# Substitution With Definite Integrals Substitution With Definite Integrals

SUGGESTED REFERENCE MATERIAL:

As you work through the problems listed below, you should reference Chapter 5.9 of the recommended textbook (or the equivalent chapter in your alternative textbook/online resource) and your lecture notes.

EXPECTED SKILLS:

Be able to evaluate definite integrals using a substitution of variables.

PRACTICE PROBLEMS:

For problems 1-3, use the given substitution to express the given integral (including the limits of integration) in terms of the variable u . Do not evaluate the integrals.

1.

Z

1

5

(3 x

4)

10 dx , u = 3 x

4

1

Z

3

11 u

10 du

− 1

2.

Z

1 e e

(ln x )

3 dx , x u = ln x

Z

1

1 u

3 du ; Detailed Solution: Here

3.

Z

0

4

1

2 x + 1 dx , u = 2 x + 1

1 Z

2

1

9

1 u du

For problems 4-19, evaluate the following integrals.

4.

Z

1

2

1 x

3

− x

4 dx

0

4

8

15

; Detailed Solution: Here

1

5.

Z

4

π sin

2

(3 x ) cos (3 x ) dx

0

2

36

6.

ln 10

Z e

4 x dx

1

2500

1

4 e

4

7.

Z

2

π

(csc

2 x

− sin x cos x ) dx

π

3

1

3

1

8

; Detailed Solution: Here

8.

Z

1

1 dx

1 + 3 x

2

1

2 π

3

; Detailed Solution:

9

Here

9.

π

Z

12 sec

2

(4 x ) dx

0

3

4

10.

Z

10

1

2 + x dx

1 ln 12

11.

Z

2

(2 x + 2)( x

2

+ 2 x

3) dx

3

25

2

2

12.

Z

6

π cos

4

(3 x ) sin (3 x ) dx

0

1

15

13.

Z

16

1

√ x e

√ x dx

1

2 e + 2 e

4

14.

Z

0

2 sin

1

1

− x x

2 dx

π

2

32

15.

Z

4

π tan ( x ) dx

π

6

1

2

(ln 3

− ln 2); Detailed Solution: Here

16.

Z

2 x

2

( x

3 −

4)

3 dx

5

5

32

17.

Z

1

1 x

2 −

4 x + 4 dx

1

2

3

18.

Z

4

5 x

√ x

4 dx

46

15

3

19.

Z

6 p

3 +

| x

| dx

− 1

70

4

3

3

20. For each of the following, express the given definite integral (including the limits of integration) in terms of u . Then, evaluate the “new” integral by using an appropriate formula from geometry.

(a)

Z

0

π

2 x

3

4

− x

8 dx (HINT: Express x

8 as the square of some term).

4

(b)

Z e

4 p

16

(ln x )

2 dx x

1

4 π

8

21. It can be shown that

4 x

2

+ 4 x

15

=

1

2 x

3

2 x

1

+ 5

.

(a) Let t be a fixed constant such that t > 2. Use these facts to evaluate

Z

2 t

4 x

2

8

+ 4 x

15 dx .

1

2 ln

2 t

3

2 t + 5

+ ln 3

(b) Evaluate lim t

+

Z

2 t

4 x

2

8

+ 4 x

15 dx ln 3

4