Contents

1 Chapter 1 1

2 Chapter 1 1

3 Chapter 1 1

4 Integration 24.1 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

4.1.1 Definition of an Antiderivative . . . . . . . . . . . . . . . . . . . . . . 24.1.2 Antiderivatives of Trigonometric Functions . . . . . . . . . . . . . . . 34.1.3 Antiderivatives of Other Functions . . . . . . . . . . . . . . . . . . . 4

4.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64.2.1 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64.2.2 Expanding Summations . . . . . . . . . . . . . . . . . . . . . . . . . 64.2.3 Area of Plane Regions . . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.3 Riemann Sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.3.1 Definition of a Riemann Sum . . . . . . . . . . . . . . . . . . . . . . 14

4.4 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.4.1 Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . 204.4.2 Evaluating Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.4.3 Integrals of Absolute Value Functions . . . . . . . . . . . . . . . . . . 214.4.4 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.4.5 The Average Value of a Function . . . . . . . . . . . . . . . . . . . . 22

4.5 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.5.1 Review of the Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . 254.5.2 Basic Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.6 The Natural Logarithm Function . . . . . . . . . . . . . . . . . . . . . . . . 294.7 Antiderivative of the Inverse Trigonometric Functions . . . . . . . . . . . . . 31

4.7.1 Basic Antiderivatives of the Inverse Trigonometric Functions . . . . . 314.7.2 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . 31

4.8 Simpson’s Rule and the Trapezoid Rule . . . . . . . . . . . . . . . . . . . . . 354.8.1 The Trapezoid Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.8.2 Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1 Chapter 1

See Math 151 Notes

2 Chapter 1

See Math 151 Notes

3 Chapter 1

See Math 151 Notes

1

4 Integration

4.1 Antiderivatives

4.1.1 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∫cxndx = cn+1x

n+1 + C

Now, let’s try doing so examples that use the power rule.

Example 1

Evaluate the following integral∫

x2dx

Solution:∫x2dx = 1

2+1x2+1 + C = 1

3x3 + C

Example 2

Evaluate the following integral∫

(x4 + 3x2 + 3x)dx

Solution:∫(x4 + 3x2 + 3x)dx = 1

4+1x4+1 + 3

2+1x2+1 + 3

1+1x1+1 + C = 1

5x5 + x3 + 3

2x2 + C

Example 3

Evaluate the following integral∫

(x3 − 5x2 + 6)dx

Solution:∫(x3 − 5x2 + 6x0)dx = 1

3+1x3+1 − 5

2+1x2+1 + 6

1+0x0+1 + C = 1

4x4 − 5

3x3 + 6x+ C

2

Example 4

Evaluate the following integral∫

5x4dx

Solution:∫5x4dx = 5

4+1x4+1 + C = x5 + C

4.1.2 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∫

(sec2x− csc2x)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

2+1x2+1 + C = −cos(x) + x3 + C

3

Example 7

Evaluate the following integral∫

(sec2x− csc2x)dx

Solution:∫(cos(x) + 4)dx = sin(x) + 4

0+1x0+1 + C = sin(x) + 4x+ C

4.1.3 Antiderivatives of Other Functions

Antiderivatives of exponential and logarithmic functions∫1xdx = ln|x|+ C∫

exdx = ex + C

Example 8

Evaluate the following integral∫

exdx

Solution:∫ex = ex + C

Example 9

Evaluate the following integral∫

( 1x+ x4)dx

Solution:∫( 1x+ x4) = ln|x|+ 1

4+1x4+1 + C = 1

5x5 + C

Example 10

Evaluate the following integral∫

x32dx

Solution:∫x

32dx = 13

2+1

x32+1 + C = 15

2

x52 + C = 2

5x

52 + C

4

Example 11

Evaluate the following integral∫

1x3dx

Solution:∫1x3dx

∫x−3dx = 1−3+1x

−3+1 + C = −12x−2 + C = − 1

2x2+ C

Example 12

Evaluate the following integral∫

( 1x2

+ cos(x))dx

Solution:∫( 1x2

+ cos(x))dx =∫

(x−2 + cos(x))dx = 1−2+1x−2+1 + sin(x) +C = −x−1 + sin(x) +C =

− 1x+ sin(x) + C

5

4.2 Area

4.2.1 Sigma Notation

The sum of the terms a1, a2, a3, a4.......an is:n∑

i=1

ai = a1 + a2 + a3 + a4 + ......+ an

where i is the index of a summation, ai is the ith term of the sum, and the upper and lowersum of the summation are 1 and n

Here are some example of summations.

Example 1

Evaluate6∑

i=1

i

Solution

6∑i=1

i = 1 + 2 + 3 + 4 + 5 + 6 = 21

Example 2

Evaluate4∑

i=0

(2i+ 1)

Solution:4∑

i=0

(2i+ 1) = 2(0) + 1 + 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + 2(4) + 1 = 1 + 3 + 5 + 7 + 9 = 25

4.2.2 Expanding Summations

Example 3

Evaluate4∑

j=0

j2

Solution:4∑

j=0

j2 = 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30

6

Example 4

Evaluate6∑

k=3

k(k − 2)

Solution:6∑

k=3

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

Example 5

Evaluate5∑

j=3

1

j

Solution:

5∑j=3

1

j=

1

3+

1

4+

1

5=

47

60

Example 6

Evaluate4∑

j=1

[(i− 1)2 + (i+ 1)3]

Solution:4∑

j=1

[(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−

1)2+(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 51+1

+ 51+2

+5

1+3+ .......+ 5

1+12

Solution:Since the series of terms is adding one to each successive term and then the dividing thistotal into 5 the summation would be written as follows:12∑j=1

[5

1 + i]

7

Example 8

Express the following series of terms into a expression using summation notation [1− (14)2]+

[1− (24)2] + 1− [(3

4)2] + [1− (4

4)2]

Solution:Since the series of terms is dividing each successive term by 4, squaring the result, and addingone the summation would be written as follows:4∑

j=1

[1 + (i

4)2]

Theorem 4.2: Summation Formulas

1.u∑

j=1

c = cn

2.u∑

j=1

i =n(n+1)

2

3.u∑

j=1

i2 =n(n+1)(2n+1)

6

4.u∑

j=1

i3 =n2(n+1)2

4

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∑j=1

(2i− 3)

Solution:15∑j=1

(2i− 3) =15∑j=1

2i−15∑j=1

3 = 2 ·15∑j=1

i−15∑j=1

3 = 2 · 15(16)2

− 15(3) = 240− 45 = 195

8

Example 10

Use the properties of Theorem 4.2 to expand out the following summation.

15∑j=1

i(i2 + 1)

Solution:10∑j=1

i(i2 + 1) =10∑j=1

i3 + i

=n2(n+1)2

4 +n(n+1)

2 (where n=10)

=102(10+1)2

4 +10(10+1)

2

= 100·1214 +10·112

= 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 toapproximate areas under the curve by dividing the region under the curve into smallerrectangles and summing up the rectangles as shown in the next diagram.

Theorem 4.3: Limits of Upper and Lower SumsLet f be continuous and nonnegative on the interval [a, b]. The limits as n → ∞ of both thelower and upper sums exist and are equal to each other.

limh→∞

S(n) = limh→∞

f(mj)△x = limh→∞

f(Mj)△x = S(n) where △x = b−an

9

Example 10

Use upper sums and lower to approximate the area of the function f(x) =√x below bounded

by the x-axis y = 0 and the values x = 0 and x = 1.

Solution:

First find △x (The length of each rectangle)

a = 0 and b = 1 ⇒ △x = 1−04

= 14

Next, compute the left and right endpoints.

Left Endpoint: mi = a+ (i− 1)△x = 0 + (i− 1) · 14 =i−14

Right Endpoint: Mi = a+ i△x = 0 + i · 14 =i4

Computing the Lower Sums

S(n) =4∑

j=1

f(mi)△x

=4∑

i=1

f(i− 14

) · (14)

=4∑

i=1

√i− 14

· 14

=4∑

i=1

√i− 12

· 14

=4∑

i=1

√i− 18

=√1−18 +

√2−18 +

√3−18 +

√4−18

=√08 +

√18 +

√28 +

√38 = 0 + .125 + .177 + .217 ≈ .519

10

Computing the Upper Sums

S(n) =4∑

j=1

f(Mi)△x

=4∑

i=1

f(i

4) · (1

4)

=4∑

i=1

√i

4· 14

=4∑

i=1

√i

2· 14

=4∑

i=1

√i

8

=√18 +

√28 +

√38 +

√48

= .125 + .177 + .217 + .25 ≈ .769

Example 11

Find the limit of S(n) as n → ∞ of the summation S(n) = 64n3[n(n+1)(2n+1)

6]

Solution:

limh→∞

S(n) = limh→∞

64

n3[n(n+ 1)(2n+ 1)

6]

= limh→∞

64

n3[(n2 + n))(2n+ 1)

6]

= limh→∞

64

n3[(2n3 + n2 + 2n2 + n

6]

= limh→∞

64[2n3 + 3n2 + n

6n3]

= limh→∞

128n3 + 192n2 + 64n

6n3

11

= limh→∞

128n3

6n3+

192n2

6n3+

64n

6n3

= limh→∞

64

3+

32

n+

32

3n2= 64

3+ 0 + 0 = 64

3

12

4.3 Riemann Sum

Example 1: A Partition with Unequal Widths

Consider the region bounded by the graph of f(x) =√x and the x-axis for 0 ≤ xi ≤ 1

Now, let’s evaluate the limit limh→∞

n∑i=1

f(ci)△xi

where ci is the right endpoint of the portion given by ci =i2

n2and △xi is the width of the

ith interval.

Let: △xi = right endpoint - left endpoint

△xi = i2

n2 −(i−1)2n2

= i2

n2 −i2−2i+1

n2

= i2−(i2+2i−1)

n2

= i2−i2−2i+1)

n2

= −2i+1n2

Now, we can evaluate the following limit

limh→∞

n∑i=1

f(ci)△xi = limh→∞

n∑i=1

f(i2

n2) · (2i− 1

n2)

= limh→∞

n∑i=1

√i2

n2· (2i− 1

n2)

= limh→∞

n∑i=1

i

n· 2i− 1

n2

= limh→∞

n∑i=1

2i2 − in3

13

=1

n3limh→∞

n∑i=1

(2i2 − i)

=1

n3limh→∞

[2(n(n+ 1)(2n+ 1)

6)− n(n+ 1)

2]

=1

n3limh→∞

(2n(n+ 1)(2n+ 1)

6− n

2 + n

2]

=1

n3limh→∞

((2n2 + 2n)(2n+ 1)

6− n

2 + n

2]

=1

n3limh→∞

(4n3 + 6n2 + 2n

6− n

2 + n

2]

=1

n3limh→∞

4n3 + 6n2 + 2n− 3n2 − 3n6

=1

n3limh→∞

4n3 + 3n2 − n6n3

= limh→∞

[4n3 + 3n2 − n

6n3]

= limh→∞

4n3

6n3+

3n2

6n3+

n

6n3

= limh→∞

2

3+

1

2n+

1

6n2=

2

3+ 0 + 0 =

2

3

4.3.1 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

ci is any point in ith subinterval,

n∑i=1

f(ci)△xi then the sum is called the Riemann sum.

Definition of a Definite Integral

If f defined on a closed interval [a, b] and the limit lim|△|→∞

n∑i=1

f(ci)△xi exist as described

above, then f is integrable on [a, b] and the limit is lim|△|→∞

n∑i=1

f(ci)△xi =∫ ba f(x)dx

14

Example 2

Evaluate the following definite integral by the limit definition.

Solution:∫ 3−2 xdx

Solution:

a = −2 b = 3 △x = b−an =3−(−2)

n

Now, rewrite the integral as a summation∫ 3−2 xdx = limn→∞

n∑i=1

f(ci)△xi

= limn→∞

n∑i=1

(− 2 + 5i

n

)· 5n

= limn→∞

n∑i=1

(−10n

+25i

n2)

= limn→∞

n∑i=1

(−10n

+25

n2· n(n+ 1)

2

)= lim

n→∞

n∑i=1

(−10n

+25n2 + 25n

2n2)

= limn→∞

n∑i=1

(−10n

+25n2

2n2+

25n

2n2)

= limn→∞

n∑i=1

(−10n

+25

2+

25

2n

)= −0 + 12.5 + 0 = 12.5

Example 3

Evaluate the following definite integral by the limit definition.

Solution:

15

∫ 31 3x

2dx

Solution:

a = 1 b = 3 △x = 3−1n =2n

ci = 1 + i · 2i = 1 +2in

Now, rewrite the integral as a summation∫ 31 xdx = limn→∞

n∑i=1

f(ci)△xi

= limn→∞

n∑i=1

f(1 +2i

n) · 2

n

= limn→∞

n∑i=1

3(1 +2i

n)2 · 2

n

= limn→∞

n∑i=1

3(4i2

n2+

4i

n+ 1) · 2

n

= limn→∞

n∑i=1

3(8i2

n3+

8i

n2+

2

n)

= limn→∞

n∑i=1

(24i2

n3+

24i

n2+

6

n)

= limn→∞

n∑i=1

(24(n(n+ 1)(2n+ 1))6n3

+24n(n+ 1)

2n2+

6n

n

)= lim

n→∞

n∑i=1

(24(2n3 + 3n2 + n)6n3

+24n2 + 24n

2n2+

6n

n

)= lim

n→∞

n∑i=1

(48n36n3

+72n2

6n3+

24n

6n3+

24n2

2n2+

24n

n2+ 6

)16

=(8 +

12

n+

4

n2+ 12 +

12

n+ 6

)= 26

Example 4

Evaluate the following definite integral using the limit definition of a Riemann Sum∫ 104 6dx

Solution:

a = 4; b = 10; △x = 10−4n =6n

ci = a+ i · △x = 4 + i · 6n = 4 +6in

f(x) = 6

Now, rewrite the integral as a summation∫ 104 6dx = limn→∞

n∑i=1

f(ci)△xi

= limn→∞

n∑i=1

f(4 +

6i

n

)· 6n

= limn→∞

n∑i=1

(6) · 6n

= limn→∞

(6n) · 6n= 36

17

Example 5

Evaluate the following integral using the limit definition of an integral∫ 21 (x

2 + 1)dx

a = 1; b = 2; △x = 2−1n =1n

ci = a+ i · △x = 1 + i · 1n = 1 +in

f(x) = x2 + 1

Now, rewrite the integral as a summation∫ 21 (x

2 + 1)dx = limn→∞

n∑i=1

f(ci)△xi

= limn→∞

n∑i=1

f(1 +

i

n

)· 1n

= limn→∞

n∑i=1

((1 +

i

n

)2+ 1

)· 1n

= limn→∞

n∑i=1

((1 +

i

n

)(1 +

i

n

)+ 1

)· 1n

= limn→∞

n∑i=1

((1 +

2i

n+

i2

n2

)+ 1

)· 1n

= limn→∞

n∑i=1

(2 +

2i

n+

i2

n2

)· 1n

= limn→∞

n∑i=1

( 2n+

2i

n2+

i2

n3

)= lim

n→∞

( 2n· (n) + 2

n2· n(n+ 1)

2+

1

n3· n(n+ 1)(2n+ 1)

6

)= lim

n→∞

(2 +

2n2 + 2n

2n2+

(n2 + n)(2n+ 1)

6n3

)= lim

n→∞

(2 +

2n2 + 2n

2n2+

2n3 + 3n2 + n

6n3

)= lim

n→∞

(2 +

2n2

2n2+

2n

2n2+

2n3

6n3+

3n2

6n3+

n

6n3

)= lim

n→∞

(2 + 1 +

1

n+

1

3+

1

2n+

1

6n2

)= 2 + 1 + 0 +

1

3+ 0 + 0 =

10

3

18

Example 6

Evaluate using the limit definition of an integral:∫ 30 5xdx

a = 0 b = 3 △x = 3−0n =3n

ci = a+ i · △x = 0 + i · 3n =3n

f(x) = 5x

Now, rewrite the integral as a summation

∫ 30 5xdx = limn→∞

n∑i=1

f(ci)△xi

= limn→∞

n∑i=1

f(3in

)· 3n

= limn→∞

n∑i=1

5(3in

)· 3n

= limn→∞

n∑i=1

45i

n2

= limn→∞

n∑i=1

45

n2· n(n+ 1)

2

= limn→∞

n∑i=1

(45n(n+ 1)2n2

)= lim

n→∞

n∑i=1

(45n2 + 45n2n2

)= lim

n→∞

n∑i=1

(45n22n2

+45n

2n2

)= lim

n→∞

n∑i=1

(452

+45

2n

)=

45

2+ 0 =

45

2

19

4.4 Definite Integrals

4.4.1 Fundamental Theorem of Calculus

If a function f is continuous on a closed interval [a, b] and F is an the antiderivative of f onthe interval [a, b], then∫ ba f(x)dx = F (b)− F (a)

When we evaluating a definite integral, we first find the antiderivative and then substitutethe limits into the resulting antiderivative.

4.4.2 Evaluating Integrals

Example 1

Evaluate the following definite integral∫ 204xdx

Solution:∫ 204xdx = 4

1+1x2∣∣20= 2x2|20 = 2(2)2 − 2(0)2 = 8− 0 = 8

Example 2

Evaluate the following definite integral∫ 20 x

3 + x2dx

Solution:∫ 21 x

3 + x2dx =(

13+1

x3+1 + 12+1

x2+1)∣∣21

=(

14x4 + 1

3x3)

∣∣21

=(1424 + 1

323)− (1

414 + 1

313)

= 4 + 83−(14+ 1

3)

= 203− 1

12

= 7912

Example 3

Evaluate the following definite integral∫ 20 6x

2dx

Solution:∫ 20 6x

2dx = 62+1

x2+1∣∣21

= 2x3|21= 2 · 23 − 2 · 03= 16− 0= 16

20

Example 4

Evaluate the following definite integral∫ 20 e

xdx

Solution:∫ 20 e

xdx = ex|20= e2 − e0= e2 − 1 ≈ 6.39

Example 5

Evaluate the following definite integral∫ 10

x+√x

2dx

= [12· x2

2+ 1

2· 12· x 32 ]

∣∣∣10

= [x2

4+ 1

3· x 32 ]

∣∣∣10

= [12

4+ 1

3· 1 32 ]− [02

4+ 1

3· 0 32 ]

14− 1

3− 0 = 7

12

Solution:∫ 10

x+√x

2dx =

∫ 10

x2+

√x2dx =

4.4.3 Integrals of Absolute Value Functions

Example 6

Evaluate the following definite integral∫ 20 |2x− 3|dx

Solution:

Note:

f(x) =

2x− 3 x ≥32

−(2x− 3) x < 32∫ 2

0 |2x− 3|dx =∫ 3

2

0−(2x− 3)dx+

∫ 232(2x− 3)dx

=∫ 3

2

0 (−2x+ 3)dx+∫ 2

32(2x− 3)dx

= (−x2 + 3x)|320 + (x

2 − 3x)|232

= [((32)2 + 3 · 3

2)− ((−0)2 + 3 · 0)] + [(22 − 3(2))− ((3

2)2 − 3 · 3

2)]

= 104

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 theclosed interval [a, b] such that:∫ ba 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 bythe integral.

1b−a

∫ ba 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:

AV = 12−(−2)

∫ 2−2 (x

2 + 2)dx

= 14

∫ 2−2 (x

2 + 2)dx

= 14· (1

3x3 + x)

∣∣2−2

= 14· [1

3(2)3 + 2]− 1

3· [1

3(−2)3 − 2]

= 14· 14

3− 1

3· −14

3

= 76+ 7

6

= 146

= 73

Example 8

Find the average value of the function f(x) = cos(x) on the interval [0, π2]

Solution:

AV = 1π2−0

∫ π2

0cos(x)dx

= 2π

∫ π2

0cos(x)dx

= 2π·(sin(x))|

π20

= 2π· [sin(π

2)− sin(0)]

= 2π· [1− 0]

= 2π· 1

= 2π

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:

Area=∫ 20 (x

2 + 2)dx

=(

x3

3+ 2x)

∣∣∣20

= [23

3+ 2(2)]− [03

3+ 2(0)]

= [83+ 4]− [0 + 0]

= 203− 0

= 203

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:

Area=∫ 20 (e

x)dx

= [ex]|20= e2 − e0= e2 − 1

Example 11

Find the value of the transcendental function.∫ π0 (1 + cos(x))dx

Solution:∫ π0 (1 + cos(x))dx =

(x+ sin(x))|π0

= (π + sin(π))− (0 + sin(0))= π + 0− 0 = π

24

4.5 Integration by Substitution

4.5.1 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+ 4Using the chain rule we get the follow derivative.f ′(x) = 6u5 · duf(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)6dx

Solution:

Let u = x2 + 3x ⇒ du = (2x+ 3)dx∫(2x+ 3)(x2 + 3x)6dx =

∫u6 · du

Now, find the antiderivative.∫u6 · du = 1

6+1u6+1 + C = 1

7u7 + C = 1

7(x2 + 3x)7 + C

Example 3

Evaluate∫

3(1 + 3x)3dx

Solution:

Let u = 1 + 3x ⇒ du = 3dx∫3(1 + 3x)3dx =

∫u3 · du

25

Now, find the antiderivative.∫u3 · du = 1

3+1u3+1 + C = 1

4u4 + C = 1

4(1 + 3x)4 + C

Example 4

Evaluate∫

x2(2 + x3)4dx

Solution:

Let u = 2 + x3 ⇒ du = 3x2dx ⇒ 13du = x2dx∫

x2(2 + x3)4dx =∫

13u4 · du

Now, find the antiderivative.∫13u4 · du = 1

3· 15u5 + C = 1

15u5 + C = 1

15(2 + x3)5 + C

26

Example 5

Evaluate∫

2t√t2 + 3dt

Solution:

Let u = t2 + 3 ⇒ du = 2tdt∫2t√t2 + 3dt =

∫ √u · du =

∫u

12 · du

Now, find the antiderivative.∫u

12 · du = 2

3· u 32 + C = 2

3(t2 + 3)

32 + C = 2

3(√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√1−4x2dx

Solution:

Let u = 1− 4x2 ⇒ du = −8xdx ⇒ −du8= xdx∫

x√1−4x2dx =

∫−1

8· du√

u

Now, find the antiderivative.∫−1

8· u− 12 = −1

8· 2u 12 + C = −1

4u

12 + C = −1

4(1− 4x2) 12 + C = −1

4

√1− 4x2 + C

27

Example 8

Evaluate∫

2xe2x2dx

Solution:

Let u = 2x2 ⇒ du = 4xdx ⇒ du2= 2xdx∫

2xe2x2dx =

∫12eudu

Now, find the antiderivative.∫12eudu = 1

2eu + C = 1

2e2x

2+ C

Example 9

Evaluate∫

2xsin(2x2)dx

Solution:

Let u = 2x2 ⇒ du = 4xdx ⇒ 12du = 2xdx∫

2xsin2x2dx =∫

12sin(u)du

Now, find the antiderivative.∫12sin(u)du = −1

2cos(u) + C = −1

2cos(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.∫

1x dx = ln|x|+ C

2.∫

1u du = ln|x|+ C

Example 1

Evaluate∫

13x dx

Solution:

Let u = 3x ⇒ du = 3dx ⇒ 13du = dx∫

13x dx =

∫13· 1udu

Now, find the antiderivative.∫13· 1udu = 1

3ln|u|+ C = 1

3ln|3x|+ C

Example 2

Evaluate∫

15x+3 dx

Solution:

Let u = 5x+ 3 ⇒ du = 5dx ⇒ 15du = dx∫

13x dx =

∫15· 1udu

Now, find the antiderivative.∫15· 1udu = 1

5ln|u|+ C = 1

5ln|5x+ 3|+ C

29

Example 3

Evaluate∫

x2

3−x3 dx

Solution:

Let u = 3− x3 ⇒ du = −3x2dx ⇒ −13du = x2dx∫

x2

3−x3 dx =∫

−13· 1udu

Now, find the antiderivative.∫−1

3· 1udu = −1

3ln|u|+ C = −1

3ln|3− x3|+ C

Example 4

Evaluate∫

sinθcosθ dθ

Solution:

Let u = cosθ ⇒ du = −sinθ ⇒ −du = sinθ∫sinθcosθ dθ =

∫− 1

udu

Now, find the antiderivative.∫− 1

udu = −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 =∫

1udu

Now, find the antiderivative.∫1udu = 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.∫

du√a2−u2 = arcsin

ua +C

2.∫

dua2+u2 =

1a arctan

ua +C

3.∫

duu√u2−a2 =

1a arcsec

|u|a +C

4.7.2 Integration by Substitution

Here are some example of integration that results in inverse trigonometric functions.

Example 1

Evaluate∫

41+9x2 ·dx

Solution:

Let u = 3x ⇒ du = 3dx ⇒ 13du = dx

Now, integrate by substitution.∫4

1+9x2 ·dx=∫

41+(3x)2 ·dx

= 13

∫4du1+u2

= 43

∫du

1+u2

= 43arctan(u) + C

= 43arctan(3x) + C

Example 2

Evaluate∫

dx4+(x+1)2

Solution:

Let u = x+ 1 ⇒ du = dx

Now, integrate by substitution.

31

∫dx

4+(x+1)2 =∫

du22+u2

= 12arctan(u

2) + C

= 12arctan(x+1

2) + C

Example 3

Evaluate∫

7√4−x2 ·dx

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∫

dxx·√x2−9

Solution:

Let u = x ⇒ du = dx

Now, integrate by substitution.∫dx

x·√x2−9 =

∫du

u√u2−9

= 13arcsec(u

3) + C

= 13arcsec( |x|

3) + C

32

Example 5

Evaluate∫

tt4+16 ·dt

Solution:

Let u = t2 ⇒ du = 2tdt ⇒ 12du = tdt

Now, integrate by substitution.∫t

t4+16 ·dt =12

∫du

u2+16

= 12

∫du

u2+42

= 12

(14arctan(u

4))+ C

= 18arctan( t

2

4) + 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

= arcsin( t2

3) + C

Example 7

Evaluate∫

exdx9+e2x

Solution:

Let u = ex ⇒ du = exdx

Now, integrate by substitution.∫exdx9+e2x =

∫du

32+u2

= 13arctan(u

3) + C

= 13arctan( e

x

3) + 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⇒ −1

2du = (x− 3)dx

Now, substitute∫x+5√

9+(x−3)2=∫

x−3√9+(x−3)2

+∫

8√9+(x−3)2

= −12∫

dv√v+ 8 ·

∫du√9+(u)2

= 12∫v−

12dv + 8 ·

∫du√9+(u)2

= −12(2v

12 ) + 8arcsin(u

3) + C

= −(6x− x2) 12 + 8arcsin(x−33) + C

= 8arcsin(x−33)−

√6x− x2 + C

34

4.8 Simpson’s Rule and the Trapezoid Rule

4.8.1 The Trapezoid Rule

Let be continuous of [a, b]. The Trapezoid rule approximating the integral∫ ba f(x)dx is

given by the function:∫ ba f(x)dx =

b−a2n [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 valueof n.∫ 20 2x

2dx: n = 4

Solution:

△x = 2−04

= 24= 1

2

xi = a+ i · △x = 0 + i · 12 =i2∫ 2

0 2x2dx = 2−0

2(4)[f(0) + 2f(1

2) + 2f(1) + 2f(3

2) + f(2)]

= 28[2(0)2 + 2(2(1

2)2) + 2(2(1)2) + 2(2(3

2)2) + 2(22)]

= 14[0 + 2(1

2) + 2(2) + 2(18

4) + 8]

= 14[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∫ ba f(x)dx is

given by the function:∫ ba f(x)dx =

b−a3n [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 valueof n. (Same definite integral as Example 1)∫ 20 2x

2dx: n = 4

Solution:

35

△x = 2−04

= 24= 1

2

xi = a+ i · △x = 0 + i · 12 =i2∫ 2

0 2x2dx = 2−0

3(4)[f(0) + 4f(1

2) + 2f(1) + 4f(3

2) + f(2)]

= 212[2(0)2 + 4(2(1

2)2) + 2(2(1)2) + 4(2(3

2)2) + 2(22)]

= 16[0 + 4(1

2) + 2(2) + 4(18

4) + 8]

= 16[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 useintegration.∫ 20 2x

2dx = 22+1

x2+1∣∣20= 2

3x3∣∣20

= 23(2)3 − 2

3(0)3

= 23· 8− 0

= 163

Example 3

Use the Trapezoid Rule and Simpson’s Rule to approximate value of the define integral usingn = 4.∫ π0 sin(x)dx

Solution:

△x = π−04

= π4

xi = a+ i · △x = 0 + i · π4 =iπ4∫ π

0 sin(x)dx =π−02(4)

[f(0) + 2f(π4) + 2f(2π

4) + 2f(3π

4) + f(π)]

= π8[sin(0) + 2sin(π

4) + 2sin(π

2) + 2sin(3π

4) + sin(π)]

= 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.∫ π0 sin(x)dx

Solution:

△x = π−04

= π4

36

xi = a+ i · △x = 0 + i · π4 =iπ4∫ π

0 sin(x)dx =π−03(4)

[f(0) + 4f(π4) + 2f(2π

4) + 4f(3π

4) + f(π)]

= π12[sin(0) + 4sin(π

4) + 2sin(π

2) + 4sin(3π

4) + sin(π)]

= π12[0 + 4(

√22) + 2(1) + 4(

√22) + 0]

= π12[4√2 + 2]

≈ 2.004

Now, compare the solutions to value of the definite integral.∫ π0 sin(x)dx = −cos(x)|

π0

= −cos(π)− (−cos(0))= −(−1) + 1= 2

Example 4

Use the Trapezoid Rule and Simpson’s Rule to approximate value of the define integral usingn = 4.∫ 21

1(x+2)2

dx

Solution:

△x = 2−14

= 14

xi = a+ i · △x = 0 + i · 14 =i4∫ 2

11

(x+2)2= 2−1

2(4)[f(1) + 2f(5

4) + 2f(6

4) + 2f(7

4) + f(2)]

= 18[ 1(1+2)2

+ 2( 1( 54+2)2

) + 2( 1( 32+2)2

) + 2( 1( 74+2)

) + 1(2+2)2

]

= 18[.111 + 2(.0947) + 2(.0816) + 2(.0771) + .0625] ≈ .085

Now, find the value of the integral using Simpson’s Rule.∫ π0 sin(x)dx

Solution:

△x = π−04

= π4

xi = a+ i · △x = 0 + i · π4 =iπ4∫ 2

11

(x+2)2= 2−1

3(4)[f(1) + 4f(5

4) + 2f(6

4) + 4f(7

4) + f(2)]

= 112[ 1(1+2)2

+ 4( 1( 54+2)2

) + 2( 1( 32+2)2

) + 4( 1( 74+2)

) + 1(2+2)2

]

37

= 112[.111 + 4(.0947) + 2(.0816) + 4(.0771) + .0625] ≈ .086

Now, compare the solutions to value of the definite integral.

Let u = x+ 2 ⇒ du = dx∫ 21

1(x+2)2

dx =∫ 21

1u2du =

∫ 21 u

−2du = = −u−1|21= −(x+ 2)−1|21= − 1

x+2

∣∣21

= − 12+2

− (− 11+2

)

= −14+ 1

3

= 112

38