# evaluate the integral: $\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$

$$\int_0 ^\sqrt5 \frac{4x}{\sqrt{x^2+4}}dx$$

I got $4(5^\frac{1}{4})$ but I'm not sure if thats right

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(i) Please make the question self-contained; don't rely on the "title", anymore than a book relies on the spine to begin the narrative. (ii) If you show your work, rather than just your result, there might be more of a chance of telling you where, if anywhere, you went wrong. – Arturo Magidin Jul 26 '12 at 3:28
I just put the question in the title because previously I've had people edit my questions to where they put the question in the title for me. So I just thought that's what I was supposed to do. – user69 Jul 26 '12 at 3:33
Based on my calculation, it is 4. – Lei Hao Jul 26 '12 at 3:38
I'm not saying don't put the question in the title (the title should be clear and informative, not generic); I'm saying, don't put the question just in the title. The only place in your post that said you wanted to evaluate the integral was in the title. If someone starts reading your post at the first character of the post, they should get the whole information without having to start reading the post at the title. Just like a novel starts at the first letter of the first page, not at the spine. – Arturo Magidin Jul 26 '12 at 3:56

The substitution $u=x^2+4$ gives $du = 2x\,dx$. When $x=0$, we have $u=4$; when $x=\sqrt{5}$, we get $u=9$. So \begin{align*} \int_0^{\sqrt{5}}\frac{4x}{\sqrt{x^2+4}}\,dx &= 2\int_0^{\sqrt{5}}\frac{2x\,dx}{\sqrt{x^2+4}}\\ &= 2\int_4^9\frac{du}{\sqrt{u}}\\ &= 2 \int_4^9 u^{-1/2}\,du\\ &= 4u^{1/2}\Bigm|_4^9\\ &= 4(9^{1/2} - 4^{1/2})\\ &= 4(3-2)\\ &= 4. \end{align*}

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oh, I forgot to get the new limits for u! :] – user69 Jul 26 '12 at 3:32
@Kudla69: And we could have told you as much if you had shown your work instead of just timidly putting forth your answer. – Arturo Magidin Jul 26 '12 at 3:55

There are two approaches to the definite integral when you use substitution.

$1$. Use substitution as usual, in this case $u=x^2+4$ (or maybe better $u^2=x^2+4$), to find that one indefinite integral is $4u^{1/2}$. Then replace $u$ by $x^2+4$ to find that $4(x^2+4)^{1/2}$ is an indefinite integral of your original function, and "plug in" as usual.

$2$. Use substitution, but substitute for everything. So our original integral is $$\int_{x=0}^{\sqrt{5}}\frac{4x\,dx}{\sqrt{x^2+4}}.$$ Note that I put in the $x$ in the limits of integration to remind myself that there is an $x$ hidden there.

Substitute, remembering to substitute for everything. So we get $$\int_{u=2}^3 2u^{-1/2}\,du.$$ Now the substitution has been fully done. Integrate. We get $4u^{1/2}$. Do the plugging in. We get $4$.

In general I prefer the second approach. Use one or the other, and not, as you did, a hybrid of both.

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Since it's an elementary integral, I'd solve it directly:

$$I=4\int_{0}^{\sqrt{5}} \frac{x}{\sqrt{4+x^2}} dx=4\int_{0}^{\sqrt{5}} \left({\sqrt{4+x^2}}\right)' dx = 4 \sqrt{4+x^2}\Bigm|_0^{\sqrt{5}}=4.$$

Q.E.D.

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