I am given the equation y= x(16-x^2)^.5

and need to find the area bounded between the equation and the x-axis between (-4,0) and (4,0).

I do not know how to set up an integral for this problem.

I suspect a u substitution where

u = 16-x^2 and du = -2x dx which would give:

(-1/2)u^(1/2) du = (-1/2)(2/3)u^(3/2)

but this yields: (-1/3)(16-x^2)^(3/2) which, when I enter -4 and 4 for x, results in zero.

I'm missing some step, apparently. Could someone please explain?