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):

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

  • $\begingroup$ what are the other limits in your book? $\endgroup$ – Math-fun Jun 18 '15 at 10:18
  • $\begingroup$ 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? $\endgroup$ – Urgje Jun 18 '15 at 10:34
  • $\begingroup$ @Math-fun Added the answer in the book $\endgroup$ – JaVaPG Jun 18 '15 at 10:42
  • $\begingroup$ @Urgje You'r right typo, edited. I also added the answer in the book, $\endgroup$ – JaVaPG Jun 18 '15 at 10:43
  • 1
    $\begingroup$ Then it is a little mystery. Maybe @Math-fun can shed some light on this matter. $\endgroup$ – Urgje Jun 18 '15 at 11:05

In both approaches, first the inner integral is evaluated and then the outer. Since there is no dispute about the outer integral, we look at how the inner integrals are done.

Let look at your approach first. First you look at the inner integral and write \begin{align} \int_0^1 \frac{u}{\sqrt{u^2+4v}} du&=\frac12\int_{4v}^{1+4v} \frac{1}{\sqrt{t}} dt\\ &=\sqrt{1+4v}-2\sqrt{v} \end{align} Now let look at how your book handle the inner integral: \begin{align} \int_0^1 \frac{u}{\sqrt{u^2+4v}} du&=\sqrt{u^2+4v}\Big|_{u=0}^{u=1}\\ &=\sqrt{1+4v}-2\sqrt{v} \end{align}

Therefore both are identical.

  • $\begingroup$ 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 $t$. $\endgroup$ – JaVaPG Jun 18 '15 at 12:16
  • $\begingroup$ when we change the integration variable, we can go back to the original variable once we finish the integration in the indefinite integrals, for the definite integrals we do not generally go back to the original variable. We change variables to make the calculation easier, hence going back would make the effort of change the variable just useless. $\endgroup$ – Math-fun Jun 18 '15 at 12:23
  • $\begingroup$ 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? $\endgroup$ – JaVaPG Jun 18 '15 at 12:34
  • $\begingroup$ exactly, cheers :-) $\endgroup$ – Math-fun Jun 18 '15 at 12:35

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