The question:
Evaluate the following by inspection:
$\displaystyle \int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0$
I'm stumped. Substitution doesn't seem to work, and I have no idea what 'by inspection' means. Any help would be greatly appreciated!
The question:
Evaluate the following by inspection:
$\displaystyle \int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0$
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 $\displaystyle u = a^3-x^3~\implies~\dfrac{du}{dx}=-3x^2~\implies~du = -3x^2 \cdot dx$
Now re-write the integral to
$\displaystyle -\dfrac13 \cdot \int_{-a}^a\left(-3x^2 \cdot \sqrt{a^3-x^3} \right)dx~\implies~-\dfrac13 \cdot \int \sqrt{u} du$
I'll leave the rest for you.