# Thread: Another integration by inspection question

1. ## Another integration by inspection question

The question:
Evaluate the following by inspection:

$\int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0$

I'm stumped. Substitution doesn't seem to work, and I have no idea what 'by inspection' means. Any help would be greatly appreciated!

2. Originally Posted by Glitch
The question:
Evaluate the following by inspection:

$\int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0$

I'm stumped. Substitution doesn't seem to work, and I have no idea what 'by inspection' means. Any help would be greatly appreciated!

Hint: $\int f'(x)\sqrt{f(x)} dx= \frac{2}{3}f(x)^{3/2}+C\,,\,C=$ constant , so...

Tonio

3. I'm failing to see how you've replaced $x^2$ with f'(x) and made f(x) = $a^3 - x^3$. It doesn't look like its derivative to me. :/

4. Originally Posted by Glitch
The question:
Evaluate the following by inspection:

$\int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0$

I'm stumped. Substitution doesn't seem to work (?), and I have no idea what 'by inspection' means. Any help would be greatly appreciated!
Let $u = a^3-x^3~\implies~\dfrac{du}{dx}=-3x^2~\implies~du = -3x^2 \cdot dx$

Now re-write the integral to

$-\dfrac13 \cdot \int_{-a}^a\left(-3x^2 \cdot \sqrt{a^3-x^3} \right)dx~\implies~-\dfrac13 \cdot \int \sqrt{u} du$

I'll leave the rest for you.

5. Oh wow, I actually had that but didn't think to divide by 3. >_<

Let's see if I can complete it. Thank you.

6. Originally Posted by Glitch
I'm failing to see how you've replaced $x^2$ with f'(x) and made f(x) = $a^3 - x^3$. It doesn't look like its derivative to me. :/

But $x^2=-\frac{1}{3}\cdot (-3x^2)$ ...!

Tonio

7. Sorry dude, my brain is half dead from coding a parallel Sudoku solver all week. D: