
If f has a continuous second derivative on [a,b], then the error E in approximating
 f xdx by the trapezoidal rule is:
b
a
E
b  a 3 max
12n 2
f x   where a ≤ x ≤ b
Two types of problems 1) find the maximum error and 2) find
n (the number of sub-intervals) given a specific error.
Examples:
Use the error formula to find the maximum possible error in approximating the integral, with n = 4.

1
0
1
dx
x 1
Use the error formula to find n so that the error in the approximation of the definite integral is less than .00001.
 1  xdx
1
1
0
Homework – Problems: pg 527 – 529: probs: 7, 8, 9
Read: section 7.8
Chapter 7: Techniques of Integration
Section 7.8: Improper Integrals
SOLs: APC.13: The student will find the indefinite integral of algebraic, exponential, logarithmic, and trigonometric
functions. The special integration techniques of substitution (change of variables) and integration by parts will be included.
Objectives: Students will be able to:
Solve improper integrals
Vocabulary:
None
Key Concept:
Infinite Limits of Integration
Type I: One infinite limit
1.


0
1
dx
1 x2
2.

1

xe x dx
2
You must rewrite these as limits:

c
f x dx  lim

c
a  a


f x dx

f x dx  lim
 f xdx
a
a  c
c
If the limit exists, then the integral is said to converge to the limit value. If the limit fails to exist, then the integral is said to
diverge.
Type II: Two infinite limits – you must rewrite this as the sum of 2 limits:


f x dx  lim
a  a

3.



x
x2  4

0
dx
f x dx  lim
 f xdx
b
b 0
4.


1
dx
 x  2 x  5
2
Chapter 7: Techniques of Integration
Infinite Discontinuities
f x  is continuous on a, b and is discontinuous at b , then
Type I: If

b
a
If
f x  is continuous on a, b and is discontinuous at a , then
f x dx  lim
t b

b
a

1
1.
0
1
dx
x
Type II: If
2.

2
0
 f xdx .
f x dx  lim
t a
t
a
 f xdx .
b
t
dx
4  x2
f x  is continuous on a, b except at c , where a  c  b , then

b
a
f x dx 

c
a
provided both integrals converge.

 x  1
3
1
3.
1
dx
2
2 x
4.
0
dx
2
3
Comparison Theorem: Suppose that f(x) and g(x) are continuous functions with f(x) ≥ g(x) for x ≥ a.

a.

 g xdx also converges.
If g x dx diverges, then

 f xdx also diverges.
a

b.

f x dx converges, then
If
a
a

a
See page 530.
Homework – Problems: example problem 2 and 4
Read: review and study chapter 7
f x dx 
 f xdx ,
b
c
Chapter 7: Techniques of Integration
Chapter 7: Review
SOLs: None
Objectives: Students will be able to:
Know material presented in Chapter 7
Vocabulary: None new
Key Concept:
The book review problems are on page 534.
Homework – Problems: pg 468-469: 2, 7, 13, 25, 27, 30
Read: Study for Chapter 7 Test
Non-Calculator Multiple Choice
1.
 cos  2x  dx 
2
x sin(4 x)

c
2
8
x sin(4 x)

c
D.
4
16
A.
2.
x
2
x sin(4 x)

c
2
8
x sin(4 x )

c
E.
2
4
B.
1
ln | x 2  2 x  10 | c
B.
2
1
1
ln | x 2  2 x  10 |   tan 1
D.
2
3
x
2
1
ln | x 2  2 x  10 | c
B.
2
1
1
ln | x 2  2 x  10 |   tan 1
D.
2
3
5.
1 1 x  1
1 x  1
tan
c
c
C. tan
3
3
3
x 1
1
x 1
c
ln | x 2  2 x  10 | 2 tan 1
c
E.
3
2
3
1
dx 
 2 x  10
A.
4.
x sin(4 x)

c
2
4
x 1
dx 
 2 x  10
A.
3.
C.
1 1 x  1
1 x  1
tan
c
c
C. tan
3
3
3
x 1
1
x 1
c
ln | x 2  2 x  10 | 2 tan 1
c
E.
3
2
3
 x e dx 
2 x
e
x
A.
e x ( x 2  2 x)  c
B.
e x ( x 2  2 x  2)  c
D.
e x ( x 2  1)2  c
E.
e x ( x 2  1)2  c
C.
e x ( x 2  2 x  2)  c
C.
1 x
e  cos x  sin x   c
2
cos xdx 
A.
ex  cos x  sin x   c
B. e sin x  c
D.
2ex  cos x  sin x   c
E.
x
1 x
e  cos x  sin x   c
2
Chapter 7: Techniques of Integration
6.
 sin
1
xdx 
 sin x 
1
sin x  1  x2  c
A.
1
x cos x  1  x  c
D.
7.
2
B.
E.
2
c
2
C.
sin 1 x  1  x2  c
1
x sin x  1  x2  c
dx
 ( x  3)( x  4) 
2
A. ln|(x-3)(x+4)|+c
1
x3
ln
c
x 4
D. 7
B. (1/7)ln|(x-3)(x+4)|+c
C. 7
x 4
c
x3
ln
E. none of these
8. Which one of the following improper integrals diverges?

dx
B.  x
e
0
dx
A.  2
x
1

D.

2
9. 1
1
1x
t 1
2
 sin
3
dx
1
3
x
E. none of these

A. 2/3
10.
C.
1
dx
dt
3


B. 3/2
C. 3
D. 1
E. none of these
x cos 2 xdx 
1
1
cos3 x  cos5 x  c
3
5
1
1
3
5
D.  cos x  cos x  c
3
5
A.
1
sin 4 x cos3 x  c
12
1 4
1 6
E.  sin x  sin x  c
4
6
B.
7
11. Use the Trapezoidal Rule, with n = 4, to approximate
x
2
C.
1 4
1
sin x  sin 6 x  c
4
6
 10 x  21 dx
3
A. -10.67
B. 10.67
Free Response: (1995-BC5) Let f ( x ) 
C. -10.00
D. 10.00
E. -10.33
1
for 0  x  1 and let R be the region between the graph of f and the x-axis.
x
a. Determine whether region R has finite area. Justify your answer using calculus.
b. Determine whether the solid generated by revolving region R about the y-axis has finite volume.
Justify your answer using calculus.
c. Determine whether the solid generated by revolving region R about the x-axis has finite volume.
Justify your answer using calculus.
Chapter 7: Techniques of Integration
Answers
1. A
2. A
3. B
4. B
5. C
6. E
7. D
8. D
9. B
10. D
11. C
Free Response:
1
a. The area integral is
1
0 x dx which is an improper integral.
1
1
1
1
dx  lim ln x |1c  lim (ln1  ln c)   .
0 x dx  clim
 
0
c 0
c 0
x
c
Thus, the integral diverges and there is not finite area.
1

1
 dx  2  dx  2 .
x
0
1
b. There is finite volume. 2 x 
0
2
1
dx
1
1
 1

c. There is infinite volume.     dx  lim   2  lim      lim   1    
c 0
c 0
x
x
c
 x  c c 0 
0
c
1
1
After Chapter 7 Test:
Homework – Problems: None
Read: Chapter 7 to see what’s coming next