MA 124 Fall-07, Exam 1 Answers
- 1) (1/2) (cos(\sqrt(x)) / x)
- 2) -2/(pi)^2
- 3) ln|x - 1| - (1/2)ln(x^2 + 9) - (1/3)tan^{-1}(x/3) + C
- 4) (-1/2)(1/(x^2 + 1)) + C
- 5) 1/e
- 6) Since \sqrt(1 + x^6) > \sqrt(x^6) = x^3 , we get (1 \over \sqrt(1 +
x^6)) < 1/x^3 , hence (x \over \sqrt(1 +
x^6)) < x/x^3 = 1/x^2. So \integral_1^\infinity (x \over \sqrt(1 +
x^6)) is less than or equal to \integral_1^\infinity 1/x^2 . The latter is
a p-integral with p = 2 > 1 , hence it converges. So our original integral
converges by comparison since it is less than or equal to an integral which converges.
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