Given the Leibniz rule:

$$\frac{d}{dy}\int^a_b f(x,y) dx = \int_b^a \frac{\partial f}{\partial y} (x,y) dx$$

How do I prove a more general case using the chain rule and the above:

$$\frac{d}{dy} \int_{g_1(y)}^{g_2(y)} f(x,y) dx =?$$

From the fundamental theorem of calculus we have that: $$\frac{d}{dt} \int_{f_2(t)}^{f_1(t)} g(s) ds = g(f_1(t))f'_1(t) - g(f_2(t))f'_2(t)$$

But when I just apply the fundamental theorem of calculus I get an incorrect answer (because I also need to use Leibniz rule itself...).


Do the linear change of variables $$x = g_1(y)+(g_2(y)-g_1(y))u.$$

The integral becomes $$\int_{g_1(y)}^{g_2(y)} f(x,y) dx =(g_2(y)-g_1(y)) \int_0^1 f(g_1(y)+(g_2(y)-g_1(y))u,y) du$$ and now you can apply the traditionnal 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.