Calculus Volume 1

546

Chapter 5 | Integration

− π π

π /2

t 1+cos t

sin tdt ≥ π

Hint : sin t ≥ 2 t

95. ∫

108. Show that ∫ 0

dt

π over

4 . (

⎡ ⎣ 0, π 2

⎤ ⎦ )

In the following exercises, find the net signed area between f ⎛ ⎝ x ⎞ ⎠ and the x-axis.

− π /4 π /4

109. Show that ∫

cos tdt ≥ π 2/4.

3 (2− x ) dx ( Hint: Look at the graph of f .)

96. ∫

1

In the following exercises, find the average value f ave of f between a and b , and find a point c , where f ( c ) = f ave .

4 ( x −3) 3 dx ( Hint: Look at the graph of f .)

97. ∫

2

110. f ( x ) = x 2 , a =−1, b =1 111. f ( x ) = x 5 , a =−1, b =1 112. f ( x ) = 4− x 2 , a =0, b =2 113. f ( x ) = (3− | x |), a =−3, b =3 114. f ( x ) = sin x , a =0, b =2 π 115. f ( x ) =cos x , a =0, b =2 π

In the following exercises,

given that

1

1

1

2 , ∫

x 3 dx = 1

x 2 dx = 1

xdx = 1

and

3 ,

4 ,

0

0

0

compute the integrals.

1 ⎛ ⎝ 1+ x + x 2 + x 3 ⎞

98. ∫

⎠ dx

0

1 ⎛ ⎝ 1− x + x 2 − x 3 ⎞

99. ∫

⎠ dx

0

1 (1− x ) 2 dx

In the following exercises, approximate the average value using Riemann sums L 100 and R 100 . How does your answer compare with the exact given answer? 116. [T] y = ln( x ) over the interval [1, 4]; the exact solution is ln(256) 3 −1. 117. [T] y = e x /2 over the interval [0, 1]; the exact solution is 2( e −1).

100. ∫

0

1 (1−2 x ) 3 dx

101. ∫

0

1 ⎛

x 2 ⎞

102. ⌠

⎝ 6 x − 4 3

⎠ dx

⌡ 0

1 ⎛

⎝ 7−5 x 3 ⎞

103. ∫

⎠ dx

118. [T] y = tan x over the interval ⎡

⎤ ⎦ ; the exact

⎣ 0, π 4

0

solution is 2ln(2) π . 119. [T] y = x +1 4− x 2 exact solution is π 6 .

In the following exercises, use the comparison theorem . 104. Show that ∫ 0 3 ⎛ ⎝ x 2 −6 x +9 ⎞ ⎠ dx ≥0. 105. Show that ∫ −2 3 ( x −3)( x +2) dx ≤0. 106. Show that ∫ 0 1 1+ x 3 dx ≤ ∫ 0 1 1+ x 2 dx . 107. Show that ∫ 1 2 1+ xdx ≤ ∫ 1 2 1+ x 2 dx .

over the interval [−1, 1]; the

In the following exercises, compute the average value using the left Riemann sums L N for N = 1, 10, 100. Howdoes the accuracy compare with the given exact value? 120. [T] y = x 2 −4 over the interval [0, 2]; the exact solution is − 8 3 .

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