Question

Isn't solution for 3A-2e in Integration Problem Set false?

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-c-mean-value-theorem-antiderivatives-and-differential-equations/problem-set-5/MIT18_01SC_pset3prb.pdf

If I derive the solution given, I get "x * (8-2x^2)^1/2 dx" instead of "x / (8-2x^2)^1/2 dx"

In additon the calculation showed with advanced guessing seems false.

Isn't the right solution "-1/2 * (8-2x^2)^1/2 + c", instead of "-1/6 * (8 − 2x^2)^3/2 + c"?

Answer 1

For the problem
\[ \int (8-2x^2)^{- \frac{1}{2}}\ x\ dx \]
let u = 8-2x^2
du = -4x dx
insert -4 (times -¼), and sub in u and du
\[ -\frac{1}{4} \int (8-2x^2)^{- \frac{1}{2}} -4 x dx \\
-\frac{1}{4} \int u^{- \frac{1}{2}}\ du
\]
integrate to get
\[ -\frac{1}{4} \cdot 2 u^{\frac{1}{2}} \\
-\frac{1}{2} \sqrt{8-2x^2} +c\]

which matches your solution. It looks like the answer key used u^(+½)

btw, though I would not use wolfram to do the problems (if you don't do the problems on your own you are wasting your time), you can use it as a check
http://www.wolframalpha.com/input/?i=integrate++x+dx+%2F+sqrt%288-2x%5E2%29

Answer 2

Thank you! An the tool seems quite awesome for checking answers indeed. oO