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

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by Quixotic View Post
    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}}?
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by Quixotic View Post
    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.
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  5. #5
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    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.)
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