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TEST 2 ANSWERS
BRIEF ANSWERS FOR TEST 2
- The integral for the volume can be set up in many ways.
One is: Integral from 0 to 1 of [Pi*y dy]. The answer is
Pi/2.
- Integral from 0 to Pi of [2*Pi*sin(x)*square_root(1+cos^2(x)) dx].
- The substitution u = cos(x) reduces the integral
to the one evaluated at the top of the page. The final answer is
2*Pi*[square_root(2) + ln(1+square_root(2))].
- Integrate ...
- The substitution u = e^x reduces this integral to the
integral of 1/(u*(u-1)). After expanding by partial fractions, this
is easy to integrate. The final answer is -x + ln(e^x - 1) + C.
- Integrate by parts using u = tan^(-1)(x) and dv = dx.
The answer is x*tan^(-1)(x) - (1/2)*ln(x^2+1) + C.
- This integral converges. Using integration by parts you
can show that the integral of ln(x)/x^2 is
-ln(x)/x - 1/x. Using this it is not hard to show that
the improper integral converges to 1.
- The series converges to 1/(1-u^2) whenever |u| is less than 1.
Last modified on Mar 31, 1998.