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Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$

This is what I did:

$u = x^2 + a^2 $

$du/dx = 2x$

$du = 2xdx$

$1/2 du = x dx$

$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$ from $0$ to $a$.

$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.

I eventually got:

$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$

but this was incorrect.

The correct answer was:

$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$

Any help?

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    $\begingroup$ You didn't change the bounds of integral after replacing $u=x^2+a^2$. The new bounds are $a^2$ and $2a^2$ then ur answer would be correct $\endgroup$ – Mostafa Ayaz Jan 26 '18 at 6:48
  • $\begingroup$ Yes it seems like you are right. Please excuse my ignorance but when do we change the bounds? Why do we change the bounds here? Thank you. $\endgroup$ – sktsasus Jan 26 '18 at 6:52
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    $\begingroup$ That always happens when jumping from one variable to another. This makes sense since a replacement like this changes the function (compresses, expands or disperses it at general case) so would be the bounds. $\endgroup$ – Mostafa Ayaz Jan 26 '18 at 6:55
  • $\begingroup$ OK I understand. Thank you. $\endgroup$ – sktsasus Jan 26 '18 at 6:57
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    $\begingroup$ My guess is that through sloppy writing, the "a" became a "9". $\endgroup$ – Mike Jan 26 '18 at 10:35
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Hint

You made mistakes here:

$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}\big |_0^a=\frac 1 3 (2a^2)^{\frac 3 2}-\frac {a^3}{3}=\frac 1 3 (\sqrt{2^3}|a|^3)-\frac {a^3}{3}=....$$

You were almost done...

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    $\begingroup$ Why $a^3$ and not $|a|^3$? $\endgroup$ – Rodrigo de Azevedo Jan 26 '18 at 10:30
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You calculated the antiderivative correctly.

$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$$ from $0$ to $a$ is $$\frac{1}{3}\cdot \left(a^2+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\cdot \left(0^2+a^2\right)^{\frac{3}{2}}=\\\frac{1}{3}\cdot((2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2})=\frac{1}{3}\cdot(2^\frac{3}{2}a^3-a^3)=\frac{1}{3}(2^\frac{3}{2}-1)a^3$$

Which explains the answer in your book. I don't know how you got $$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$$ from the previous step.

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  • $\begingroup$ Sloppy writing perhaps? The "a" in the bounds becomes a "9"? $\endgroup$ – Mike Jan 26 '18 at 10:30
  • $\begingroup$ @Mike Hmm, potentially. I have been guilty of worse sloppy writing offenses than turning 'a's into '9's. :) $\endgroup$ – Lug Gian Jan 26 '18 at 21:24
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You can simply use the reverse chain rule for integration and you get: $$\int_0^a x\sqrt{x^2+a^2}dx=\frac 12\int_0^a 2x\sqrt{x^2+a^2}dx=\left[\frac{(x^2+a^2)^\frac 32}3\right]_0^a=\\=\frac{(2a^2)^\frac 32}3-\frac{(a^2)^\frac 32}3=\color{red}{\frac{(2\sqrt 2-1)}3|a|^3}$$

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  • $\begingroup$ Why $a^3$ and not $|a|^3$? $\endgroup$ – Rodrigo de Azevedo Jan 26 '18 at 10:30
  • $\begingroup$ Because we are integrating between 0 and $a\ge 0$ $\endgroup$ – user507623 Jan 26 '18 at 10:31
  • $\begingroup$ Where is $a \geq 0$ stated? $\endgroup$ – Rodrigo de Azevedo Jan 26 '18 at 10:31
  • $\begingroup$ Ok, you’re right. @RodrigodeAzevedo $\endgroup$ – user507623 Jan 26 '18 at 10:33

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