Question

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?!

Answer 1

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

Answer 2

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.

Answer 3

odd function is because the highest degree of x is 3

Answer 4

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

Answer 5

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

Answer 6

no unless it's an improper integral