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justinfung Group Title

solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt

  • one year ago
  • one year ago

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  1. satellite73 Group Title
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    derivative of the integral is the integrand, plus chain rule replace \(t\) by \(x^2\) and multiply by the derivative of \(x^2\) get \[\sin(x^2)\sqrt{x^2}\times 2x\] for the top part bottom part is the same, but make it negative

    • one year ago
  2. justinfung Group Title
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    I think I am suppose to break it up. integral from sqrt x to 0 plus 0 to x^2. But i have no idea why this is the way to do it

    • one year ago
  3. satellite73 Group Title
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    yes but you can break up anyway you like

    • one year ago
  4. satellite73 Group Title
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    the top part gives what i wrote above, the bottom part is analogous, just make it negative

    • one year ago
  5. justinfung Group Title
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    the bottom can be any number??

    • one year ago
  6. justinfung Group Title
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    I know I'm suppose to use the second fundamental theorem

    • one year ago
  7. satellite73 Group Title
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    yes \[\int_{\sqrt{x}}^{x^2}f(t)dt=\int_{\sqrt{x}}^af(t)dt+\int_a^{x^2}f(t)dt\]

    • one year ago
  8. justinfung Group Title
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    My teacher chose 0 though. Is there a reason for that? should I just choose 0 everytime?

    • one year ago
  9. satellite73 Group Title
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    take the derivative of the second part, get \(f(x^2)\times 2x\)

    • one year ago
  10. satellite73 Group Title
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    the derivative of the first part is \(-f(\sqrt{x})\times \frac{1}{2\sqrt{x}}\)

    • one year ago
  11. satellite73 Group Title
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    if you like to pick zero, go ahead a pick zero makes no difference

    • one year ago
  12. justinfung Group Title
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    Ok I understand it now. Thanks!!!! :)

    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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