# Double integral finding new limit $\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv$

I don't understand how to find new limits when I use substitution method.

I have this integral :

$$\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv=$$

$$t=u^2+4v, dt=2udu, \frac{dt}{2}=udu$$

The new limits for $du$ are $\int_{4v}^{1+4v}$ after I substitute $u=1,u=0$ for $t$.

But for some reason in my book they get $\int_{0}^{1}$, but I don't understand what about $4v$

Any ideas?

Any help will be appreciated, Thanks in advance!

EDIT (The full answer in the book):

$$\int_1^2\int_0^1 \frac{u}{\sqrt{u^2+4v}} dudv=\int_1^2(\sqrt{u^2+4v}|_{u=0}^{u=1})dv=\int_1^2(\sqrt{4v+1}-2\sqrt{v})dv=\\\frac{2}{3}(\frac{1}{4}\sqrt{(4v+1)^3}-2\sqrt{v^3} |_{1}^{2}=\frac{35}{6}-\frac{5\sqrt{5}}{5}-\frac{8\sqrt{2}}{3}$$

• what are the other limits in your book? – Math-fun Jun 18 '15 at 10:18
• It seems that your calculation is right except that $4v$ should be down and $1+4v$ up (the integral should be positive). You can calculate the full double integral. What do they obtain in your book? – Urgje Jun 18 '15 at 10:34
• @Math-fun Added the answer in the book – JaVaPG Jun 18 '15 at 10:42
• @Urgje You'r right typo, edited. I also added the answer in the book, – JaVaPG Jun 18 '15 at 10:43
• Then it is a little mystery. Maybe @Math-fun can shed some light on this matter. – Urgje Jun 18 '15 at 11:05

• Thank you for you answer, I have another simple question if I may, In case I change the bounds I shouldn't go back to $v$ and in case the bounds remain the same I should go back to $v$. In the first approach you left it as variable $t$ however in second apporach you went back to $\sqrt{u^2+4v}$ from $t$. – JaVaPG Jun 18 '15 at 12:16
• Thank you for the answer, I'd like to make sure I get it right. because this is very important, in definite integrals, we can calculate the integral of $t$ (assume $t$ is our substitution variable) and then we can go back to the original variable after we calculate the integral using $t$ but we must go back to the original bounds or else we can just leave it as $t$ but we must use the new bounds that we found for $t$, is that statement correct? – JaVaPG Jun 18 '15 at 12:34