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Math Help - Sum of two integrals

  1. #1
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    Sum of two integrals

    I'm confused here...I'm guessing it's just using one that's above the x axis and subtracting the one that's below the x axis but I'm not too sure and the function is throwing me off.

    Evaluate (x+3)sqrt[4-x^2] from -2 to 2 by writing it as a sum of two integrals and interpreting one of those in terms of an area.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sfgiants13 View Post
    I'm confused here...I'm guessing it's just using one that's above the x axis and subtracting the one that's below the x axis but I'm not too sure and the function is throwing me off.

    Evaluate (x+3)sqrt[4-x^2] from -2 to 2 by writing it as a sum of two integrals and interpreting one of those in terms of an area.
    ok, to begin: just expand, right?

    (x + 3) \sqrt{4 - x^2} = x \sqrt{4 - x^2} + 3 \sqrt{4 - x^2}

    further note that y = \sqrt{4 - x^2} is the equation for the upper half of a circle
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    Quote Originally Posted by Jhevon View Post
    ok, to begin: just expand, right?

    (x + 3) \sqrt{4 - x^2} = x \sqrt{4 - x^2} + 3 \sqrt{4 - x^2}

    further note that y = \sqrt{4 - x^2} is the equation for the upper half of a circle
    So it would be the derivatives of the 2nd equation subtracted from the 1st correct?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sfgiants13 View Post
    So it would be the derivatives of the 2nd equation subtracted from the 1st correct?
    no, we're adding. there is a + sign between them

    to do the problem as they said, you will need to use integration by parts for the first integral. otherwise, a simple substitution of u = 4 - x^2 would have finished this problem nicely.
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  5. #5
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    Ok thanks I figured it out...after getting a nice messy equation I got 6pi.
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