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  1. #1
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    antiderivative of ...

    I'm trying to show the calculation of the area under the graph of semicircle f(x) = - (4-x^2)^1/2 from [-2,2] by integration.

    I understand the calculation can be done by taking antiderivative F(2) - F (-2), but I can't remember how to find F for this type.

    Thanks!
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  2. #2
    Super Member 11rdc11's Avatar
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    Do you mean how to setup the integral?

    2 \int_0^2 \sqrt{4-x^2}~dx

    or how to actually integrate it?
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  3. #3
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    Sorry, yes how to integrate it. I remember how to integrate simple polynomials, but not square roots with exponents.
    Last edited by 2clients; Sep 7th 2008 at 09:54 PM.
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  4. #4
    Super Member 11rdc11's Avatar
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    2 \int_0^2 \sqrt{4-x^2}~dx

    = 2 \int_0^2 2\sqrt{1-\bigg(\frac{x}{2}\bigg)^2}

    = 4 \int_0^2 \sqrt{1-\bigg(\frac{x}{2}\bigg)^2}


    Ok now we need some substitutions and a trig identity.

    1- sin^2(u) = cos^2(u)

    sin(u) = \frac{x}{2}

    dx = 2cos(u)

    Also have to change upper and lower bounds

    new lower bound

    \frac{0}{2} = sin(u)

    new upper bound

    \frac{2}{2} = sin(u)

    so now with the substitutions

    8 \int_0^{\frac{\pi}{2}} cos^2(u)~du = 8 \int_0^{\frac{\pi}{2}} \bigg(\frac{1 + cos(2u)}{2}\bigg)~du

    = 4u + 2sin(2u)\bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{2}
    Last edited by 11rdc11; Sep 7th 2008 at 10:52 PM.
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  5. #5
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    Sub x = 2\sin{\theta}
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  6. #6
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    Quote Originally Posted by 11rdc11 View Post
    2 \int_0^2 \sqrt{4-x^2}~dx

    = 2 \int_0^2 2\sqrt{1-\bigg(\frac{x}{2}\bigg)^2}

    = 4 \int_0^2 \sqrt{1-\bigg(\frac{x}{2}\bigg)^2}


    Understand so far?
    Nope, sorry. I'm so rusty that I would really be grateful for some sentences. I can integrate polynomials no problem, but that's about it.
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  7. #7
    Super Member 11rdc11's Avatar
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    What part do you need me to explain? Feel free to ask questions.
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  8. #8
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    Quote Originally Posted by 2clients View Post
    Nope, sorry. I'm so rusty that I would really be grateful for some sentences. I can integrate polynomials no problem, but that's about it.
    If you sub x = 2\sin{\theta}, dx = 2\cos{\theta} d\theta, then;

    4 \int 2\cos{\theta}\sqrt{4-4\sin^2{\theta}}~d\theta

    16 \int \cos^2{\theta}~d\theta

    And use the half angle identity \cos^2{\theta} = \frac{1}{2}(1+\cos{2\theta}). Problem solved.
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  9. #9
    Super Member 11rdc11's Avatar
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    Oops didn't think about that, guess that what happens when your sleepy
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