# Math Help - How does one find area using definite integrals?

1. ## How does one find area using definite integrals?

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?

2. Originally Posted by Quixotic
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.

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, I end up with zero.

I'm missing some step, apparently. Could someone please explain?
Are you sure that the equation is

$y = x(16 - x^2)^{\frac{1}{2}}$

and not

$y = (16 - x^2)^{\frac{1}{2}}$?

3. $y = x(16 - x^2)^{\frac{1}{2}}$

This is the equation I am given. Sorry, didn't know how to write it in this cleaner way.

4. Originally Posted by Quixotic
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?
$y=x$ is an odd function
$y=\sqrt{16-x^2}$ is an even function
The product of an odd function and an even function is an odd function
Since an odd function is symmetric with respect to the origin,
$\int_{-a}^af(x)\,dx=0$ because the area above the x-axis and the area below the x-axis is the same.

5. Be careful. Usually when asked to find the area like in this problem, you are really calculating $\int |f(x)| dx$ 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:

$A~=~2 \int_0^4 x\sqrt{16-x^2} \, dx$

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:

$A ~=~ \int_{-4}^0 (-x)\sqrt{16-x^2} \, dx + \int_0^4 x\sqrt{16-x^2} \, dx$

(Both methods should give the same result.)