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anonymous

  • 5 years ago

Sketch the regions enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. y = sqrt of x y = (1/3)x x=16

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  1. anonymous
    • 5 years ago
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    It doesn't matter, you can do it either way. Either way you end up splitting it up into 2 integrals. If you integrate with respect to x you would do: \[\int\limits_{0}^{9}(\sqrt(x)-x/3)dx+\int\limits_{9}^{16}(x/3-\sqrt(x))dx\] To find the place where you would split the integral up you need to set x/3 equal to sqrtx. Once you solve for x, the answers end up being 0 and 9. Pretty obvious how to split it up at that point.

  2. myininaya
    • 5 years ago
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  3. anonymous
    • 5 years ago
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    thank you guys both soo much :)

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