Answer Key
841
⎛ ⎝ 1 b
⎞ ⎠ = ∫
1/ a dx x
b dx x = ln( b )−ln( a ) = ln ⎛ ⎝ 1 a ⎞
379 . ∫
⎠ −ln
a
1/ b
381 . 23 383 . We may assume that x >1, so 1 x <1. Then, ∫ 1 1/ x dt
du = − dt t 2
t . Now make the substitution u = 1 t , so
and
1/ x dt
x du
du u = −
dt t , and change endpoints: ∫ 1
t =− ∫
u =−ln x .
1
⎠ . Then, 1= E '(ln x ) x
387 . x = E ⎛
⎝ ln( x ) ⎞
or x = E '(ln x ). Since any number t can be written t = ln x for some x , and for such t
we have x = E ( t ), it follows that for any t , E '( t ) = E ( t ). 389 . R 10 =0.6811, R 100 =0.6827
391 . sin −1 x | 0 393 . tan −1 x |
3/2
= π 3
1
= − π 12
3
395 . sec −1 x | 1 2
= π 4
⎛ ⎝ x 3
⎞ ⎠ + C
397 . sin −1
−1 ⎛ −1 ⎛
⎞ ⎠ + C ⎞ ⎠ + C
⎝ x 3 ⎝ x 3
399 . 1 401 . 1
3 tan
3 sec
⎛ ⎝ π
θ ⎞ ⎠ = sin θ . So, sin
−1 t = π
−1 t .
403 . cos
They differ by a constant.
2 −cos
2 −
405 . 1− t 2 is not defined as a real number when t >1. 407 .
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