f5

CHAPTER 5

Integration SECTION 5.1 5.1.1 Estimate the area under the curve y = x2 by dividing the interval [0, 2] into 4 subintervals of equal length and constructing rectangles over the intervals.

5.1.2 Estimate the area under the curve y = 1/x by dividing the interval [1, 2] into 4 subintervals of equal length and constructing rectangles over the intervals.

5.1.3 Estimate the area under the curve y = x3 + 2 by dividing the interval [1, 4] into 3 subintervals of equal length and constructing rectangles over the intervals.

5.1.4 Estimate the area under the curve y = 1/x2 by dividing the interval [1, 4] into 6 subintervals of equal length and constructing rectangles over the intervals. √ 5.1.5 Estimate the area under the curve y = x by dividing the interval [0, 4] into 4 subintervals of equal length and constructing rectangles over the intervals.

5.1.6 Estimate the area under the line y = 2x + 3 by dividing the interval [1, 9] into 4 subintervals of equal length and constructing rectangles over the intervals.

5.1.7 Estimate the area of the graph of a function f (x) = there are 4 rectangles.

√

2x and the interval [a, b] = [0, 1] where

5.1.8 Estimate the area of the graph of a function f (x) = x3 and the interval [a, b] = [0, 1] where there are 4 rectangles.

5.1.9 Estimate the area of the graph of a function f (x) = 4x3 and the interval [a, b] = [0, 1] where there are 4 rectangles.

5.1.10 Estimate the area of the graph of a function f (x) = where there are 4 rectangles.

√

2x + 4 and the interval [a, b] = [0, 1]

5.1.11 Use a simple formula from geometry to find the area function A(x) that gives the area between the function f (x) = 5 and the interval [a, x] = [0, x].

5.1.12 Use a simple formula from geometry to find the area function A(x) that gives the area between the function f (x) = x + 1 and the interval [a, x] = [−1, x].

5.1.13 Use a simple formula from geometry to find the area function A(x) that gives the area between the function f (x) = x + 4 and the interval [a, x] = [0, x].

1

SOLUTIONS SECTION 5.1 k 2−0 1 = , x∗k = 0 + k∆x = 4 2 2 1 15 (1 + 2 + 3 + 4) = 2 4

5.1.1 ∆x =

1 3+k 2−1 = , x∗k = 1 + (k − 1)∆x = 5.1.2 ∆x = 4 4 4

1 1 1 1 319 1 + + + = 16 20 24 28 4 420 5.1.3 ∆x =

4−1 = 1, x∗k = 1 + k∆x = 1 + k 3

(8 + 2 + 27 + 2 + 64 + 2)(1) 1+k 4−1 1 5.1.4 ∆x = = , x∗k = 1 + (k − 1)∆x = 6 2 2

4 1 4 1 4 1 + + ≈ 1.0236 1+ + + 2 9 4 25 9 49 4−0 = 1, x∗k = 0 + k∆x = k 4 √ √ 1 + 2 + 3 + 2 ≈ 6.1463

5.1.5 ∆x =

9−1 = 2, x∗k = 1 + k∆x = 1 + 2k 4 4 4

f (x∗k )∆x = [2(1 + 2k) + 3](2) = 120

5.1.6 ∆x =

k=1

k=1

(9 + 13 + 17 + 21)2 = 120 √ √ √ √ 5.1.7 0.25[ 0.25 + 0.75 + 1.25 + 1.75] = 0.951734 5.1.8 0.25[0.1253 + 0.3753 + 0.6253 + 0.8753 ] = 0.242188 5.1.9 0.25(4)[0.1253 + 0.3753 + 0.6253 + 0.8753 ] = 0.96875 √ √ √ √ 5.1.10 0.25[ 0.25 + 4 + 0.75 + 4 + 1.25 + 4 + 1.75 + 4] = 4.951734 5.1.11 Rectangle 5 by x A(x) = 5x 5.1.12 Right triangle with base x + 1 and height x + 1 A(x) =

1 (x + 1)2 2

5.1.13 Trapezoid with bases 4 and x + 4 and height x 1 1 (x + 8)x = x2 + 4x 2 2

2

Questions, Section 5.2

3

SECTION 5.2

(1 + x)2 dx. x1/2

3 2 5.2.3 Evaluate x + 8x + 2 dx. x

5.2.5 Evaluate (x2 + 1)2 dx.

5.2.1 Evaluate

5.2.7 Evaluate

5.2.9 5.2.11 5.2.13 5.2.15 5.2.17 5.2.19

x4 + 1 dx. x3

dx Evaluate . cos2 x

3 2 Evaluate x − 4 dx. x

√ 1 dx.

Evaluate x+ √ x

7 Evaluate dt. t4

3 x Evaluate + cos x dx.

4

Evaluate (x−2 + sec2 x + 3) dx.

5.2.21 Evaluate

√ 2 x dx.

x4 − 3x2 + 2 dx.

5.2.23 Evaluate

5.2.25 Evaluate

dx . sec x

(x2 + 2x + 5)dx.

5.2.2 Evaluate

5.2.4 Evaluate

5.2.6 Evaluate

7 dx √ . x √ (3 x + 1)dx.

(x3 + 2)2 dx.

5.2.8 Evaluate

(x3 − x + 5)dx.

5.2.10 Evaluate

5.2.12 Evaluate

5.2.14 Evaluate

5.2.16 Evaluate

x2 − 4 √ dx.

3 x2 √ (x + 1) x dx. √ ( x + 2)2 dx.

(x − 2)2 x dx.

5.2.18 Evaluate

5.2.20

(x3 − csc x cot x + 7) dx.

sin x dx. cos3 x

3 dx.

5.2.24 Evaluate x+ cos2 x

1 + 4t4 dt.

5.2.26 Evaluate t4 5.2.22 Evaluate

SOLUTIONS SECTION 5.2

5.2.1

(1 + x)2 dx = x1/2

(1 + 2x + x2 ) dx = x1/2

(x−1/2 + 2x1/2 + x3/2 )dx

4 2 = 2x1/2 + x3/2 + x5/2 + C 3 5 x3 + x2 + 5x + C 3 √ 5.2.4 14 x + C

5.2.3

5.2.2

(x4 + 2x2 + 1)dx =

5.2.5

x3 3 + 4x2 − + C 3 x

x5 2 + x3 + x + C 5 3

5.2.6 2x3/2 + x + C

5.2.7

(x + x−3 )dx =

x2 1 x−2 x2 +C = − 2 +C − 2 2 2x 2

(x6 + 4x3 + 4)dx =

5.2.8

x7 + x4 + 4x + C 7

sec2 x dx = tan x + C

5.2.9

5.2.11

5.2.12

(x2 − 3x−4 ) dx =

5.2.10

1 x3 + 3 +C 3 x

(x4/3 − 4x−2/3 )dx =

3 7/3 x − 12x1/3 + C 7

2 3/2 x + 2x1/2 + C 3

2 2 5.2.14 (x3/2 + x1/2 )dx = x5/2 + x3/2 + C 5 3 5.2.13

5.2.15 −

5.2.16

7 +C 3t3 √ x2 8 (x + 4 x + 4)dx = + x3/2 + 4x + C 2 3

x4 + sin x + C 16

4 x4 − x3 + 2x2 + C 5.2.18 (x3 − 4x2 + 4x)dx = 4 3 5.2.17

4

x4 x2 − + 5x + C 4 2

Solutions, Section 5.2

5.2.19 −

1 + tan x + 3x + C x

5

5.2.20

x4 + csc x + 7x + C 4

5.2.21

x21 +C 21

5.2.22

4x3/2 +C 3

5.2.23

1 +C 2 cos2 x

5.2.24

x5 − x3 + 2x + C 5

5.2.25

x2 cos x + 6 sin x +C 2 cos x

5.2.26 sin x + C

5.2.27 4t −

1 +C 3t3

6

Questions, Section 5.3

SECTION 5.3

5.3.1 Evaluate

3x 1 − 2x2 dx.

√ 3

5.3.3 Evaluate

5.3.5 Evaluate

5.3.7 Evaluate

5.3.11 Evaluate

5.3.13 Evaluate

5.3.17 Evaluate

5.3.8 Evaluate

5x4 − 18 dx.

tan3 5x sec2 5x dx.

5.3.10 Evaluate 5.3.12 Evaluate

√ 1 √ sin x dx. x

5.3.14 Evaluate

x2 dx √ . x+1

x2 dx √ .

1 + x3

√ x x − 5 dx.

sin x dx . cos3 x

x−2 dx. (x2 − 4x + 4)2

n 5.3.21 Evaluate x a + bx2 dx.

(2 + sin 3t)1/2 cos 3t dt.

5.3.19 Evaluate

5.3.23 Evaluate

csc 2t cot 2t dt.

tan 3x sec 3x dx.

sin 2x cos 2x dx.

2

5.3.15 Evaluate

5.3.6 Evaluate

x3

t2 (2 − 3t3 )3 dt.

5.3.4 Evaluate

dx . cos2 2x

5.3.9 Evaluate

3x dx .

3 − 7x2

5.3.2 Evaluate

[tan(tan θ)] sec2 θ dθ.

(x2 + 1)(x3 + 3x)10 dx.

(x3 − x)(x4 − 2x2 )15 dx.

5.3.16 Evaluate

5.3.18 Evaluate

4 dx. (x + 4)3 x sec2 x2 dx.

5.3.20 Evaluate

x3 sin(x4 + 2) dx.

5.3.22 Evaluate

5.3.24 Evaluate

5.3.25 Evaluate the integral by making the indicated substitution.

2x x2 + 3 dx, u = x2 + 3 5.3.26 Evaluate the integral by making the indicated substitution.

x cos(x2 )dx, u = x2 5.3.27 Evaluate the integral by making the indicated substitution.

2x √ dx, u = x2 + 2 x2 + 2 5.3.28 Evaluate the integral by making the indicated substitution.

cos x √ dx, u = sin x sin x

√ x 3 x + 1 dx.

Questions, Section 5.3

5.3.29 Evaluate the integral by making the indicated substitution.

√ x 2 + x dx, u = 2 + x 5.3.30 Evaluate the integral by making the indicated substitution.

tan(3x) dx, u = 3x 5.3.31 Evaluate the integral by making the indicated substitution.

sin3 (4x) cos(4x) dx, u = sin(4x)

7

SOLUTIONS SECTION 5.3 du 5.3.1 u = 1 − 2x2 , du = −4x dx, x dx = − 4

1 1 3 u1/2 du = − u3/2 + C = − (1 − 3x2 )3/2 + C − 4 2 2 du 5.3.2 u = 2 − 3t3 , du = −9t2 dt, t2 dt = − 9

1 1 1 u3 du = − u4 + C = − (2 − 3t3 )4 + C − 9 36 36 du 5.3.3 u = 3 − 7x2 , du = −14x dx, x dx = − 14

du 3 9 2/3 9 − = − u + C = − (3 − 7x2 )2/3 + C 14 28 28 u1/3 du 5.3.4 u = sin 2x, du = 2 cos 2x dx, cos 2x dx = 2

1 2 1 1 2 u du = u + C = sin 2x + C 4 4 2

du dx = sec2 2x dx, u = 2x, du = 2 dx, dx = 5.3.5 cos2 2x 2

1 1 1 sec2 u du = tan u + C = tan 2x = C 2 2 2 du 5.3.6 u = 2 + sin 3t, du = 3 cos 3t dt, cos 3t dt = 3

2 2 1 u1/2 du = u3/2 + C = (2 + sin 3t)3/2 + C 9 9 3 du 5.3.7 u = 2t, du = 2dt, dt = 2

1 1 1 csc u cot u du = − csc u + C = − csc 2t + C 2 2 2 du 5.3.8 u = tan 5x, du = 5 sec2 5x dx, sec2 5x dx = 5

1 1 1 u3 du = u4 + C = tan4 5x + C 5 20 20 du 5.3.9 u = 5x4 − 18, du = 20x3 dx, x3 dx = 20

1 1 1 u1/2 du = u3/2 + C = (5x4 − 18)3/2 + C 20 30 30 5.3.10 u = x − 5, du = dx, x = u + 5

2 10 1/2 (u + 5)u du = (u3/2 + 5u1/2 )du = u5/2 + u3/2 + C 5 3 =

10 2 (x − 5)5/2 + (x − 5)3/2 + C 5 3

8

Solutions, Section 5.3

9

5.3.11 u = cos x, du = − sin x dx, sin x dx = −du

1 1 1 − u−3 du = 2 + C = + C = sec2 x + C 2u 2 cos2 x 2 5.3.12 u = tan θ, du = sec2 θ dθ

tan u du = − ln | cos u| + C = − ln | cos(tan θ)| + C √ dx 1 5.3.13 u = x, du = √ dx, √ = 2 du x 2 x

√ 2 sin u du = −2 cos u + C = −2 cos x + C du 5.3.14 u = x3 + 3x, du = 3(x2 + 1)dx, (x2 + 1)dx = 3

1 11 1 3 1 10 11 u +C = (x + 3x) + C u du = 33 33 3 du = sec2 3x dx 5.3.15 u = tan 3x, du = 3 sec2 3x dx, 3

1 u2 1 1 u du = · + C = tan2 3x + C 3 3 2 6 du 5.3.16 u = x4 − 2x2 , du = 4(x3 − x)dx, = (x3 − x)dx 4

1 1 16 1 4 u15 du = u +C = (x − 2x2 )16 + C 4 64 64 5.3.17 u = x + 1, du = dx, x = u − 1

(u − 1)2 du = u1/2

(u2 − 2u + 1) du = u1/2

(u3/2 − 2u1/2 + u−1/2 )du

=

2 5/2 4 3/2 u − u + 2u1/2 + C 5 3

=

4 2 (x + 1)5/2 − (x + 1)3/2 + 2(x + 1)1/2 + C 5 3

5.3.18 u = x + 4, du = dx

4 u−3 du = −2u−2 + C = −

2 +C (x + 4)2

du 5.3.19 u = x2 − 4x + 4, du = 2(x − 2)dx, (x − 2)dx = 2

du 1 1 1 +C =− C =− 2 u2 2u 2(x2 − 4x + 4) du 5.3.20 u = x2 , du = 2x dx, x dx = 2

1 1 1 sec2 u du = tan u + C = tan x2 + C 2 2 2

10

Solutions, Section 5.3

du = x dx 5.3.21 u = a + bx2 , du = 2bx dx, 2b

1 n n u(n+1)/n + C = (a + bx2 )(n+1)/n + C u1/n du = 2b(n + 1) 2b(n + 1) 2b du 5.3.22 u = x4 + 2, du = 4x3 dx, x3 dx = 4

1 1 1 sin u du = − cos u + C = − cos(x4 + 2) + C 4 4 4 du 5.3.23 u = 1 + x3 , du = 3x2 dx, x2 dx = 3

2 2 1 u−1/2 du = u1/2 + C = 1 + x3 + C 3 3 3 5.3.24 u = x + 1, du = dx, x = u − 1

3 3 3 3 (u − 1)u1/3 du = (u4/3 − u1/3 )du = u7/3 − u4/3 + C = (x + 1)7/3 − (x + 1)4/3 + C 7 4 7 4

5.3.25

u

u

−1/2

5.3.27

2(x2 + 2)3/2 +C du, 3

1/2

du, 2 x2 + 2 + C

(u − 2)u

5.3.29

5.3.30

1/2

5.3.26

5.3.28

cos u sin(x2 ) du, +C 2 2 √ u−1/2 du, 2 sin x + C

√ 2x(x − 2) 2 − x 8(2 − x)3/2 − +C du, 5 15

tan u − ln | cos(3x)| du, +C 3 3

5.3.31

u3 sin4 (4x) du, +C 4 16

Questions, Section 5.4

11

SECTION 5.4 5.4.1 Evaluate

4

(i2 + 2).

5.4.2 Evaluate

i=1

5.4.3 Evaluate

(2j − 3).

j=1

2

2

3j .

5.4.4 Evaluate

j=−2

5.4.5 Evaluate

4

4

4

k=1

sin

n=1

k . k+1

nπ .

2

1 2 3 4 5 + + + + in sigma notation.

2 3 4 5 6

5.4.6 Express

5.4.7 Express 1 +

5.4.8 Evaluate

2 3 4 5 + + + in sigma notation.

3 5 7 9

10

(k + 2)2 by first changing f (k) = (k + 2)2 to f (k) = k 2 and then, an appropriate

k=1

change in the limits of summation.

5.4.9 Evaluate

30

(k 2 + 2).

k=1

5.4.10 Evaluate

10

(k + 3)3 by first changing f (k) = (k + 3)3 to f (k) = k 3 and then by making an

k=1

appropriate change in the limits of summation.

5.4.11 Evaluate

30

(k 2 + 3k − 5).

k=1

5.4.12 Express 1 + 3 + 32 + 33 + 34 + 35 in sigma notation with j = 3 as the lower limit.

5.4.13 Express

10

1 in sigma notation with k = 1 as the lower limit. k

k=5

30

5.4.14 Express

k=10

5.4.15 Evaluate

1 in sigma notation with k = 4 as the lower limit. k(k + 1)

n 2

k 1 k=1

5.4.17 Evaluate

n

k=1

n

1+

n 2k n

.

5.4.16 Evaluate

n

1 k k=1

1 . n

5.4.18 Evaluate

n

n

.

2 n

k+2 1 . n n

k=1

12

Questions, Section 5.4

20

1 .

5.4.19 Evaluate k 1− k k=2

5.4.20 Write the expression in sigma notation. 5 · 2 + 5 · 3 + 5 · 4 + 5 · 5 + 5 · 6 5.4.21 Write the expression in sigma notation. 1 − 2 + 3 − 4 + 5 − 6 + 7 − 8

SOLUTIONS SECTION 5.4 5.4.1 3 + 6 + 11 + 18 = 38

5.4.2 −1 + 1 + 3 + 5 = 8

5.4.3 12 + 3 + 0 + 3 + 12 = 30

5.4.4

5.4.5 1 + 0 − 1 + 0 = 0

5.4.6

163 1 2 3 4 + + + = 2 3 4 5 60 k

k=1

5.4.7

5

k=1

5.4.8

10

k 2k − 1 (k + 2)2 =

30

k2 +

k=1

5.4.10

10

30

2=

k=1

(k + 3)3 =

k=1

5.4.11

30

8

k2 =

k2 + 3

30

12

k2 −

2

k2 =

k=1

k=1

12(13)(25) 2(3)(5) − = 645 6 6

1 (30)(31)(61) + 2(30) = 9515 6

13

k3 =

k=4

k=1

5.4.12

12

k=3

k=1

5.4.9

k−

k=1

13

k3 −

k=1 30

5=

k=1

3

k3 =

k=1

(13)(14) 2

24

k=4

2

−

(3)(4) 2

2 = 8245

1 3 (30)(31)(61) + (30)(31) − 5(30) = 10700 6 2

3j−3

5.4.13

j=3

5.4.14

k k+1

6

k=1

1 k+4

1 (k + 6)(k + 7)

n n

(n)(n + 1)(2n + 1) k2 1 2 1 (n + 1)(2n + 1) 5.4.15 · = 3 k = = 3 3 n n n 6 6n2 k=1

5.4.16

k=1

n n

k 1 1 n(n + 1) n+1 = k = = n2 n2 n2 2 2n

k=1

5.4.17

n

k=1

k=1

2k 1+ n

1 = n n

k=1

1 2k + 2 n n

=

n n 1 2 1+ 2 k n n k=1

k=1

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

13

14

5.4.18

Solutions, Section 5.4 n

(k + 2)2 k=1

n3

n n 1 1 2 2 = 3 (k + 2) = 3 (k + 4k + 4) n n k=1 k=1 n

n n

1 2 = 3 k +4 k+4 1 n k=1 k=1 k=1

1 n(n + 1)(2n + 1) n(n + 1) + (4)(n) = 3 + (4) n 6 2

=

5.4.19

20

k=2

5.4.20

5

k=1

(k − 1) =

2n2 + 15n + 37 6n2

19

k=1

5 · (k + 1)

k=

(19)(20) = 190 2

5.4.21

8

k=1

(−1)k+1

Questions, Section 5.5

15

SECTION 5.5

1 7

5.5.1 Evaluate

2

(x + 1)(3x + 1) dx.

π/4

(x + 2 sec2 x) dx.

1

(

√ 3

t2

+

√

t ) dt.

0 π/2

5.5.7 Evaluate π/4

1 dθ. sin2 θ

2

5.5.11 Evaluate 1

1 1 + 3 t2 t

−3

1 1 − 3 t2 t

−1

2

f (x) dx if f (x) = −2

5

dt.

x > −1 x3 , .

1 − x2 , x ≤ −1

f (x)dx = 3 and

g(x) dx = 10.

π

5.5.18

2

5.5.19

5

1

4x3 + 6 dx =

sin x + sin x cos x dx = 0

π

cos x + sin x cos x dx = 0

5.5.21 Find

f (x)dx where f (x) = 0

3

x5 dx =

5.5.20

2

|2x + 1| dx.

x>2 x2 , .

1 − x, x ≤ 2

4

5.5.17

3

−3

1

1

5

[2f (x) − g(x)]dx if

5.5.16 Find

(θ − csc θ cot θ) dθ.

5.5.13 Evaluate

f (x)dx if f (x) =

5.5.15 Evaluate

1 dφ. cos2 φ

π/3

|x + 1| dx.

3

5.5.14 Evaluate

dt.

π/2

5.5.10 Evaluate

5

5.5.12 Evaluate

(cos θ − csc θ cot θ) dθ.

sec θ dθ.

0

√ 1 1+t t− 2 t

dx.

π/6

π/4

5.5.9 Evaluate

π/3

5.5.8 Evaluate

2

π/3

5.5.6 Evaluate

0

1

5.5.5 Evaluate

4

5.5.4 Evaluate

0

1 2x + 2 2x 2

1

5.5.3 Evaluate

5.5.2 Evaluate

−1

3

−3

x, x ≤ 1 .

1, x > 1

SOLUTIONS SECTION 5.5 1 x8 3x3 3x10 + + +x =4 10 8 3 −1

5.5.1

3

2x2 +

5.5.2 1

x−2 2

dx =

2x3 1 − 3 2x

3

53 3

= 1

π/4 π2 x2 = + tan x +2 2 32 0

5.5.3

4

5.5.4 1

4

293 2t5/2 1 = (1 + t3/2 − t−2 )dt = t + + 5 t 1 20

1

(t2/3 + t1/2 )dt =

5.5.5 0

π/3

5.5.6 0

−1

−3

−3

−2

π/3

−1

(x + 1)dx = −

−(2x + 1)dx +

3

2

1

csc2 θ dθ = − cot θ π/4

tan θ]0

5

π/2

π/4

5.5.11

2 = 1

+ [x2 + x]3−1/2 =

(2x + 1)dx = −(x + x) −3

3

2 2 3 x x 47 x2 dx = x − + = 2 −1 3 6 2

2

3

−1

5

47 480

−1/2

2 −1

x3 x4 29 (1 − x )dx + x dx = x − + = 3 −2 4 12 −1

2

5

f (x)dx − 1

π/4

=1

t−5 t−3 − −3 −5

2

−1/2

π/2

1 5 2 x2 x +x +x + = 18 2 2 −1 −3

3

2[f (x) − g(x)]dx = 2

5.5.16

5.5.7

2 5π 2 − √ +1 72 3

(1 − x)dx +

−1

5.5.15

=

−1

π/2

2

5.5.14

0

5.5.9

−(x + 1)dx +

−1/2

5.5.13

19 15

=

√ 7 3 − 15 = 6

π/3 csc θ]π/6

θ2 + csc θ 2

5.5.12

1

π/3 √ sec2 φ dφ = tan φ 0 = 3

5.5.8 [ − sin θ +

5.5.10

3 5/3 2 3/2 t + t 5 3

5

g(x)dx = 2(3) − 10 = −4 1

16

37 2

=1

Solutions, Section 5.5

17

5.5.17

4 x4 + 6x 2 = 252

5.5.19

sin x −

5.5.21

π cos2 x

=0 2 0

1

x dx + 0

1

2

2

1

x2

3

dx = + x

=

2 0 2 1

π cos2 x

=2 2 0

5.5.18

− cos x −

5.5.20

3 x6

=0 6 −3

18

Questions, Section 5.6

SECTION 5.6 5.6.1 Find the area under the curve y = x2 + 2 for 2 ≤ x ≤ 3. Make a sketch of the region.

5.6.2 Find the area of the region between y = 16 − x2 and the x-axis. Make a sketch of the region.

5.6.3 Find the area of the region between y = x2 − x − 6 and the x-axis for 0 ≤ x ≤ 2. Make a sketch of the region.

5.6.4

5.6.6

5.6.8

d dx d dx d dx

x

5

t dt = sin2 t

5.6.5

sec2 t dt =

5.6.7

x

1

d dx d dx

x

t dt = t3 − 2t

x

t2 + 3 dt =

3

7

5

sin t dt = x

5.6.9 Find all values of x∗ in the interval [0, 2] that satisfy the Mean-Value Theorem for Integrals, where f (x) = x3 .

5.6.10 Find all values of x∗ in the interval [0, 2] that satisfy the Mean-Value Theorem for Integrals, where f (x) = x − 2.

SOLUTIONS SECTION 5.6

3

(x2 + 2)dx =

5.6.1 A = 2

3 x3 25 + 2x = 3 3 2

4 x3

256 (16 − x )dx = 16x − 5.6.2 A =

= 3 3 −4 −4

4

2

5.6.3 A = 0

2

3 2 x 34 x2 −(x2 − x − 6)dx = − + + 6x = 3 2 3 0

19

20

Solutions, Section 5.6

x sin2 x

5.6.5

5.6.6 sec2 x

5.6.7

5.6.4

x x3 − 2x √

x2 + 3

5.6.8 − sin x

5.6.9

2

x3 dx = f (x∗ )(2 − 0)

0

4 = 2f (x∗ ) 2 = f (x∗ ) √ x= 32

5.6.10

2

x − 2 dx = f (x∗ )(2 − 0)

0

−2 = 2f (x∗ ) −1 = f (x∗ ) x=1

Questions, Section 5.7

21

SECTION 5.7 5.7.1 A stone is thrown downward from the top of a 160 foot high cliff with an initial velocity of 48 feet per second. What is the speed of the stone upon impact with the ground?

5.7.2 A projectile is fired downward from a height of 128 feet and reaches the ground in 2 seconds.

What was its initial velocity?

5.7.3 A projectile is launched vertically upward from the ground with an initial velocity of 80 feet per second. (a) How long does it take the projectile to reach the ground? (b) When is the projectile 64 feet above the ground? (c) What is the velocity of the projectile when it is 64 feet above the ground?

5.7.4 A playful student drops your math book from a dormitory window and it hits the ground in 3 seconds. How high up is the window?

5.7.5 A particle is moving so that at any time, its acceleration is equal to 10t for t ≥ 0. At the end of 3 seconds, the particle has moved 105 feet. What is its velocity at the end of 3 seconds?

5.7.6 A projectile fired upward from the ground is to reach 144 feet. (a) What must be its initial velocity? (b) What is the velocity of the projectile when it is 80 feet above the ground?

5.7.7 Find the position, velocity, speed, and acceleration at time t = 1 second of a particle if v(t) = 2t − 4; s = 3 when t = 0.

5.7.8 A ball is rolled across a level floor with an initial velocity of 28 feet per second. How far will the ball roll if the speed diminishes by 4 feet/sec2 due to friction?

5.7.9 A particle, initially moving at 16 ft/sec. is slowing down at the rate of 0.8 ft/sec2 . How far will the particle travel before coming to rest?

5.7.10 A projectile is fired vertically upward from a point 20 feet above the ground with a velocity of 40 feet per second. Find the speed of the projectile when it is 36 feet above the ground.

5.7.11 A rapid transit trolley moves with a constant acceleration and covers the distance between two points 300 feet apart in 8 seconds. Its velocity as it passes the second point is 50 feet per second. (a) What is its acceleration? (b) What is the velocity of the trolley as it passes the first point?

5.7.12 A jet plane travels from rest to a velocity of 300 feet per second in a distance of 450 feet. What is its constant acceleration?

5.7.13 A particle is moving so that its velocity, v(t) = t2 − t − 2 for 0 ≤ t ≤ 3. Find the displacement and total distance travelled by the particle.

5.7.14 A particle is moving so that its velocity, v(t) = 4 − t for 0 ≤ t ≤ 6. Find the displacement and total distance travelled by the particle.

22

Questions, Section 5.7

5.7.15 A particle is moving so that its velocity, v(t) = 8 − 2t for 0 ≤ t ≤ 5. Find the displacement and total distance travelled by the particle.

5.7.16 A particle is moving so that its velocity, v(t) = t2 − 3t + 2 for 0 ≤ t ≤ 3. Find the displacement and total distance travelled by the particle.

5.7.17 A particle is moving so that its velocity, v(t) = t2 − 4t + 3 for 0 ≤ t ≤ 4. Find the displacement and total distance travelled by the particle.

8 5.7.18 A particle is moving so that its velocity, v(t) = t − 2 for 1 ≤ t ≤ 3. Find the displacement t and total distance travelled by the particle.

5.7.19 The (a) (b) (c)

graph of a velocity function over the interval [t1 , t2 ] is as shown.

Is the acceleration positive or is it negative?

Is the acceleration increasing or is it decreasing?

Is the displacement positive or is it negative?

5.7.20 Find the average value of f (x) =

√

4x + 1 over the interval 0 ≤ x ≤ 2.

5.7.21 Find the average value of f (x) = x sec x for 0 ≤ x ≤ 2

2

3

3

π .

4

5.7.22 (a) Find the average value of f (x) = 3x + 1 over [0, 6]. (b) Find a point x∗ in [0, 6] such that f (x∗ ) = fave . (c) Sketch the graph of f (x) = 3x + 1 over [0, 6] and construct a rectangle over the interval whose area is the same as the area under the graph of f over the interval.

5.7.23 (a) Find the average value of f (x) = (x + 1)2 over [−1, 2]. (b) Find a point x∗ in [−1, 2] such that f (x∗ ) = fave . (c) Sketch the graph of f (x) = (x+1)2 over [−1, 2] and construct a rectangle over the interval whose area is the same as the area under the graph of f over the interval.

5.7.24 Find the average value of f (x) = x cos x2 for 0 ≤ x ≤

π .

2

√ 5.7.25 Find the average value of f (x) = x3 3x4 + 1 for −1 ≤ x ≤ 2.

5.7.26 Find the average value of f (x) = x3 + 1 for 0 ≤ x ≤ 2 and find all values of x∗ described in the Mean Value Theorem for Integrals.

5.7.27 A particle moves along the s-axis. Use the given information to find the position of the particle. v(t) = t4 ; s(0) = 2 5.7.28 A particle moves along the s-axis. Use the given information to find the position of the particle. v(t) = 3t4 − 4t + 1; s(1) = 5 5.7.29 A particle moves along with a velocity of v(t) = t2 − 8t m/s along an s-axis. Find the distance travelled by the particle during the period 0 ≤ t ≤ 4.

Questions, Section 5.7

23

5.7.30 A particle moves along with a velocity of v(t) = t2 − 8t m/s along an s-axis. Find the displacement of the particle during the period 0 ≤ t ≤ 4.

5.7.31 A particle moves along with a velocity of v(t) = t2 − t4 m/s along an s-axis. Find the distance travelled by the particle during the period 0 ≤ t ≤ 2.

5.7.32 A particle moves along with a velocity of v(t) = t2 − t4 m/s along an s-axis. Find the displacement of the particle during the period 0 ≤ t ≤ 2.

SOLUTIONS SECTION 5.7 5.7.1 s(t) = 0 upon impact with the ground. s(t) = −16t2 − 48t + 160 = −16(t2 + 3t − 10) = −16(t + 5)(t − 2) s(t) = 0 when t = 2 sec. v(t) = −32t − 48; v(2) = −32(2) − 48 = −112, the speed at impact is 112 ft/sec.

5.7.2 s = 128 when t = 0, so s(t) = −16t2 + v0 t + 128 = 0. s = 0 when t = 2, −16(2)2 + v0 (2) + 128 = 0. v0 = −32 ft/sec 5.7.3 (a) s(t) = 0 when the projectile hits the ground. s(t) = −16t2 + 80t = −16t(t − 5) s(t) = 0 when t = 0 and when t = 5 seconds. (b) −16t2 + 80t = 64 −16(t2 − 5t + 4) = −16(t − 4)(t − 1) The projectile is 64 feet above the ground when t = 1 and t = 4 seconds. (c) v(t) = −32t + 80 v(1) = −32(1) + 80 = 48 ft/sec v(4) = −32(4) + 80 = −48 ft/sec 5.7.4 Let h = height of the dormitory window, then, s = h and v = 0 when t = 0, thus, s(t) = −16t2 + h. s(3) = 0, so −16(3)2 + h = 0 h = 144 feet 5.7.5 a(t) = 10t

v(t) = 10t dt = 5t2 + C1 , v(0) = v0 , so C1 = v0 and v(t) = 5t2 + v0

5 s(t) = (5t2 + v0 )dt = t3 + v0 t + C2 3 s(0) = 0, so C2 = 0 5 s(3) = 105, so (3)3 + 3v0 = 105 3 v0 = 20 Thus, v(t) = 5t2 + 20 v(3) = 5(3)2 + 20 = 65 ft/sec 5.7.6 (a) s(t) = −16t2 + v0 t v(t) = −32t + v0 v = 0 when the projectile is at its maximum height, thus, −32t + v0 = 0 v0 t= 32 v 2 v v0 0 0 sec so −16 = 144 s(t) = 144 feet when t = + v0 32 32 32 2 v0 = (64)(144) v0 = 96 ft/sec (positive since ďŹ red upward).

24

Solutions, Section 5.7

25

(b) −16t2 + 96t = 80 −16(t2 − 6t + 5) = −16(t − 5)(t − 1) = 0 The projectile is 80 feet above the ground when t = 1 and t = 5 seconds. v(t) = −32t + 96 v(1) = −32(1) + 96 = 64 ft/sec v(5) = −32(5) + 96 = −64 ft/sec

(2t − 4)dt = t2 − 4t + C1 ; s(0) = 3; c1 = 3

5.7.7 s(t) =

s(t) = t2 − 4t + 3, s(1) = (1)2 − 4(1) + 3 = 0 v(1) = 2(1) − 4 = −2 |v(1)| = | − 2| = 2 dv = 2, a(1) = 2 a(t) = dt

5.7.8 v(t) = −4dt = −4t + C1 ; v(0) = 28 so C1 = 28, v(t) = −4t + 28

s(t) = (−4t + 28)dt = −2t2 + 28t + C2 , if s(0) = 0, C2 = 0 and s(t) = −2t2 + 28t The ball comes to rest when v = 0, so −4t + 28 = 0, t = 7 sec, thus s(7) = −2(7)2 + 28(7) = 98. The ball rolls 98 feet before coming to rest.

−0.8dt = −0.8t + C1 , v(0) = 16 so C1 = 16

5.7.9 v(t) =

v(t) = −0.8t + 16

s(t) = v(t)dt = (−0.8t + 16)dt = −0.4t2 + 16t + C2 , if s(0) = 0 C2 = 0, s(t) = −0.4t2 + 16t The particle comes to rest when v(t) = 0, so −0.8t + 16 = 0, t = 20 sec; thus, s(20) = −0.4(20)2 + 16(20) = 160. The particle travels 160 feet before coming to rest.

5.7.10 s = 20 when t = 0, so, s(t) = −16t2 + 40t + 20. When s = 36, s(t) = −16t2 + 40t + 20 = 36 or −8(2t − 1)(t − 2) = 0; thus, the projectile is 36 feet above the ground when t = 1/2 second and t = 2 second. v(t) = −32t + 40 so speed = | − 32(1/2) + 40| = | − 32(2) + 40| = 24 ft/sec.

5.7.11 Let a be the acceleration of the trolley, so that v(t) = v(0) = v0 so that c1 = v0 and v(t) = at + v0 .

s(t) =

v(t)dt =

(at + v0 )dt =

Let s(0) = 0 so c2 = 0 and s(t) =

at2 + v0 t + C2 2 at2 + v0 t.

2

a dt, v(t) = at + c1 . When t = 0,

26

Solutions, Section 5.7

When t = 8 secs, v = 50 ft/sec and s = 300 ft so   a 2 8a + 2v0 = 75  (8) + v0 (8) = 300 25 2 so that a = or and v0 = 25.   8 8a + v0 = 50 8a + c0 = 50 25 ft/sec2 . (a) The acceleration of the trolley is 8 (b) The velocity of the trolley as it passes the first point is 25 ft/sec.

a dt, v(t) = at + C1 . When t = 0, v = 0

5.7.12 Let a = acceleration of the jet plane so that v(t) = 2

at2 at , thus, v = at + C2 . Let s(0) = 0 so C2 = 0 and s(t) = 2 2

2 300 a 300 at2 . When s = 450, v = 300, so 300 = at, t = and 450 = and s = or a = 100.

2 a 2 2 2 The acceleration of the jet plane is 100 ft/sec . so C1 = 0 and v(t) = at, s(t) =

3

t2 t3 3 − − 2t

= − 5.7.13 displacement = (t − t − 2)dt = 3 2 2 0 0 2 3 3 31 2 2 −(t − t − 2)dt + (t2 − t − 2)dt = |t − t − 2|dt = distance = 6 0 2 0

3

2

6 t2 (4 − t)dt = 4t − =6 5.7.14 displacement = 2 0 0 6 4 6 distance = |4 − t|dt = (4 − t)dt + −(4 − t)dt = 10

6

0

0

4

5 2t2 5.7.15 displacement = (t − 2t)dt = 8t − = 15 2 0 0 5 4 5 distance = |8 − 2t|dt = (8 − 2t)dt + −(8 − 2t)dt = 17

5

0

0

4

3 3t2 t3 3 5.7.16 displacement = − + 2t = (t − 3t + 2)dt = 3 2 2 0 0 2 3 1 11 2 2 (t − 3t + 2)dt + −(t − 3t + 2)dt + (t2 − 3t + 2)dt = distance = 6 0 1 2

3

2

t3 4t2 (t − 4t + 3)dt = − + 3t 3 2

4

2

5.7.17 displacement =

4

0

=

4 3

4

0

4

|t2 − 4t + 3|dt

distance = 0

1 2

2

(t − 4t + 3)dt +

= 0

3

(t2 − 4t + 3)dt = 4

−(t − 4t + 3)dt + 1

3

Solutions, Section 5.7

27

3

8 8 t2 4 + t − 2 dt = =− 5.7.18 displacement = t 2 t 1 3 1

3 2 3

8 11 8

8

t − 2 dt = − t − 2 dt + distance =

t − t2 dt = t t 3 1 1 2

3

5.7.19 (a) acceleration is positive (b) acceleration is increasing (c) displacement is positive 5.7.20

5.7.21

1 2−0

2

4x + 1 dx =

0

1

π −0 3 4

5.7.22 (a)

√

3 √ π/4

1 6−0

3 √ 13 4 1 π/4 3 = x sec x dx = tan x 0 3 π π 33 4 2

0

2 13 1 (4x + 1)3/2 = 12 6 0 2

3

6

(3x + 1)dx = 10 0

(b) 3x∗ + 1 = 10, x∗ = 3 (c)

5.7.23 (a)

1 2 − (−1)

2

(x + 1)2 dx = −1

(b) (x + 1) = 3 only x∗ = ∗

(c)

2

√

2

1 (x + 1)3

= 3 9 −1

3 − 1 is in [−1, 2]

28

5.7.24

5.7.25

Solutions, Section 5.7

1

π −0 2 1 2 − (−1)

5.7.26 fav

2

√π/2 0

2

x3 −1

1 = 2−0

2

√π/2 1 1 x cos x2 dx = =√ sin x2 0 π 2π 2 2

2

1 4 335 (3x + 1)3/2 3x4 + 1 dx = = 54 54 −1

2 1 x4 (x + 1)dx = + x = 3; 2 4 0 3

0

(x3 + 1)dx = f (x∗ )(2 − 0) = (x∗3 + 1)(2)

0

6 = 2(x∗3 + 1) √ x∗ = 3 2

5.7.27

s(t) =

v(t) dt =

t4 dt =

t5 +C 5

s(0) = 2, so C = 2 t5 +2 5

s(t) = v(t) dt = 3t2 − 4t + 1 dt = t3 − 2t2 + t + C s(t) =

5.7.28

s(1) = 5, so C = 5 s(t) = t3 − 2t2 + t + 5

4

0

128 m 3

5.7.30

56 m 15

5.7.32

t2 − 8t dt = −

5.7.29

4

t2 − 8t dt = − 0

128 m 3

128 m 3

2

t2 − t4 dt = −

5.7.31 0

56 m 15

2

t2 − t4 dt = − 0

56 m 15

Questions, Section 5.3

29

SECTION 5.8

sin

2/π

5.8.1 Evaluate 4/π

2

5.8.3 Evaluate

1 t dt. t2

(x2 + 1) 2x3 + 6x dx.

1

x 9x2 + 16 dx.

0

3

5.8.7 Evaluate

x 9 − x2 dx.

tan2 x sec2 x dx.

cos2 3t sin 3t dt.

0

2

5.8.11 Evaluate

√ x2 x − 1 dx.

5

5.8.12 Evaluate

1

√ x 2x − 1 dx.

1 10

5.8.13 Evaluate 5

5.8.15 Evaluate √

π/3

1

5.8.17 Evaluate 0

√π/6

x √ dx. x−1

√π/2

θ

sin2

t dt. cos2 2t2

5.8.14 Evaluate 0

θ2 2

dθ.

x2 3 2x + 7 dx.

2

4 5.8.19 Given that f (x) dx = 4, find −2 4 f (t) dt (a)

5.8.16 Evaluate

√ π/3

√ π/6

x csc x2 cot x2 dx.

π/8

5.8.18 Evaluate

(2x + sec 2x tan 2x) dx.

0

4

(b)

f (u) du. −2

1

5.8.20 Given that f (x) dx = −2, find −3 1 f (t) dt (a)

1

(b)

f (w) dw. −3

−3

x

t3 dt.

5.8.21 Evaluate

x

5.8.22 Evaluate

sin t dt. −2π

3

5.8.23 Differentiate 1

x dx.

9 + x2

π/9

5.8.10 Evaluate

0

−2

√

π/4

x4 + 2x2 + 1 dx.

x dx.

(1 + x2 )2

0

5.8.9 Evaluate

4

5.8.8 Evaluate

0

1

5.8.6 Evaluate

0

3

5.8.4 Evaluate

dx . x+1

0

5.8.5 Evaluate

√ 0

1

1

5.8.2 Evaluate

x3

d2 y 5.8.24 Find if y = dx2

sin t dt. t

x

sin2 θ dθ.

5.8.25 Find F (x) if F (x) = 0

x

t3 cos2 t dt.

3

30

Questions, Section 5.3

5.8.26 Find F (x) if F (x) = differentiating.

5.8.27 Find F (x) if F (x) =

x

(t3 + 1) dt. Check your work by first integrating and then 1

sin x

1 − t3 dt.

0

x

5.8.28 If F (x) = 1

sin 2t dt, find lim F (x). x→0 t

5.8.29 Find F (x) if F (x) =

√ 0

5.8.30 Find

d dx

x

x

1 dt.

1 − 3t2

(t + 1)1/2 dt . Check your work by first integrating, then differentiating.

0

5.8.31 Express the antiderivative of an integral, F (x).

1 on the interval (−∞, +∞) whose value at x = −2 is 0 as 4 + x2

π/3

tan6 x dx.

5.8.32 Use a CAS to find the exact value of the integral −π/3

π/3

sin5 x cos4 x dx.

5.8.33 Use a CAS to find the exact value of the integral 0

π/3

6x tan(x2 ) dx.

5.8.34 Use a CAS to find the exact value of the integral 0

0

4x tan(x2 ) dx.

5.8.35 Use a CAS to find the exact value of the integral −π/3

SOLUTIONS SECTION 5.8 1 1 dt , du = 2 dt, 2 = −du t t t

5.8.1 u =

π/2

−

sin u du = π/4

π/2 [ cos u]π/4

2

5.8.2 u = x + 1, du = dx,

√

√

2/π 1 2 2 or cos =− =− 2 t 4/π 2

2

√ u−1/2 du = 2u1/2 = 2 2 − 2 or 1

1

√ 1 √ 2 x+1 =2 2−2 0

5.8.3 u = 2x3 + 6x, du = 6(x2 + 1)dx, (x2 + 1)dx = √ √ 56 7 − 16 2 u = or 9 8 8 √ √ 2 56 7 − 16 2 1 3 (2x + 6x)3/2 = 9 9 1

1 6

28

1/2

3

5.8.4

1 du = u3/2 9

28

(x2

+

1)2

3 x3 (x + 1)dx = + x = 12 3 0 2

dx =

0

3

0

5.8.5 u = 9x2 + 16, du = 18x dx, x dx = 1 18

25

u1/2 du = 16

du , 6

du , 18

1 1 2 1 3/2 25 61 61 u or (9x + 16)3/2 = = 27 27 27 27 16 0

du , 2 1 2

1 2 −2 1 1 1 1 u du = − = or − = 2 1 2u 1 4 2(1 + x2 ) 0 4

5.8.6 u = 1 + x2 , du = 2x dx, x dx =

du , 2

3 1 9 1/2 1 3/2 9 1 u u du = = 9 or − (9 − x2 )3/2 = 9 2 0 3 3 0 0

5.8.7 u = 9 − x2 , du = −2x dx, x dx = −

du 2 4 √ = 2 or 9 + x2 0 = 2

5.8.8 u = 9 + x2 , du = 2x dx, x dx =

1 2

25

25

u−1/2 du = u1/2 9

0

5.8.9 u = tan x, du = sec2 x dx,

1

1

π/4 tan3 x

u3

1 1 u du = = or = 3 0 3 3 0 3 2

0

31

32

Solutions, Section 5.8

du , 3 π/9 7 7 1 = or − cos3 3t 0 = 72 9 72

5.8.10 u = cos 3t, du = −3 sin 3t dt, sin 3t dt = −

1 3

1

u2 du = 1/2

1 3 1 u 1/2 9

5.8.11 u = x − 1, du = dx, x = u + 1,

1

0

1

(u5/2 + 2u3/2 + u1/2 ) du =

(u + 1)2 u1/2 du = 0

or

4 2 2 (x − 1)7/2 + (x − 1)5/2 + (x − 1)3/2 5 3 7

2 = 1

2 7/2 4 5/2 2 3/2 u + u + u 7 5 3

1 = 0

184 105

184 105

u+1 du ,x= , 2 2

9

1 9 u+1 1 2 5/2 2 3/2 1 9 3/2 428 1/2 1/2 u du = or (u + u ) du = = u + u 4 1 4 5 3 2 15 2 1 1

5 1 2 2 428 5/2 3/2 (2x − 1) + (2x − 1) = 4 5 3 15 1

5.8.12 u = 2x − 1, du = 2 dx, dx =

5.8.13 u = x − 1, du = dx, x = u + 1,

9

4

u+1 du = u1/2

9

(u1/2 + u−1/2 ) du =

4

2 (x − 1)3/2 + 2(x − 1)1/2 3

10 = 5

2 3/2 u + 2u1/2 3

9 = 4

44 or 3

44 3

du , 5.8.14 u = 2t2 , du = 4t dt, t dt = 4 √ √π/6

3 1 π/3 1 π/3 2 2 2 t sec 2t dt = sec u du = [ tan u]0 = or 4 4 4 0 0 √

√π/6

3 1 2

tan 2t

= 4 4 0 θ2 5.8.15 u = , du = θ dθ, 2 √π/2 π/4 2 √ π/4 2 θ θ csc csc2 u du = − [ cot u]π/6 = 3 − 1 or dθ = √ 2 π/6 π/3 √ π/2

√ θ2 − cot = 3−1 √ 2 π/3

Solutions, Section 5.8

33

5.8.16 u = x2 , du = 2x dx, x dx =

du , 2

π/3 √

1 1 2 −3 − 3

√ −2 =− − csc u cot u du = − csc u

=− or

2 2 3 3 π/6 π/6 √

√π/3

3− 3 1 2

− − csc x √ 3 2 π/6

1 − 2

π/3

du , 6 √ √ 9 1 1 3 27 − 7 7 27 − 7 7 1 3/2 9 1 u , or (2x + 7)3/2 = = u1/2 du = 18 18 18 18 12 7 0 7

5.8.17 u = 2x3 + 7, du = 6x2 dx, x2 dx =

5.8.18 u = 2x, dx =

1 2

du , 2

π/4

√

π/4 2 1 π2 1 u2 + − = + sec u 64 2 2 2 2 0 √ 2 1 π2 + − = 64 2 2

(u + sec u tan u)du = 0

or

π/8 1 2 2x + sec 2x 0 2

5.8.19 (a) 4

(b) 4

5.8.20 (a) −2

(b) −2

x

x 81 t4

x4 − t dt = = 4 3 4 4 3

5.8.21 3

x

5.8.22 −2π

5.8.23

x

sin t dt = − cos t

sin x3 x3

−2π

= − cos x − (− cos(−2π)) = − cos x + 1 = 1 − cos x

d 3 sin x3 3 sin x3 (3x2 ) = [x ] = 3 dx x x

5.8.24 sin2 x

5.8.25

dy = x3 cos2 x dx d2 y = 3x2 cos2 x − 2x3 sin x cos x dx2

t4 5.8.26 F (x) = +t 4 5.8.27 5.8.28

x 1

x4 d x4 5 5

= x3 + 1 = + x − ; so F (x) = +x− 4 4 dx 4 4

d [sin x] = cos x 1 − sin3 x 1 − sin3 x dx lim

x→0

sin 2x sin 2x sin 2x = lim 2 = 2 lim = 2(1) = 2; (2x → 0 as x → 0) x→0 2x→0 2x x 2x

34

Solutions, Section 5.8

1 1 = (1 − 3x2 )−1/2 ; F (x) = − (1 − 3x2 )−3/2 (−6x) = 3x(1 − 3x2 )−3/2 2 2 1 − 3x x

d d 2 2 1/2 3/2 (x + 1) − = (x + 1)1/2 (t + 1) dt = dx 0 dx 3 3

5.8.29 F (x) = √

5.8.30

x

5.8.31 F (x) = −2

5.8.33

1 dt 4 + t2

3413 161280

2

π

5.8.35 2 ln cos 9

5.8.32

√ 2(37/2 − 5 · 33/2 + 15 3 − 5π) 15

2

π

5.8.34 −3 ln

cos 9

Chapter 5

35

SUPPLEMENTARY EXERCISES, CHAPTER 5 In Exercises 1–10, evaluate the integrals and check your results by differentiation.

1.

1 1 + √ − 5 sin x dx x3 x

2.

√ ( 5 + 2)8 √ 3. dx x √

x sin 2x2 − 5 √ 5. dx 2x2 − 5

√ √ 3 7. x(3 + x4 ) dx

x3 cos(2x4 − 1) dx

4.

6.

√

8.

sec2 (sin 5t) cos 5t dt

9.

10.

11. Evaluate

2t4 − t + 2 dt t3

cos θ sin(2θ) dθ

x1/3 dx x8/3 + 2x4/3 + 1 cot2 x dx sin2 x

y(y 2 + 2)2 dy two ways: (a) by multiplying out and integrating term by term; and (b)

by using the substitution u = y 2 + 2. Show that your answers differ by a constant.

In Exercises 12–17, evaluate the definite integral by making the indicated substitution and changing the x-limits of integration to u-limits.

0

12.

√ 5

1 − 2x dx, u = 1 − 2x

1

sin4 x cos x dx, u = sin x 0

−3

14.

0

π/2

13.

π/4

16. π/6

x dx √ , u = x2 + 16 x2 + 16

5

15.

2

sin 2x dx 3

, u = 1 − cos 2x 2 3 1 − 2 cos 2x

4

17.

1

x−2 √ dx, u = x − 1 x−1 √ √ π x π x 1 √ cos dx, u = 2 2 x

2

f (x) dx.

In Exercises 18 and 19, evaluate −2

18. f (x) =

x3 , for x ≥ 0 −x, for x < 0

In Exercises 20–22, solve for x. x 1 √ dt = 3 20. t 1

x

(4t − 1) dt = 9

22.

2

19. f (x) = |2x − 1|

21.

0

x

1 1 dt = (3t + 1)2 6

36

Supplementary Exercises

23. (a) (c) (e) (g)

6

i=3 n+3

i=n 4

k=2 4

5 n

k=0

k=1 4

(f )

kĎ&#x20AC; 4

n=4 4

(h)

k=1

24. Express in sigma notation and evaluate: (a) 3 ¡ 1 + 4 ¡ 2 + 5 ¡ 3 + ¡ ¡ ¡ + 102 ¡ 100

2

i=n 3

(d)

6 k2 sin

n+3

(b)

kâ&#x2C6;&#x2019;1 k+3

(2n + 1) sink

Ď&#x20AC; 4

(b) 200 + 198 + ¡ ¡ ¡ + 4 + 2.

25. Express in sigma notation, ďŹ rst starting with k = 1, and then with k = 2. (Do not evaluate.)

1 4 81 Ď&#x20AC; 12 64 Ď&#x20AC;3 Ď&#x20AC;4 Ď&#x20AC;2 9 (a) + â&#x2C6;&#x2019; ¡¡¡ + â&#x2C6;&#x2019; ¡¡¡ â&#x2C6;&#x2019; â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + (b) 81 100 3 11 1 2 4 9 16 In Exercises 26â&#x20AC;&#x201C;29, use the partition of [a, b] into n subintervals of equal length, and ďŹ nd a closed form for the sum of the areas of (a) the inscribed rectangles and (b) the circumscribed rectangles. (c) Use your answer in either part (a) or part (b) to ďŹ nd the area under the curve y = f (x) over the interval [a, b]. (Check your answer by integration.)

26. f (x) = 6 â&#x2C6;&#x2019; 2x; a = 1, b = 3

27. f (x) = 16 â&#x2C6;&#x2019; x2 ; a = 0, b = 4

28. f (x) = x2 + 2; a = 1, b = 4

29. f (x) = 6; a = â&#x2C6;&#x2019;1, b = 1

30. Given that

5

5

P (x) dx = â&#x2C6;&#x2019;1 1

P (x) dx = 3 3

and

5

Q(x) dx = 4 3

evaluating the following: 5 [2P (x) + Q(x)] dx (a) 3 â&#x2C6;&#x2019;5 (c) Q(â&#x2C6;&#x2019;x) dx

1

(b)

P (t) dt 5

1

(d)

â&#x2C6;&#x2019;3

P (x) dx.

3

31. Suppose that f is continuous and x2 â&#x2030;¤ f (x) â&#x2030;¤ 6 for all x in [â&#x2C6;&#x2019;1, 2]. Find values of A and B such that

Aâ&#x2030;¤

2

â&#x2C6;&#x2019;1

f (x) dx â&#x2030;¤ B

Chapter 5

37

In Exercises 32–35, find the average value of f (x) over the indicated interval and all values of x∗ described in the Mean-Value Theorem for Integrals. x

32. f (x) = 3x2 ; [−2, −1]

33. f (x) = √

34. f (x) = 2 + |x|; [−3, 1]

35. f (x) = sin2 x; [0, π].

x2

+9

[Hint: sin2 x =

; [0, 4]

1 (1 − cos 2x).]

2

In Exercises 36 and 37, find the area of the surface generated by revolving the given curve about the indicated axis.

36. y = x3 between (1, 1) and (2, 8); x-axis.

37. y =

√

2x −

x2

√ 3 1 , and (1, 1); x-axis. between 2 2

SOLUTIONS SUPPLEMENTARY EXERCISES, CHAPTER 5 √ −x−2 + 2 x + 5 cos x + C 2

2 1 1 1 2.

2t − 2 + 3 dt = t2 + − 2 + C t t t t

1.

3. u =

√

u8 du =

x + 2, 2

2 9 2 √ u + C = ( x + 2)9 + C 9 9

1 sin(2x4 − 1) + C 8

1 2x 1 sin u du = − cos 2x2 − 5 + C 5. u = 2x2 − 5, du = √ dx, 2 2 2x2 − 5 1 4. u = 2x − 1, 8 4

6.

√

cos u du =

cos θ (2 sin θ cos θ)dθ = 2

(3x1/2 + x11/6 )dx = 2x3/2 +

7.

8.

4 cos3/2 θ sin θ dθ = − cos5/2 θ + C 5 6 17/6 +C x 17

(x4/3 + 1)−2 x1/3 dx, u = x4/3 + 1,

3 4

u−2 du =

− 34 +C x4/3 + 1

1 tan(sin 5t) + C 5

1 10. cot2 x csc2 x dx, u = cot x, − u2 du = − cot3 x + C 3

1 11. (a) (y 5 + 4y 3 + 4y)dy = y 6 + y 4 + 2y 2 + C 6 9. u = sin 5t,

1 5

sec2 u du =

1 2 (y + 2)3 + C 6 [answer to (b)] – [answer to (a)]

(b)

1 = (y 6 + 6y 4 + 12y 2 + 8) + C − 6 1 12. − 2

1

u

1/5

−1

1

u4 du =

13.

0

5 du = − u6/5 12

1 6 y + y 4 + 2y 2 + C 6

1

4

4 3

=0 −1

14.

=

1

1 5

15. u = x − 1, x = u + 1,

u−1 √ du = u

4

1 2

25

u−1/2 du = u1/2

16

(u1/2 − u−1/2 )du =

1

38

25 =1 16

2 3/2 u − 2u1/2 3

4 = 1

8 3

Chapter 5

16.

39

1 3

1

u

â&#x2C6;&#x2019;1/2

1/4

2 du = u1/2 3

0

2

(â&#x2C6;&#x2019;x)dx +

18. â&#x2C6;&#x2019;2

â&#x2C6;&#x2019;2

x

20.

1

x

21.

0

x

22.

1 = 3

1/4

1 x3 dx = â&#x2C6;&#x2019; x2 2

1/2

19.

0

1

17.

0

1 + x4 4 â&#x2C6;&#x2019;2

Ï&#x20AC;

4 cos u du = sin u Ï&#x20AC; Ï&#x20AC;/2

Ï&#x20AC; =â&#x2C6;&#x2019; Ï&#x20AC;/2

4 Ï&#x20AC;

=6 0

2

1/2

+ (x2 â&#x2C6;&#x2019; x)

(2x â&#x2C6;&#x2019; 1)dx = (â&#x2C6;&#x2019;x2 + x) 1/2

x

2

2

â&#x2C6;&#x2019;(2x â&#x2C6;&#x2019; 1)dx + â&#x2C6;&#x161; 1 â&#x2C6;&#x161; dt = 2 t t

4 Ï&#x20AC;

â&#x2C6;&#x2019;2

=

17 2

1/2

â&#x2C6;&#x161; â&#x2C6;&#x161; 5 25 = 2( x â&#x2C6;&#x2019; 1) = 3 x = , x = 2 4

1

1 1 dt = â&#x2C6;&#x2019; (3t + 1)2 3(3t + 1)

x =â&#x2C6;&#x2019; 0

1 1 1 1 + = , 3x + 1 = 2, x = 3(3x + 1) 3 6 3

x (4t â&#x2C6;&#x2019; 1)dt = (2t2 â&#x2C6;&#x2019; t) = 2x2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 6 = 9, 2x2 â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 15 = 0,

2

2

(2x + 5)(x â&#x2C6;&#x2019; 3) = 0, x = â&#x2C6;&#x2019;

5 and x = 3.

2

23. (a) 5 + 5 + 5 + 5 = 20

(b) 2 + 2 + 2 + 2 = 16

(c) n + n + n + n = 4n

(d) 0 + 1/5 + 2/6 = 8/15

(e) 6/4 + 6/9 + 6/16 = 61/24 (f ) 9 â&#x2C6;&#x161; â&#x2C6;&#x161;

Ï&#x20AC; Ï&#x20AC; â&#x2C6;&#x161; 3Ï&#x20AC; 2 2 + sin + sin + sin(Ï&#x20AC;) = 0 + (g) sin(0) + sin +1+ +0=1+ 2 4 2 4 2 2 â&#x2C6;&#x161; 2 â&#x2C6;&#x161; 3 â&#x2C6;&#x161; 4 â&#x2C6;&#x161; â&#x2C6;&#x161; 2 2 2 2 3 2 3 + + (h) + + = 2 2 2 4 4 2

24. (a)

100

(k + 2)k =

k=1

(b)

100

9

(202 â&#x2C6;&#x2019; 2k) =

(â&#x2C6;&#x2019;1)k+1

11

k k+1

100

k=

k=1

100

202 â&#x2C6;&#x2019; 2

k=1

k=1

(b)

k2 + 2

k=1

k=1

25. (a)

100

100

k=1

2 =

10

1 1 (100)(101)(201) + 2 · (100)(101) = 348, 450 6 2

1 k = (100)(202) â&#x2C6;&#x2019; 2 · (100)(101) = 10, 100 2

(â&#x2C6;&#x2019;1)k

k=2

kâ&#x2C6;&#x2019;1 k

2

Ï&#x20AC; k+1 Ï&#x20AC;k = (â&#x2C6;&#x2019;1)k k kâ&#x2C6;&#x2019;1 12

(â&#x2C6;&#x2019;1)k+1

k=1

k=2

2k 2 , ck = 1 + n n

n n n n

2 8 2k 8 n+1 = 6â&#x2C6;&#x2019;2 1+ f (ck )â&#x2C6;&#x2020;x = 1â&#x2C6;&#x2019; 2 k =8â&#x2C6;&#x2019;4 n n n n n

26. (a) â&#x2C6;&#x2020;x =

k=1

k=1

k=1

k=1

40

Solutions, Supplementary Exercises

2(k − 1) n

n n

2 2(k − 1) 6−2 1+ f (dk )∆x = n n

(b) dk = 1 +

k=1

k=1

=

n n 8 8 n−1 1− 2 (k − 1) = 8 − 4 n n n k=1

k=1

3 1 = 8 − 4 = 4; (c) area = lim 8 − 4 1 + (6 − 2x)dx = 4 n→+∞ n 1 4k 4 , ck = n n

n n

4 16k 2 16 − 2 f (ck )∆x = n n

27. (a) ∆x =

k=1

k=1

=

n n 64 32 (n + 1)(2n + 1) 64 2 k = 64 − 1− 3 3 n2 n n k=1

(b) dk = n

4(k − 1) n

f (dk )∆x =

k=1

n

k=1

k=1

16(k − 1)2 16 − n2

n n 64 4 64 = 1− 3 (k − 1)2 n n n k=1

k=1

32 (n − 1)(2n − 1) 3 n2

4 1 1 128 32 128 1+ 2+ = ; (16 − x2 )dx = (c) area = lim 64 − n→+∞ 3 n n 3 3 0 = 64 −

3(k − 1) 3 , ck = 1 + n n

n n

6(k − 1) 9(k − 1)2 3 + 3+ f (ck )∆x = n n2 n

28. (a) ∆x =

k=1

k=1

=

n n n 9 18 27 1+ 2 (k − 1) + 3 (k − 1)2 n n n k=1

=9+9 (b) dk = 1 + n

3k n

k = 1f (dk )∆x =

k=1

k=1

n − 1 9 (n − 1)(2n − 1) + n 2 n2

n

k=1

3+

6k 9k 2 + 2 n n

n n n 9 3 18 27 2 = 1+ 2 k+ 3 k n n n n k=1

k=1

k=1

n + 1 9 (n + 1)(2n + 1) + =9+9 n 2 n2

4 9 1 1 1 + 1− 2− = 27; (c) lim 9 + 9 1 − (x2 + 2) dx = 27 n→+∞ n 2 n n 1

Chapter 5

41

2 , because f is constant ck can be chosen anywhere in the k-th subinterval so f (ck ) = 6 n

n n

2 = 12 and f (ck )∆x = (6) n

29. (a) ∆x =

k=1

k=1

(b) same as for (a)

1

(c) area = lim 12 = 12;

6 dx = 12

n→+∞

−1

5

5

Q(x)dx = 2(3) + (4) = 10

P (x)dx +

30. (a) 2

3

3

5

(b) −

P (x)dx = −(−1) = 1 1

5

(c) − 3

(d)

1

P (x)dx + 3

−1

−2

1 = 4

P (x)dx = (3) − (−1) = 4

3

2

−1

5

P (x)dx −

5

5

P (x)dx =

31. If x2 ≤ f (x) ≤ 6 then

33. fave

Q(x)dx = −4 3

5

32. fave =

5

Q(u)du = −

x2 dx ≤

1

2

−1

f (x)dx ≤

2

−1

6 dx, 3 ≤

2

−1

f (x)dx ≤ 18

3x2 dx = 7; 3(x∗ )2 = 7, x∗ = ± 7/3 but only − 7/3 is in [−2, −1] 4 2

−1/2

x(x + 9) 0

1 dx = (x2 + 9)1/2 4

4 = 0

1 ; 2

√ √ x 1

= , 2x∗ = (x∗ )2 + 9, 4(x∗ )2 = (x∗ )2 + 9, x∗ = ± 3 but only 3 is in [0, 4].

2 (x∗ )2 + 9 ∗

34. fave

1 = 4

1

1 (2 + |x|)dx = 4 −3

0

−3

(2 − x)dx + 0

1

13 1 21 5 + = ; (2 + x)dx = 4 2 2 4

5 5 13 5 , |x∗ | = , x∗ = ± but only − is in [−3, 1] 2 + |x∗ | = 4 4 4 4

π π π 1 1 1 1 1 1 2 x − sin 2x sin x dx = = ; 35. fave = (1 − cos 2x)dx = π 0 π 0 2 2π 2 2 0 sin2 x∗ =

1 1 π 3π , sin x∗ = ± √ , x∗ = , for x∗ in [0, π] 2 4 4 2