Evaluating Double Integrals

1. Oct 17, 2014

_N3WTON_

1. The problem statement, all variables and given/known data
Evaluate the following double integral:
$V = \int\int \frac{3y}{6x^{5}+1} \,dA$
$D = [(x,y) \hspace{1 mm}|\hspace{1 mm} 0<=x<=1 \hspace{5 mm} 0<=y<=x^2]$

2. Relevant equations

3. The attempt at a solution
$V = \int_{0}^{1} \int_{0}^{x^2} \frac{3y}{6x^{5}+1}\,dy\,dx$
$= \frac{1}{2}\int_{0}^{1}\int_{0}^{x^2}\frac{y}{x^{5}+1}\,dy\,dx$
$= \frac{1}{2}\int_{0}^{1}\int_{0}^{x^2}\frac{y^2}{2x^{5}+1}\,dx$
$= \frac{1}{2}\int_{0}^{1}\frac{x^{4}}{2x^{5}+1}\,dx$
$= \frac{1}{4}\int_{0}^{1}\frac{x^{4}}{x^{5}+1}\,dx$
To integrate, I made the following substitution:
$u=x^{5}+1$
$du=5x^4$
Therefore, the new limits for integration are:
$u=1$
$u=2$
$= \frac{1}{20}\int_{1}^{2}\frac{du}{u}$
$= \frac{1}{20}[lnu]$
$= \frac{1}{20}[ln2-ln1]$
This is my final (incorrect) answer:
$\frac{ln2}{20}$
Obviously, I am making a mistake somewhere, but I can't see where so I thought a second (or third) set of eyes may help :) Thanks

Last edited: Oct 17, 2014
2. Oct 17, 2014

RUber

6x5-1 is not 6(x5-1)

3. Oct 17, 2014

SteamKing

Staff Emeritus
You need to learn how to use parentheses so that your algebra is correct:

$\frac{y^2}{2x^{5}+1}$ should be $\frac{y^2}{2(x^{5}+1)}$ after you integrate

4. Oct 17, 2014

_N3WTON_

ok, so then the answer will turn out to be:
$\frac{ln7}{20}$
no?
thanks for the help :)

5. Oct 17, 2014

_N3WTON_

Sorry I always forget to show my solution:
$V = \int_{0}^{1} \frac{3}{6x^{5}+1}\frac{y^2}{2}\,dx$
$= \frac{3}{2}\int_{0}^{1}\frac{x^4}{6x^{5}+1}\,dx$
$u = 6x^{5}+1 \hspace{5 mm} du = 30x^4$
${when} \hspace{3 mm}x=0, \hspace{3 mm} u = 1$
${when}\hspace{3 mm} x=1, \hspace{3 mm} u = 7$
$= \frac{1}{20}\int_{1}^{7}\frac{du}{u}$
$= \frac{ln(7)}{20}$

6. Oct 17, 2014

Dick

Seems ok to me.

7. Oct 17, 2014

_N3WTON_

thank you

8. Oct 18, 2014

HallsofIvy

Staff Emeritus
Your final answer is correct but $du= 30x^4 dx$ so $(1/30)du= x^4 dx$