Contents
1 Chapter 1
1
2 Chapter 1
1
3 Chapter 1
1
4 Integration
4.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . .
4.1.1 Definition of an Antiderivative . . . . . . . . . . .
4.1.2 Antiderivatives of Trigonometric Functions . . . .
4.1.3 Antiderivatives of Other Functions . . . . . . . .
4.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Sigma Notation . . . . . . . . . . . . . . . . . . .
4.2.2 Expanding Summations . . . . . . . . . . . . . .
4.2.3 Area of Plane Regions . . . . . . . . . . . . . . .
4.3 Riemann Sum . . . . . . . . . . . . . . . . . . . . . . . .
4.3.1 Definition of a Riemann Sum . . . . . . . . . . .
4.4 Definite Integrals . . . . . . . . . . . . . . . . . . . . . .
4.4.1 Fundamental Theorem of Calculus . . . . . . . .
4.4.2 Evaluating Integrals . . . . . . . . . . . . . . . .
4.4.3 Integrals of Absolute Value Functions . . . . . . .
4.4.4 Mean Value Theorem . . . . . . . . . . . . . . . .
4.4.5 The Average Value of a Function . . . . . . . . .
4.5 Integration by Substitution . . . . . . . . . . . . . . . . .
4.5.1 Review of the Chain Rule . . . . . . . . . . . . .
4.5.2 Basic Substitutions . . . . . . . . . . . . . . . . .
4.6 The Natural Logarithm Function . . . . . . . . . . . . .
4.7 Antiderivative of the Inverse Trigonometric Functions . .
4.7.1 Basic Antiderivatives of the Inverse Trigonometric
4.7.2 Integration by Substitution . . . . . . . . . . . .
4.8 Simpson’s Rule and the Trapezoid Rule . . . . . . . . . .
4.8.1 The Trapezoid Rule . . . . . . . . . . . . . . . . .
4.8.2 Simpson’s Rule . . . . . . . . . . . . . . . . . . .
1
Chapter 1
See Math 151 Notes
2
Chapter 1
See Math 151 Notes
3
Chapter 1
See Math 151 Notes
1
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2
2
2
3
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6
6
6
9
13
14
20
20
20
21
22
22
25
25
25
29
31
31
31
35
35
35
4
Integration
4.1
4.1.1
Antiderivatives
Definition of an Antiderivative
A function F is an antiderivative of f on an interval I if F ′ (x) = f (x) for all x on I.
The integral or antiderivative of a function is essentially the reverse process of differentiation.
To start will look at the most basic antiderivative which the power rule for antiderivatives.
The power rule for antiderivatives is the reverse process of the power rule for derivatives.
Power Rule for Antiderivatives
∫
cxn dx =
c
n+1
n+1 x
+C
Now, let’s try doing so examples that use the power rule.
Example 1
Evaluate the following integral
∫
x2 dx
Solution:
∫
x2 dx =
1
x2+1
2+1
+ C = 13 x3 + C
Example 2
Evaluate the following integral
∫
(x4 + 3x2 + 3x)dx
Solution:
∫
(x4 + 3x2 + 3x)dx =
1
x4+1
4+1
Example 3
Evaluate the following integral
+
∫
3
x2+1
2+1
+
3
x1+1
1+1
+ C = 15 x5 + x3 + 32 x2 + C
(x3 − 5x2 + 6)dx
Solution:
∫
(x3 − 5x2 + 6x0 )dx =
1
x3+1
3+1
−
5
x2+1
2+1
+
6
x0+1
1+0
2
+ C = 14 x4 − 53 x3 + 6x + C
Example 4
Evaluate the following integral
∫
5x4 dx
Solution:
∫
5x4 dx =
4.1.2
5
x4+1
4+1
+ C = x5 + C
Antiderivatives of Trigonometric Functions
Basic Integration Rules
∫
∫
∫
∫
∫
∫
cos(x)dx = sin(x) + C
sin(x)dx = −cos(x) + C
sec2 (x)dx = tan(x) + C
csc2 (x)dx = −cot(x) + C
sec(x)tanxdx = sec(x) + C
csc(x)cotxdx = −csc(x) + C
Here are some example of antiderivatives of trigonometric functions.
Example 5
Evaluate the following integral
∫
(sec2 x − csc2 x)dx
Solution:
∫
(sec2 (x) − csc2 (x))dx = tan(x) + cot(x) + C
Example 6
Evaluate the following integral
∫
(sin(x) + 3x2 )dx
Solution:
∫
(sin(x) + 3x2 )dx = −cos(x) +
3
x2+1
2+1
+ C = −cos(x) + x3 + C
3
Example 7
Evaluate the following integral
∫
(sec2 x − csc2 x)dx
Solution:
∫
(cos(x) + 4)dx = sin(x) +
4.1.3
4
x0+1
0+1
+ C = sin(x) + 4x + C
Antiderivatives of Other Functions
Antiderivatives of exponential and logarithmic functions
∫ 1
dx = ln|x| + C
x
∫ x
e dx = ex + C
Example 8
Evaluate the following integral
∫
ex dx
Solution:
∫
ex = ex + C
Example 9
Evaluate the following integral
∫
( x1 + x4 )dx
Solution:
∫
( x1 + x4 ) = ln|x| +
1
x4+1
4+1
+ C = 51 x5 + C
Example 10
Evaluate the following integral
∫
3
x 2 dx
Solution:
∫
3
x 2 dx =
1
3
+1
2
3
5
5
x 2 +1 + C = 15 x 2 + C = 25 x 2 + C
2
4
Example 11
Evaluate the following integral
∫
1
dx
x3
Solution:
∫
1
dx
x3
∫
x−3 dx =
1
x−3+1
−3+1
Example 12
Evaluate the following integral
Solution:
∫
( x12 + cos(x))dx =
− x1 + sin(x) + C
∫
+ C = − 12 x−2 + C = − 2x12 + C
∫
( x12 + cos(x))dx
(x−2 + cos(x))dx =
1
x−2+1
−2+1
5
+ sin(x) + C = −x−1 + sin(x) + C =
4.2
Area
4.2.1
Sigma Notation
The sum of the terms a1 , a2 , a3 , a4 .......an is:
n
∑
ai = a1 + a2 + a3 + a4 + ...... + an
i=1
where i is the index of a summation, ai is the ith term of the sum, and the upper and lower
sum of the summation are 1 and n
Here are some example of summations.
Example 1
Evaluate
6
∑
i
i=1
Solution
6
∑
i = 1 + 2 + 3 + 4 + 5 + 6 = 21
i=1
Example 2
Evaluate
4
∑
(2i + 1)
i=0
Solution:
4
∑
(2i + 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25
i=0
4.2.2
Expanding Summations
Example 3
Evaluate
4
∑
j2
j=0
Solution:
4
∑
j 2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30
j=0
6
Example 4
Evaluate
6
∑
k(k − 2)
k=3
Solution:
6
∑
k(k−2) = 3(3−2)+4(4−2)+5(5−2)+6(6−2) = 3·1+4·2+5·3+6·4 = 3+8+15+24 = 50
k=3
Example 5
Evaluate
5
∑
1
j=3
j
Solution:
5
∑
1
j=3
j
=
1 1 1
47
+ + =
3 4 5
60
Example 6
Evaluate
4
∑
[(i − 1)2 + (i + 1)3 ]
j=1
Solution:
4
∑
[(i − 1)2 + (i + 1)3 ] = (1 − 1)2 + (1 + 1)3 + (2 − 1)2 + (2 + 1)3 + (3 − 1)2 + (3 + 1)3 + (4 −
j=1
2
1) + (4 + 1)3 = 02 + 23 + 12 + 33 + 22 + 43 + 32 + 53 = 0 + 8 + 1 + 27 + 4 + 64 + 9 + 125 = 237
Example 7
Express the following series of terms into a expression using summation notation
5
5
+ ....... + 1+12
1+3
5
1+1
5
+ 1+2
+
Solution:
Since the series of terms is adding one to each successive term and then the dividing this
total into 5 the summation would be written as follows:
12
∑
5
[
]
1
+
i
j=1
7
Example 8
Express the following series of terms into a expression using summation notation [1 − ( 14 )2 ] +
[1 − ( 42 )2 ] + 1 − [( 34 )2 ] + [1 − ( 44 )2 ]
Solution:
Since the series of terms is dividing each successive term by 4, squaring the result, and adding
one the summation would be written as follows:
4
∑
i
[1 + ( )2 ]
4
j=1
Theorem 4.2: Summation Formulas
1.
u
∑
c = cn
j=1
2.
u
∑
i=
n(n+1)
2
j=1
3.
u
∑
i2 =
n(n+1)(2n+1)
6
i3 =
n2 (n+1)2
4
j=1
4.
u
∑
j=1
Now, let’s use the four properties above to expand some examples of summations.
Example 9
Use the properties of Theorem 4.2 to expand out the following summation.
15
∑
(2i − 3)
j=1
Solution:
15
15
15
15
15
∑
∑
∑
∑
∑
15(16)
− 15(3) = 240 − 45 = 195
(2i − 3) =
2i −
3=2·
i−
3=2·
2
j=1
j=1
j=1
j=1
j=1
8
Example 10
Use the properties of Theorem 4.2 to expand out the following summation.
15
∑
i(i2 + 1)
j=1
Solution:
10
10
∑
∑
2
i(i + 1) =
i3 + i
j=1
j=1
n2 (n+1)2
=
+ n(n+1)
(where n=10)
4
2
2
2
10 (10+1)
=
+ 10(10+1)
4
2
100·121
10·11
=
+ 2
4
= 25 · 121 + 5 · 11
= 3025 + 55 = 3080
4.2.3
Area of Plane Regions
Now we are prepared to use summation to find areas of plane regions. We are going to
approximate areas under the curve by dividing the region under the curve into smaller
rectangles and summing up the rectangles as shown in the next diagram.
Theorem 4.3: Limits of Upper and Lower Sums
Let f be continuous and nonnegative on the interval [a, b]. The limits as n → ∞ of both the
lower and upper sums exist and are equal to each other.
lim S(n) = lim f (mj )△x = lim f (Mj )△x = S(n) where △x = b−a
n
h→∞
h→∞
h→∞
9
Example 10
Use upper sums and lower to approximate the area of the function f (x) =
by the x-axis y = 0 and the values x = 0 and x = 1.
Solution:
First find △x (The length of each rectangle)
= 41
a = 0 and b = 1 ⇒ △x = 1−0
4
Next, compute the left and right endpoints.
Left Endpoint: mi = a + (i − 1)△x = 0 + (i − 1) ·
Right Endpoint: Mi = a + i△x = 0 + i ·
1
4
1
4
= i−1
4
= 4i
Computing the Lower Sums
S(n) =
4
∑
f (mi )△x
j=1
=
4
∑
i=1
=
4
∑
i=1
f(
√
i−1
1
)·( )
4
4
i−1 1
·
4
4
√
i−1 1
=
·
2
4
i=1
√
4
∑
i−1
=
8
i=1
4
∑
√
√
√
√
1−1
2−1
3−1
4−1
+
+
+
8
8
√8
√ 8 √
√
0
1
2
3
= 8 + 8 + 8 + 8 = 0 + .125 + .177 + .217 ≈ .519
=
10
√
x below bounded
Computing the Upper Sums
S(n) =
4
∑
f (Mi )△x
j=1
4
∑
i
1
f( ) · ( )
4
4
i=1
√
4
∑ i 1
=
·
4 4
i=1
4 √
∑
i 1
=
·
2
4
i=1
√
4
∑
i
=
8
i=1
=
√
√
√
√
= 81 + 82 + 83 + 84
= .125 + .177 + .217 + .25 ≈ .769
Example 11
Find the limit of S(n) as n → ∞ of the summation S(n) =
Solution:
64 n(n + 1)(2n + 1)
[
]
h→∞
h→∞ n3
6
2
64 (n + n))(2n + 1)
= lim 3 [
]
h→∞ n
6
64 (2n3 + n2 + 2n2 + n
= lim 3 [
]
h→∞ n
6
2n3 + 3n2 + n
= lim 64[
]
h→∞
6n3
128n3 + 192n2 + 64n
= lim
h→∞
6n3
lim S(n) = lim
11
64 n(n+1)(2n+1)
[
]
n3
6
128n3 192n2 64n
+
+ 3
h→∞ 6n3
6n3
6n
64 32
32
= lim
+
+ 2
h→∞ 3
n
3n
= 64
+ 0 + 0 = 64
3
3
= lim
12
4.3
Riemann Sum
Example 1: A Partition with Unequal Widths
Consider the region bounded by the graph of f (x) =
Now, let’s evaluate the limit lim
h→∞
n
∑
√
x and the x-axis for 0 ≤ xi ≤ 1
f (ci )△xi
i=1
where ci is the right endpoint of the portion given by ci =
ith interval.
Let: △xi = right endpoint - left endpoint
(i−1)2
n2
i
i −2i+1
n2 −
n2
i2 −(i2 +2i−1)
n2
i2 −i2 −2i+1)
n2
−2i+1
n2
2
△xi = ni 2 −
2
2
=
=
=
=
Now, we can evaluate the following limit
n
∑
i2
2i − 1
f (ci )△xi = lim
f( 2 ) · ( 2 )
h→∞
h→∞
n
n
i=1
i=1
√
n
∑
i2 2i − 1
·( 2 )
= lim
2
h→∞
n
n
i=1
n
∑
i 2i − 1
= lim
·
h→∞
n
n2
i=1
n
∑
2i2 − i
= lim
h→∞
n3
i=1
lim
n
∑
13
i2
n2
and △xi is the width of the
=
1
n3
1
n3
1
n3
1
n3
1
n3
1
n3
1
n3
lim
h→∞
n
∑
(2i2 − i)
i=1
n(n + 1)(2n + 1)
n(n + 1)
)−
]
h→∞
6
2
2n(n + 1)(2n + 1) n2 + n
=
lim (
−
]
h→∞
6
2
2
2
(2n + 2n)(2n + 1) n + n
=
lim (
−
]
h→∞
6
2
3
2
2
4n + 6n + 2n n + n
=
lim (
−
]
h→∞
6
2
4n3 + 6n2 + 2n − 3n2 − 3n
=
lim
h→∞
6
4n3 + 3n2 − n
=
lim
h→∞
6n3
3
4n + 3n2 − n
]
= lim [
h→∞
6n3
4n3 3n2
n
= lim
+ 3+ 3
3
h→∞ 6n
6n
6n
2
1
1
2
2
= lim +
+ 2 = +0+0=
h→∞ 3
2n 6n
3
3
=
4.3.1
lim [2(
Definition of a Riemann Sum
Let f be defined on a closed interval [a, b] and let △ be a partition of [a, b] given by a =
x0 < x1 < x2 < x3 < ..... < xn−1 < xn = b where △xi is the width of the ith subinterval. If
n
∑
th
ci is any point in i subinterval,
f (ci )△xi then the sum is called the Riemann sum.
i=1
Definition of a Definite Integral
If f defined on a closed interval [a, b] and the limit lim
n
∑
|△|→∞
above, then f is integrable on [a, b] and the limit is lim
|△|→∞
14
i=1
n
∑
i=1
f (ci )△xi exist as described
∫b
f (ci )△xi = a f (x)dx
Example 2
Evaluate the following definite integral by the limit definition.
Solution:
∫3
−2 xdx
Solution:
3−(−2)
a = −2 b = 3 △x = b−a
n =
n
Now, rewrite the integral as a summation
n
∫3
∑
lim
−2 xdx = n→∞
n
∑
(
f (ci )△xi
i=1
5i ) 5
·
n→∞
n
n
i=1
n
∑ ( −10 25i )
= lim
+ 2
n→∞
n
n
i=1
n
∑
( −10 25 n(n + 1) )
= lim
+ 2·
n→∞
n
n
2
i=1
n
∑ ( −10 25n2 + 25n )
= lim
+
n→∞
n
2n2
i=1
n
∑
( −10 25n2 25n )
= lim
+
+ 2
2
n→∞
n
2n
2n
i=1
n
∑ ( −10 25 25 )
= lim
+
+
= −0 + 12.5 + 0 = 12.5
n→∞
n
2
2n
i=1
= lim
−2+
Example 3
Evaluate the following definite integral by the limit definition.
Solution:
15
∫3
2
1 3x dx
Solution:
2
a = 1 b = 3 △x = 3−1
n = n
ci = 1 + i ·
2
i
=1+
2i
n
Now, rewrite the integral as a summation
n
∫3
∑
lim
1 xdx = n→∞
= lim
n→∞
= lim
n→∞
= lim
n→∞
= lim
n→∞
n
∑
f (ci )△xi
i=1
f (1 +
2i 2
)·
n n
3(1 +
2i 2 2
) ·
n
n
i=1
n
∑
i=1
n
∑
i=1
n
∑
i=1
3(
4i2 4i
2
+ + 1) ·
2
n
n
n
8i2
8i
2
3( 3 + 2 + )
n
n
n
n
∑
24i2 24i 6
= lim
( 3 + 2 + )
n→∞
n
n
n
i=1
n
∑ ( 24(n(n + 1)(2n + 1)) 24n(n + 1) 6n )
+
+
= lim
3
2
n→∞
6n
2n
n
i=1
n
∑
( 24(2n3 + 3n2 + n) 24n2 + 24n 6n )
= lim
+
+
3
2
n→∞
6n
2n
n
i=1
n
∑ ( 48n3 72n2 24n 24n2 24n
)
= lim
+
+ 3 +
+ 2 +6
3
3
2
n→∞
6n
6n
6n
2n
n
i=1
16
(
)
12
4
12
= 8+
+ 2 + 12 +
+ 6 = 26
n
n
n
Example 4
∫ 10
Evaluate the following definite integral using the limit definition of a Riemann Sum 4
Solution:
6
a = 4; b = 10; △x = 10−4
n = n
ci = a + i · △x = 4 + i ·
6
n
=4+
6i
n
f (x) = 6
Now, rewrite the integral as a summation
∫ 10
4
6dx = lim
n→∞
n
∑
f (ci )△xi
i=1
n
(
∑
6i ) 6
·
= lim
f 4+
n→∞
n
n
i=1
n
∑
6
= lim
(6) ·
n→∞
n
i=1
6
= lim (6n) · = 36
n→∞
n
17
6dx
Example 5
∫2
Evaluate the following integral using the limit definition of an integral 1 (x2 + 1)dx
1
a = 1; b = 2; △x = 2−1
n = n
ci = a + i · △x = 1 + i ·
1
n
=1+
i
n
f (x) = x2 + 1
Now, rewrite the integral as a summation
∫2
2
lim
1 (x + 1)dx = n→∞
n
∑
f (ci )△xi
i=1
n
(
∑
i) 1
f 1+
·
n→∞
n
n
i=1
n
((
)
) 1
∑
i 2
= lim
1+
+1 ·
n→∞
n
n
i=1
n ((
) 1
∑
i )(
i)
1+
= lim
1+
+1 ·
n→∞
n
n
n
i=1
n
) 1
∑ ((
2i
i2 )
1+ + 2 +1 ·
= lim
n→∞
n
n
n
i=1
n
∑(
2i
i2 ) 1
2+ + 2 ·
= lim
n→∞
n
n
n
i=1
n
∑(2
2i
i2 )
= lim
+ 2+ 3
n→∞
n n
n
(i=1
2
2 n(n + 1)
1 n(n + 1)(2n + 1) )
= lim
· (n) + 2 ·
+ 3·
n→∞ n
n
2
n
6
(
2n2 + 2n (n2 + n)(2n + 1) )
= lim 2 +
+
n→∞
2n2
6n3
(
2
3
2n + 2n 2n + 3n2 + n )
= lim 2 +
+
n→∞
2n2
6n3
(
2
2n
2n
2n3 3n2
n )
= lim 2 + 2 + 2 + 3 + 3 + 3
n→∞
2n
2n
6n
6n
6n
(
1 1
1
1 )
1
10
= lim 2 + 1 + + +
+ 2 =2+1+0+ +0+0=
n→∞
n 3 2n 6n
3
3
= lim
18
Example 6
Evaluate using the limit definition of an integral:
3
a = 0 b = 3 △x = 3−0
n = n
ci = a + i · △x = 0 + i ·
3
n
=
3
n
f (x) = 5x
Now, rewrite the integral as a summation
∫3
lim
0 5xdx = n→∞
n
∑
f (ci )△xi
i=1
n
( 3i ) 3
∑
f
·
n→∞
n
n
i=1
n
(
)
∑ 3i
3
= lim
5
·
n→∞
n
n
i=1
n
∑
45i
= lim
n→∞
n2
i=1
n
∑
45 n(n + 1)
= lim
·
2
n→∞
n
2
i=1
n
∑ ( 45n(n + 1) )
= lim
n→∞
2n2
i=1
n (
∑
45n2 + 45n )
= lim
n→∞
2n2
i=1
n (
∑
45n2 45n )
= lim
+ 2
n→∞
2n2
2n
i=1
n
(
∑ 45 45 ) 45
45
= lim
+
=
+0=
n→∞
2
2n
2
2
i=1
= lim
19
∫3
0 5xdx
4.4
4.4.1
Definite Integrals
Fundamental Theorem of Calculus
If a function f is continuous on a closed interval [a, b] and F is an the antiderivative of f on
the interval [a, b], then
∫b
a f (x)dx = F (b) − F (a)
When we evaluating a definite integral, we first find the antiderivative and then substitute
the limits into the resulting antiderivative.
4.4.2
Evaluating Integrals
Example 1
Evaluate the following definite integral
Solution:
∫2
4xdx =
0
2
4
x2 0
1+1
∫2
0
4xdx
2
= 2x2 |0 = 2(2)2 − 2(0)2 = 8 − 0 = 8
Example 2
∫2
Evaluate the following definite integral 0 x3 + x2 dx
Solution:
∫2
( 1 3+1
3
2
x
+
x
dx
=
x
+
3+1
1(
2
= 41 x4 + 13 x3 )1
(
= 41 24 + 13(23 ) − ( 41 14 + 13 13 )
= 4 + 83 − 14 + 13 )
1
= 20
− 12
3
= 79
12
2
1
x2+1 )1
2+1
Example 3
∫2
Evaluate the following definite integral 0 6x2 dx
Solution:
∫2
6
2
2+1 2
6x
dx
=
x
2+1
0
1
2
= 2x3 |1
= 2 · 23 − 2 · 03
= 16 − 0
= 16
20
Example 4
∫2
Evaluate the following definite integral 0 ex dx
Solution:
∫2
x
x
0 e dx = e |0
2
= e2 − e0
= e2 − 1 ≈ 6.39
Example 5
∫1
Evaluate the following definite integral 0
3 1
1 x2
1 1
= [ 2 · 2 + 2 · 2 · x 2 ]
0
3 1
x2
1
= [ 4 + 3 · x 2 ]
3
2
0
√
x+ x
dx
2
3
2
= [ 14 + 13 · 1 2 ] − [ 04 + 13 · 0 2 ]
1
7
− 13 − 0 = 12
4
Solution:
∫1
0
√
x+ x
dx
2
4.4.3
=
∫1
√
x
0 2
+
x
dx
2
=
Integrals of Absolute Value Functions
Example 6
∫2
Evaluate the following definite integral 0 |2x − 3|dx
Solution:
Note: 

f (x) =

∫2
2x − 3
x≥
3
2
−(2x − 3) x <
3
2
∫2
−(2x − 3)dx + 3 (2x − 3)dx
2
∫3
∫2
2
= 0 (−2x + 3)dx + 3 (2x − 3)dx
2
3
2
2
2
2
= (−x + 3x)|0 + (x − 3x)| 3
2
= [(( 32 )2 + 3 · 32 ) − ((−0)2 + 3 · 0)] + [(22 − 3(2)) − (( 32 )2 − 3 · 32 )]
= 10
4
0 |2x − 3|dx =
∫3
2
0
21
4.4.4
Mean Value Theorem
If f is a continuous function on the closed interval [a, b], then there exist a number c in the
closed interval [a, b] such that:
∫b
a
f (x)dx = f (c)(b − a)
4.4.5
The Average Value of a Function
If f is integrable on the closed interval [a, b], the average value of the function f is given by
the integral.
1
b−a
∫b
a f (x)dx
In the next examples, we will average value of a function on a given interval.
Example 7
Find the average value of the function f (x) = x2 + 2 on the interval [−2, 2]
Solution:
∫2
1
AV = 2−(−2)
(x2 + 2)dx
−2
∫
2
= 41 −2 (x2 + 2)dx
2
= 1 · ( 1 x3 + x)
=
=
=
=
=
4
3
−2
1
1
3
·
[
(2)
+
2]
− 13
4
3
1 14
· 3 − 13 · −14
4
3
7
7
+
6
6
14
6
7
3
· [ 13 (−2)3 − 2]
Example 8
Find the average value of the function f (x) = cos(x) on the interval [0, π2 ]
Solution:
∫π
AV = π 1−0 02 cos(x)dx
∫ π2
2
= π 02 cos(x)dx
π
(
= π2 · sin(x))|02
= π2 · [sin( π2 ) − sin(0)]
= π2 · [1 − 0]
= π2 · 1
= π2
22
Example 9
Find the area of the region bounded by the following graphs of equations.
y = x2 + 2 : y = 0, x = 0, x = 2
Solution:
∫2
Area= 0 (x2 + 2)dx
2
( x3
= 3 + 2x)
3
0
3
= [ 23 + 2(2)] − [ 03 + 2(0)]
= [ 83 + 4] − [0 + 0]
= 20
−0
3
20
= 3
23
Example 10
Find the area of the region bounded by the following graphs of equations.
y = ex : y = 0, x = 0, x = 2
Solution:
∫2
Area= 0 (ex )dx
= [ex ]|20
= e2 − e0
= e2 − 1
Example 11
Find the value of the transcendental function.
∫π
0 (1 + cos(x))dx
Solution:
∫π
(
x + sin(x))|π0
= (π + sin(π)) − (0 + sin(0))
=π+0−0=π
0 (1 + cos(x))dx =
24
4.5
4.5.1
Integration by Substitution
Review of the Chain Rule
Recall that with the chain rule we used a basic substitution to find the derivative.
For example let’s look at the following derivative.
Example 1
Find the derivative of the following function. f (x) = (x2 + 4x)6
Solution:
f (x) = (x2 + 4x)6
let f (x) = u6 where u = x2 + 4x ⇒ du = 2x + 4
Using the chain rule we get the follow derivative.
f ′ (x) = 6u5 · du
f (x) = 6(x2 + 4x)5 (2x + 4)
Now, we will reverse the process of the chain rule which is integration using substitution.
Let’s start by examining the following example.
4.5.2
Basic Substitutions
Example 2
Evaluate
∫
(2x + 3)(x2 + 3x)6 dx
Solution:
Let u = x2 + 3x ⇒ du = (2x + 3)dx
∫
∫
(2x + 3)(x2 + 3x)6 dx =
u6 · du
Now, find the antiderivative.
∫
u6 · du =
1
u6+1
6+1
+ C = 17 u7 + C = 17 (x2 + 3x)7 + C
Example 3
Evaluate
∫
3(1 + 3x)3 dx
Solution:
Let u = 1 + 3x ⇒ du = 3dx
∫
3(1 + 3x)3 dx =
∫
u3 · du
25
Now, find the antiderivative.
∫
u3 · du =
1
u3+1
3+1
+ C = 14 u4 + C = 14 (1 + 3x)4 + C
Example 4
Evaluate
∫
x2 (2 + x3 )4 dx
Solution:
1
2
3
2
Let
∫ u = 2 + x ⇒ du = 3x dx ⇒ 3 du = x dx
x2 (2 + x3 )4 dx =
∫
1 4
u
3
· du
Now, find the antiderivative.
∫
1 4
u
3
· du =
1
3
· 15 u5 + C =
1 5
u
15
+C =
1
(2
15
+ x3 )5 + C
26
Example 5
Evaluate
∫
√
2t t2 + 3dt
Solution:
Let u = t2 + 3 ⇒ du = 2tdt
∫ √
∫ √
∫ 1
2t t2 + 3dt =
u · du =
u 2 · du
Now, find the antiderivative.
∫
1
u 2 · du =
2
3
√
3
3
· u 2 + C = 23 (t2 + 3) 2 + C = 23 ( t2 + 3)3 + C
Example 6
Evaluate
∫
πcos(πx)dx
Solution:
Let u = πx ⇒ du = πdx
∫
πcosπxdx =
∫
cos(u)du
Now, find the antiderivative.
∫
cos(u)du = sin(u) + C = sin(πx) + C
Example 7
Evaluate
∫
√ x
dx
1−4x2
Solution:
Let u = 1 − 4x2 ⇒ du = −8xdx ⇒ − du
= xdx
8
∫
√ x
dx
1−4x2
=
∫
− 18 ·
du
√
u
Now, find the antiderivative.
∫
√
1
1
1
1
− 18 · u− 2 = − 18 · 2u 2 + C = − 14 u 2 + C = − 14 (1 − 4x2 ) 2 + C = − 14 1 − 4x2 + C
27
Example 8
Evaluate
∫
2
2xe2x dx
Solution:
Let u = 2x2 ⇒ du = 4xdx ⇒
∫
2
2xe2x dx =
∫
du
2
= 2xdx
1 u
e du
2
Now, find the antiderivative.
∫
1 u
e du
2
2
= 12 eu + C = 12 e2x + C
Example 9
Evaluate
∫
2xsin(2x2 )dx
Solution:
Let u = 2x2 ⇒ du = 4xdx ⇒ 12 du = 2xdx
∫
2xsin2x2 dx =
∫
1
sin(u)du
2
Now, find the antiderivative.
∫
1
sin(u)du
2
= − 12 cos(u) + C = − 12 cos(2x2 ) + C
28
4.6
The Natural Logarithm Function
Definition of the antiderivative of the natural function.
Let u be a differentiable function of x.
1.
2.
∫
1
x dx = ln|x| + C
∫
1
u du = ln|x| + C
Example 1
Evaluate
∫
1
3x dx
Solution:
Let u = 3x ⇒ du = 3dx ⇒ 13 du = dx
∫
1
3x dx =
∫
1
3
·
1
u
du
Now, find the antiderivative.
∫
1
3
·
1
u
du = 13 ln|u| + C = 13 ln|3x| + C
Example 2
Evaluate
∫
1
5x+3 dx
Solution:
Let u = 5x + 3 ⇒ du = 5dx ⇒ 15 du = dx
∫
1
3x dx =
∫
1
5
·
1
u
du
Now, find the antiderivative.
∫
1
5
·
1
u
du = 15 ln|u| + C = 15 ln|5x + 3| + C
29
Example 3
Evaluate
∫
x2
3−x3 dx
Solution:
Let u = 3 − x3 ⇒ du = −3x2 dx ⇒ − 13 du = x2 dx
∫
x2
3−x3 dx =
∫
− 13 ·
1
u
du
Now, find the antiderivative.
∫
− 13 ·
du = − 13 ln|u| + C = − 13 ln|3 − x3 | + C
1
u
Example 4
Evaluate
∫
sinθ
cosθ dθ
Solution:
Let u = cosθ ⇒ du = −sinθ ⇒ −du = sinθ
∫
sinθ
cosθ dθ =
∫
− u1 du
Now, find the antiderivative.
∫
− u1 du = −ln|u| + C = −ln|cosθ| + C
Example 5
Evaluate
∫
cos(t)
3+sin(t) dt
Solution:
Let u = 3 + sin(t) ⇒ du = cos(t)dt
∫
cos(t)
3+sin(t) dθ =
∫
1
u
du
Now, find the antiderivative.
∫
1
du
u
= ln|u| + C = ln|3 + sin(t)| + C
30
4.7
Antiderivative of the Inverse Trigonometric Functions
4.7.1
Basic Antiderivatives of the Inverse Trigonometric Functions
Theorem 4.8
Integrals involving inverse trigonometric functions.
Let u be a differentiable function of x, and let a > 0
1.
2.
3.
∫
∫
∫
4.7.2
√ du
= arcsin ua +C
a2 −u2
du
1
u
a2 +u2 = a arctan a +C
|u|
√ du
= a1 arcsec a +C
u u2 −a2
Integration by Substitution
Here are some example of integration that results in inverse trigonometric functions.
Example 1
Evaluate
∫
4
1+9x2 ·dx
Solution:
Let u = 3x ⇒ du = 3dx ⇒ 13 du = dx
Now, integrate by substitution.
∫
4
∫1+9x2 4 ·dx
= 1+(3x)2 ·dx
∫ 4du
= 31 1+u
∫ du 2
4
= 3
1+u2
= 43 arctan(u) + C
= 43 arctan(3x) + C
Example 2
Evaluate
∫
dx
4+(x+1)2
Solution:
Let u = x + 1 ⇒ du = dx
Now, integrate by substitution.
31
∫
dx
4+(x+1)2 =
∫
du
22 +u2
= 12 arctan( u2 ) + C
= 12 arctan( x+1
)+C
2
Example 3
Evaluate
∫
√ 7
·dx
4−x2
Solution:
Let u = x ⇒ du = dx
Now, integrate by substitution.
∫
∫
√ 7
=
4−x2
√ 7du
22 −u2
= 7 · arcsin( u2 ) + C
= 7arcsin( x2 ) + C
Example 4
Evaluate
∫
√dx
x· x2 −9
Solution:
Let u = x ⇒ du = dx
Now, integrate by substitution.
∫
√dx
=
x· x2 −9
∫
√du
u u2 −9
= 13 arcsec( u3 ) + C
= 13 arcsec( |x|
)+C
3
32
Example 5
Evaluate
∫
t
t4 +16 ·dt
Solution:
Let u = t2 ⇒ du = 2tdt ⇒ 12 du = tdt
Now, integrate by substitution.
∫
t
·dt = 12
∫
du
u2 +16
∫
= 12 u2du
+42
(
)
1 1
u
= 2 4 arctan( 4 ) + C
t4 +16
2
= 18 arctan( t4 ) + C
Example 6
Evaluate
∫
√2tdt
9−t4
Solution:
Let u = t2 ⇒ du = 2tdt
Now, integrate by substitution.
∫
√2tdt =
9−t4
∫
√ du
9−u2
= arcsin( u3 ) + C
2
= arcsin( t3 ) + C
Example 7
∫
x
e dx
Evaluate 9+e
2x
Solution:
Let u = ex ⇒ du = ex dx
Now, integrate by substitution.
∫
ex dx
9+e2x =
∫
du
32 +u2
= 13 arctan( u3 ) + C
x
= 13 arctan( e3 ) + C
33
Example 8
Evaluate
∫
√
x+5
9+(x−3)2
Solution:
Let u = x − 3 ⇒ du = dx
Also let v = 9 − (x − 3)2
⇒ v = 9 − (x2 − 6x + 9)
⇒ v = 6x − x2
⇒ dv = (−2x + 6)dx
⇒ − 21 du = (x − 3)dx
Now, substitute
∫
√
x+5
=
9+(x−3)2
= − 12
=
1
2
∫
∫
dv
√
+
v
∫
8·
√
∫
∫
x−3
√ 8 2
+
2
9+(x−3)
9+(x−3)
√ du
9+(u)2
∫
du
v − 2 dv + 8 · √9+(u)
2
1
1
= − 12 (2v 2 ) + 8arcsin( u3 ) + C
1
= −(6x − x2 ) 2 + 8arcsin( x−3
)+C
3
= 8arcsin( x−3
)−
3
√
6x − x2 + C
34
4.8
4.8.1
Simpson’s Rule and the Trapezoid Rule
The Trapezoid Rule
Let be continuous of [a, b]. The Trapezoid rule approximating the integral
given by the function:
∫b
a f (x)dx is
∫b
b−a
a f (x)dx = 2n [f (x0 ) + 2f (x1 ) + 2f (x2 ) + 2f (x3 ) + ..... + 2f (xn−1 + f (xn )]
Example 1
Use the Trapezoid Rule to the approximate value of the definite integral for the given value
of n.
∫2
2
0 2x dx: n = 4
Solution:
△x =
2−0
4
=
2
4
=
1
2
xi = a + i · △x = 0 + i ·
∫2
1
2
=
i
2
2−0
1
3
2
0 2x dx = 2(4) [f (0) + 2f ( 2 ) + 2f (1) + 2f ( 2 ) + f (2)]
= 82 [2(0)2 + 2(2( 21 )2 ) + 2(2(1)2 ) + 2(2( 32 )2 ) + 2(22 )]
= 41 [0 + 2( 21 ) + 2(2) + 2( 18
) + 8]
4
1
= 4 [0 + 1 + 4 + 9 + 8]
= 14 [22]
= 5.5
4.8.2
Simpson’s Rule
Let be continuous of [a, b]. The Simpson’s rule approximating the integral
given by the function:
∫b
a f (x)dx is
∫b
b−a
a f (x)dx = 3n [f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + ..... + 4f (xn−1 + f (xn )]
Example 2
Use the Trapezoid Rule to the approximate value of the definite integral for the given value
of n. (Same definite integral as Example 1)
∫2
2
0 2x dx: n = 4
Solution:
35
△x =
2−0
4
=
2
4
=
1
2
xi = a + i · △x = 0 + i ·
∫2
1
2
=
i
2
2−0
1
3
2
0 2x dx = 3(4) [f (0) + 4f ( 2 ) + 2f (1) + 4f ( 2 ) + f (2)]
2
= 12
[2(0)2 + 4(2( 12 )2 ) + 2(2(1)2 ) + 4(2( 23 )2 ) + 2(22 )]
) + 8]
= 16 [0 + 4( 12 ) + 2(2) + 4( 18
4
1
= 6 [0 + 2 + 4 + 18 + 8]
= 16 [32]
= 5.3333
Now, lets find the exact value of the same definite integral in Example 1 and Example 2 use
integration.
∫2
2
2 3 2
2
2+1 2
2x
dx
=
x
=
x
2+1
3
0
0
0
= 23 (2)3 − 32 (0)3
= 23 · 8 − 0
= 16
3
Example 3
Use the Trapezoid Rule and Simpson’s Rule to approximate value of the define integral using
n = 4.
∫π
0 sin(x)dx
Solution:
△x =
π−0
4
=
π
4
xi = a + i · △x = 0 + i ·
∫π
π
4
=
iπ
4
π−0
π
2π
3π
0 sin(x)dx = 2(4) [f (0) + 2f ( 4 ) + 2f ( 4 ) + 2f ( 4 ) + f (π)]
π
) + 2sin( 3π
) + sin(π)]
= π8 [sin(0) √+ 2sin( π4 ) + 2sin(
4
√ 2
= π8 [0 + 2( 22 ) + 2(1) + 2( 22 ) + 0]
√
= π8 [2 2 + 2]
≈ 2.003
Now, find the value of the integral using Simpson’s Rule.
∫π
0 sin(x)dx
Solution:
△x =
π−0
4
=
π
4
36
xi = a + i · △x = 0 + i ·
∫π
π
4
=
iπ
4
π−0
π
2π
3π
0 sin(x)dx = 3(4) [f (0) + 4f ( 4 ) + 2f ( 4 ) + 4f ( 4 ) + f (π)]
π
π
= 12
[sin(0) √+ 4sin( π4 ) + 2sin(
) + 4sin( 3π
) + sin(π)]
4
√ 2
2
2
π
= 12 [0 + 4( 2 ) + 2(1) + 4( 2 ) + 0]
√
π
= 12
[4 2 + 2]
≈ 2.004
Now, compare the solutions to value of the definite integral.
∫π
π
0 sin(x)dx = −cos(x)|0
= −cos(π) − (−cos(0))
= −(−1) + 1
=2
Example 4
Use the Trapezoid Rule and Simpson’s Rule to approximate value of the define integral using
n = 4.
∫2
1
1
dx
(x+2)2
Solution:
△x =
2−1
4
=
1
4
xi = a + i · △x = 0 + i ·
∫2
1
4
i
4
=
1
2−1
5
6
7
1 (x+2)2 = 2(4) [f (1) + 2f ( 4 ) + 2f ( 4 ) + 2f ( 4 ) + f (2)]
1
1
1
1
1
= 18 [ (1+2)
2 + 2( ( 5 +2)2 ) + 2( ( 3 +2)2 ) + 2( ( 7 +2) ) + (2+2)2 ]
4
2
4
= 18 [.111 + 2(.0947) + 2(.0816) + 2(.0771) + .0625] ≈ .085
Now, find the value of the integral using Simpson’s Rule.
∫π
0 sin(x)dx
Solution:
△x =
π−0
4
=
π
4
xi = a + i · △x = 0 + i ·
∫2
π
4
=
iπ
4
2−1
5
6
7
1
1 (x+2)2 = 3(4) [f (1) + 4f ( 4 ) + 2f ( 4 ) + 4f ( 4 ) + f (2)]
1
1
1
1
1
1
) + 2( ( 3 +2)
= 12
[ (1+2)
2 + 4( 5
2 ) + 4( ( 7 +2) ) + (2+2)2 ]
( +2)2
4
2
4
37
=
1
[.111
12
+ 4(.0947) + 2(.0816) + 4(.0771) + .0625] ≈ .086
Now, compare the solutions to value of the definite integral.
Let
= dx
∫ 2 u = x + 2 ⇒∫ du
2 1
1
1 (x+2)2 dx = 1 u2 du =
2
= −(x + 2)−1 |1
1 2
= − x+2
1
1
1
= − 2+2
− (− 1+2
)
1
1
= −4 + 3
1
= 12
∫2
−2
−1
1 u du = = −u |1
2
38