## anonymous 5 years ago From my calc book: "Since the integrand is an odd function, integral_(-2)^(2) (x^3)*(16-x^2)^1/2 dx = 0" My question: Why?!

1. anonymous

$\int\limits_{-2}^{2}x^3\sqrt{16-x^2}dx = 0$ <-----A little easier to view

2. anonymous

Also, I understand the trig substitution for the indefinite integral but I don't know why this definite integral is 0 and what an odd function is referring to.

3. anonymous

odd function is because the highest degree of x is 3

4. anonymous

and it is equal to 0 since the are from -2 to 0 "cancels" out the area from 0 to 2

5. anonymous

Will this usually happen for odd functions where it will "cancel" out?

6. anonymous

no unless it's an improper integral