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anonymous

  • 5 years ago

From my calc book: "Since the integrand is an odd function, integral_(-2)^(2) (x^3)*(16-x^2)^1/2 dx = 0" My question: Why?!

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  1. anonymous
    • 5 years ago
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    \[\int\limits_{-2}^{2}x^3\sqrt{16-x^2}dx = 0\] <-----A little easier to view

  2. anonymous
    • 5 years ago
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    Also, I understand the trig substitution for the indefinite integral but I don't know why this definite integral is 0 and what an odd function is referring to.

  3. anonymous
    • 5 years ago
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    odd function is because the highest degree of x is 3

  4. anonymous
    • 5 years ago
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    and it is equal to 0 since the are from -2 to 0 "cancels" out the area from 0 to 2

  5. anonymous
    • 5 years ago
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    Will this usually happen for odd functions where it will "cancel" out?

  6. anonymous
    • 5 years ago
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    no unless it's an improper integral

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