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Math Help - Another integration by inspection question

  1. #1
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    Another integration by inspection question

    The question:
    Evaluate the following by inspection:

    \int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0

    I'm stumped. Substitution doesn't seem to work, and I have no idea what 'by inspection' means. Any help would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by Glitch View Post
    The question:
    Evaluate the following by inspection:

    \int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0

    I'm stumped. Substitution doesn't seem to work, and I have no idea what 'by inspection' means. Any help would be greatly appreciated!

    Hint: \int f'(x)\sqrt{f(x)} dx= \frac{2}{3}f(x)^{3/2}+C\,,\,C= constant , so...

    Tonio
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  3. #3
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    I'm failing to see how you've replaced x^2 with f'(x) and made f(x) = a^3 - x^3. It doesn't look like its derivative to me. :/
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  4. #4
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    Quote Originally Posted by Glitch View Post
    The question:
    Evaluate the following by inspection:

    \int_{-a}^a{x^2\sqrt{a^3-x^3}dx}, a > 0

    I'm stumped. Substitution doesn't seem to work (?), and I have no idea what 'by inspection' means. Any help would be greatly appreciated!
    Let u = a^3-x^3~\implies~\dfrac{du}{dx}=-3x^2~\implies~du = -3x^2 \cdot dx

    Now re-write the integral to

    -\dfrac13 \cdot \int_{-a}^a\left(-3x^2 \cdot \sqrt{a^3-x^3} \right)dx~\implies~-\dfrac13 \cdot \int \sqrt{u} du

    I'll leave the rest for you.
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  5. #5
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    Oh wow, I actually had that but didn't think to divide by 3. >_<

    Let's see if I can complete it. Thank you.
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  6. #6
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    Quote Originally Posted by Glitch View Post
    I'm failing to see how you've replaced x^2 with f'(x) and made f(x) = a^3 - x^3. It doesn't look like its derivative to me. :/

    But x^2=-\frac{1}{3}\cdot (-3x^2) ...!

    Tonio
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  7. #7
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    Sorry dude, my brain is half dead from coding a parallel Sudoku solver all week. D:
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