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.

  • $\begingroup$ 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

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...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.