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mr570

How do I find the integral from -4 to 4 of (3*sqrt(16-x^2)^2 dx ? I think you have to use u-substitution but I don't entirely understand how that works in this case.

  • one year ago
  • one year ago

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  1. satellite73
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    use nothing

    • one year ago
  2. Umangiasd
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    well the sqrt gets eliminated with the power 2, so...Its a simple integral

    • one year ago
  3. satellite73
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    \[y=\sqrt{16-x^2}\] is the upper half of a circle centered at the origin with radius 4 so \[6\int_{-4}^4\sqrt{16-x^2}dx\] is 6 times the area of the upper half of a circle with radius 4

    • one year ago
  4. satellite73
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    oh sorry i didn't see the square at the end if it is really the square root squared, then it is easy enough, although a rather strange way to write it

    • one year ago
  5. satellite73
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    it is really \[3\int_{-4}^4\sqrt{16-x^2}^2dx\]?

    • one year ago
  6. Dido525
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    Assuming it's what sattelite said then this becomes very simple.

    • one year ago
  7. Dido525
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    |dw:1354850130324:dw| Then this become easy to integrate.

    • one year ago
  8. RadEn
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    i think the typo from asker

    • one year ago
  9. Dido525
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    I think so too.

    • one year ago
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