Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

All I know is that it uses the fundamental theorem of calculus.

$$\large\frac{d}{dx}\int_{x^2}^{\sin x} e^{xt^2}dt = e^{x\;\sin^2 x}\cos x - e^{x^5}2x+\int_{x^2}^{\sin x} t^2e^{xt^2}dt$$

share|cite|improve this question
Zev Chonoles, thanks for the edit. I have to learn how to write those formulas. – Dokkat Jan 16 '12 at 9:51

This is more an application of differentiation under the integral sign, which is a generalization of the fundamental theorem (and can be proved using it). $$ \frac{d}{dx}\,\int_{a(x)}^{b(x)}f(x,t)\,dt = f(x,b(x))\,b'(x) - f(x,a(x))\,a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}\, f(x,t)\; dt $$

In this case, $a(x) = x^2$, $b(x) = \sin(x)$, and $f(x, t) = e^{x t^2}$. $$ a'(x) = 2x $$ $$ b'(x) = \cos(x) $$ $$ \frac{\partial}{\partial x} f(x, t) = t^2 e^{x t^2} $$ So $$ \large\frac{d}{dx}\int_{x^2}^{\sin x} e^{xt^2}dt = e^{x\;\sin^2 x}\cos x - 2xe^{x^5}+\int_{x^2}^{\sin x} t^2e^{xt^2}dt $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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