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Find the derivative of the following integral

$$ F(x)=\int_x^{x^2}e^{t^7}dt $$

Find F′(x) given F(x).

Normally I would show my attempt in working out the problem: however, I don't even know where to start with this question. I am thrown of by the variables acting as the lower and upper boundaries of the integral.

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    $\begingroup$ Wouldn't the result of the integral be in the form $F(a)-F(b)$ for endpoints $a,b$? Perhaps there is a chain rule that could apply? $\endgroup$ – abiessu Jan 26 '15 at 1:55
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Let $$U(t) = \int_{0}^{x} e^{t^7}dt$$ $$F(x) = U(x^2) - U(x)$$ $$F'(x) = (2x)U'(x^2) - (1)U'(x) = 2xU'(x^2) - U'(x)$$ Of course, $U'(t) = e^{t^7}$ so $$F'(x) = 2xe^{x^{14}}-e^{x^7}$$

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  • $\begingroup$ Did you get 2x from the derivative of $x^2$? And if so, is there suppose to be a hidden 1 in $e^{x^7}$ which comes from differentiating x? $\endgroup$ – amundi12 Jan 26 '15 at 2:12
  • $\begingroup$ Yes and yes. Updated to be more clear, thanks. $\endgroup$ – charlotte Jan 26 '15 at 2:16

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