Double Integration-may someone check my work for me?

[pockystix]

hihi,
I was working through one of my hw problems and I got stuck...I did all of the work, but I still feel like my answer is wrong. May someone check over this for me, please? My integration is a little rusty, having not touched cal.for almost 2 years. Thanks!

This is the original problem: Evaluate the double integral ??R y/x+y^2dA, where
R =[1,2]x[1,2]

Here is what I did-->http://img23.imageshack.us/img23/7452/hwproblem.jpg

 

[galactus]

\(\displaystyle \int_{1}^{2}\int_{1}^{2}\frac{y}{x+y^{2}}dxdy\)

Your solution is \(\displaystyle \frac{ln(\frac{432}{3125})}{2}\)

Should be \(\displaystyle \frac{ln(\frac{6912}{3125})}{2}\)

\(\displaystyle \int\frac{y}{x+y^{2}}dx=yln(x+y^{2})\Rightarrow yln(y^{2}+2)-yln(y^{2}+1)\)

\(\displaystyle \int yln(y^{2}+2)dy-\int yln(y^{2}+1)dy=\frac{(y^{2}+1)(ln(y^{2}+1)-1)}{2}-\frac{(y^{2}+2)(ln(y^{2}+2)-1)}{2}\)

Using the limits, we get \(\displaystyle \frac{3ln(12)}{2}-\frac{3}{2}-\left(\frac{ln(\frac{3125}{4})}{2}-\frac{3}{2}\right)=\frac{ln(\frac{6912}{3125})}{2}\)

There is a multiple of 16 somewhere. 6912/432=16. It always some little algebra error.

 

[pockystix]

oooh your answer makes more sense.I've been trying to figure this out for hours. Thanks galactus!