Answer Key
833
125 . Use tan 2 θ +1= sec 2 θ . Then, B − A = ∫ − π /4 π /4
1 dx = π
2 .
127 . ∫ 2 π cos 2 tdt = π , so divide by the length 2 π of the interval. cos 2 t has period π , so yes, it is true. 129 . The integral is maximized when one uses the largest interval on which p is nonnegative. Thus, A = − b − b 2 −4 ac 2 a and B = − b + b 2 −4 ac 2 a . 131 . If f ( t 0 ) > g ( t 0 ) for some t 0 ∈ ⎡ ⎣ a , b ⎤ ⎦ , then since f − g is continuous, there is an interval containing t 0 such that f ( t ) > g ( t ) over the interval ⎡ ⎣ c , d ⎤ ⎦ , and then ∫ d d f ( t ) dt > ∫ c d g ( t ) dt over this interval. 133 . The integral of f over an interval is the same as the integral of the average of f over that interval. Thus, ∫ a b f ( t ) dt = ∫ a 0 a 1 f ( t ) dt + ∫ a 1 a 2 f ( t ) dt +⋯+ ∫ a N +1 a N f ( t ) dt = ∫ a 0 a 1 1 dt + ∫ a 1 a 2 1 dt +⋯+ ∫ a N +1 a N 1 dt = ( a 1 − a 0 )+( a 2 − a 1 )+⋯+( a N − a N −1 ) = a N − a 0 = b − a . Dividing through by b − a gives the desired identity. 135 . ∫ 0 N f ( t ) dt = ∑ i =1 N ∫ i −1 i f ( t ) dt = ∑ i =1 N i 2 = N ( N +1)(2 N +1) 6 137 . L 10 =1.815, R 10 =1.515, L 10 + R 10 2 =1.665, so the estimate is accurate to two decimal places. 139 . The average is 1/2, which is equal to the integral in this case. 141 . a. The graph is antisymmetric with respect to t = 1 2 over [0, 1], so the average value is zero. b. For any value of a , the graph between [ a , a +1] is a shift of the graph over [0, 1], so the net areas above and below the axis do not change and the average remains zero. 0
143 . Yes, the integral over any interval of length 1 is the same. 145 . Yes. It is implied by the Mean Value Theorem for Integrals. 147 . F ′(2) =−1; average value of F ′ over [1, 2] is −1/2. 149 . e cos t 151 . 1 16− x 2 153 . x d dx x = 1 2 155 . − 1−cos 2 x d dx cos x = | sin x | sin x 157 . 2 x | x | 1+ x 2 159 . ln( e 2 x ) d dx e x =2 xe x 161 . a. f is positive over [1, 2] and ⎡ ⎣ 5, 6 ⎤ maximum value is 2 and the minimum is −3. c. The average value is 0. 163 . a. ℓ is positive over [0, 1] and ⎡ ⎣ 3, 6 ⎤
⎦ , negative over [0, 1] and [3, 4], and zero over [2, 3] and ⎡ ⎣ 4, 5 ⎤
⎦ . b. The
⎦ , and negative over [1, 3]. b. It is increasing over [0, 1] and ⎡ ⎣ 3, 5 ⎤
⎦ , and it is
⎤ ⎦ . c. Its average value is 1 3 .
constant over [1, 3] and ⎡
⎣ 5, 6
−2 3 x 3 +6 x 2 + x −5 dx =48
165 . T 10 =49.08, ∫
167 . T 10 =260.836, ∫ 1 9 ⎛
⎝ x + x 2 ⎞
⎠ dx =260
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