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Math Help - Integral with square root

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    Integral with square root

    Evaluate \int_{-8}^8\sqrt{64-x^2}\,dx
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by liyi View Post
    Evaluate \int_{-8}^8\sqrt{64-x^2}\,dx
    the easy way: you could realize that this integral gives you the area of the upper half of the circle with radius 8

    the hard way: use the trig substitution x = 8 \sin \theta ....since we have an even function here, you may also wish to change the problem to 2 \int_0^8 \sqrt{64 - x^2}~dx, and change the limits based on the trig substitution
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