Substitution With Definite Integrals

advertisement

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

5

(3 x − 4)

10 dx , u = 3 x − 4

1

1 Z

11 u

10 du

3

− 1

2.

Z e

(ln x ) 3 dx , u = ln x

1 e x

Z

1

− 1 u

3 du ; Detailed Solution: Here

3.

Z

4

0

1

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

1 Z

9

2

1

1 u du

For problems 4-19, evaluate the following integrals.

4.

Z

1

2 x 3

1 − 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: Here

9

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

; Video Solution: http://www.youtube.com/watch?v=0plDXxeFNvQ

14.

Z

0

2 sin

− 1 x

1 − 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

5 x

√ x − 4 dx

4

46

15

; Video Solution: https://www.youtube.com/watch?v=LE52v9SV po

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

1

2 x + 5

.

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

Z t

2

8

4 x 2 + 4 x − 15 dx .

1

2 ln

2 t − 3

2 t + 5

+ ln 3

(b) Evaluate lim t → + ∞

Z t

2

8

4 x 2 + 4 x − 15 dx ln 3

4

Download