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?
Be careful. Usually when asked to find the area like in this problem, you are really calculating over the interval, so instead what we are integrating is an even function.
But, we can assume that the area left of the y-axis and right of the y-axis are equal, and it's clearly positive everywhere to the right of the y-axis, so:
Using this will give you a non-zero solution.
If you didn't recognize this, you still have to break the integral into two separate pieces however, because the function is negative left of the y-axis. Like so:
(Both methods should give the same result.)