832
Answer Key
1
65 . ∫ 67 . ∫
xdx
0
6
xdx
3
2
69 . ∫ ⎞ ⎠ dx 71 . 1+2·2+3·3=14 73 . 1−4+9=6 75 . 1−2 π +9=10−2 π 77 . The integral is the area of the triangle, 1 2 79 . The integral is the area of the triangle, 9. 81 . The integral is the area 1 2 πr 2 =2 π . 1 x log ⎛ ⎝ x 2
83 . The integral is the area of the “big” triangle less the “missing” triangle, 9− 1 2 . 85 . L =2+0+10+5+4=21, R =0+10+10+2+0=22, L + R 2 =21.5 87 . L =0+4+0+4+2=10, R =4+0+2+4+0=10, L + R 2 =10 89 . ∫ 2 4 f ( x ) dx + ∫ 2 4 g ( x ) dx =8−3=5 91 . ∫ 2 4 f ( x ) dx − ∫ 2 4 g ( x ) dx =8+3=11 93 . 4 ∫ 2 4 f ( x ) dx −3 ∫ 2 4 g ( x ) dx =32+9=41 95 . The integrand is odd; the integral is zero. 97 . The integrand is antisymmetric with respect to x =3. The integral is zero. 99 . 1− 1 2 + 1 3 − 1 4 = 7 12 101 . ∫ 0 1 ⎛ ⎝ 1−6 x +12 x 2 −8 x 3 ⎞ ⎠ dx = ⎛ ⎝ ⎜ x - 3 x 2 +4 x 3 - 2 x 4 ⎞ ⎠ ⎟ = ⎛ ⎝ ⎜ 1 - 3+4 - 2 ⎞ ⎠ ⎟
⎛ ⎝ ⎜ 0 - 0+0 - 0
⎞ ⎠ ⎟ =0
103 . 7− 5 4 = 23 4 105 . The integrand is negative over [−2, 3]. 107 . x ≤ x 2 over [1, 2], so 1+ x ≤ 1+ x 2 over [1, 2]. 109 . cos( t ) ≥ 2
2 . Multiply by the length of the interval to get the inequality.
111 . f ave =0; c =0 113 . 3 2 when c = ± 3 2 115 . f ave =0; c = π 2 ,
3 π 2 117 . L 100 =1.294, R 100 =1.301; the exact average is between these values. 119 . L 100 × ⎛ ⎝ 1 2 ⎞ ⎠ =0.5178, R 100 × ⎛ ⎝ 1 2 ⎞ ⎠ =0.5294 121 . L 1 =0, L 10 × ⎛ ⎝ 1 2 ⎞ ⎠ = 8.743493, L 100 × ⎛ ⎝ 1 2 ⎞
⎠ = 12.861728. The exact answer ≈26.799, so L 100 is not accurate.
⎛ ⎝ 1 π
⎞ ⎠ =1.352, L 10 × ⎛ ⎝ 1 π ⎞
⎠ =−0.1837, L 100 × ⎛ ⎝ 1 π ⎞
123 . L 1 ×
⎠ =−0.2956. The exact answer ≈ −0.303, so L 100 is not
accurate to first decimal.
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