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$$\int_0^{y^2}e^{-xy}dx$$

The Leibnitz integral rule states that: $$\frac{d}{dy}\int_{x_1}^{x_2} f(x,y)dx=\int_{x_1}^{x_2} (\frac{\partial f}{\partial y})dx$$

However I can't seem to get this to help me. I find that I have to integrate by parts which gives me the original problem back.

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  • $\begingroup$ why you have to use this rule? you can simply integrate it $\endgroup$ – Kamil Jarosz Nov 26 '15 at 18:34
  • $\begingroup$ I was asked to in my problem sheet $\endgroup$ – RobChem Nov 26 '15 at 18:35
  • $\begingroup$ the result is $\frac{e^{y^3}-1}{y}$ $\endgroup$ – Dr. Sonnhard Graubner Nov 26 '15 at 18:37
  • $\begingroup$ It is not clear to me how the Leibnitz rule helps here. Sometimes it results in a simple ODE. Incidentally. you need a slightly different rule here since the bounds are functions of $y$. $\endgroup$ – copper.hat Nov 26 '15 at 18:43
  • $\begingroup$ The Leibniz integral rule is for differentiating an integral with respect to a variable that appears both in the limits of integration and the integrand. This is not the situation you have. $\endgroup$ – JohnD Nov 26 '15 at 18:52
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As mentioned in the comment, the Leibniz integral rule is not for this situation. Instead, just integrate it outright:

$$ \int_0^{y^2} e^{-xy}\,dx={e^{-xy}\over -y}\Bigg|_{x=0}^{x=y^2}={e^{-y^3}\over -y}-{1\over -y}={1-e^{-y^3}\over y}.$$


Perhaps you were asked to find ${d\over dy}\int_0^{y^2} e^{-xy}\,dx$? Now that would summon the Leibniz rule...

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