1
$\begingroup$

I'm trying to solve the following question:

Evaluate $$\frac{\mathrm{d} }{\mathrm{d} s} \int^s_0 e^{st^2} dt $$

My thinking was that by the fundamental theorem of calculus, we have $ F(s) = \int^s_0 e^{st^2} dt $ and thus $ \frac{\mathrm{d} }{\mathrm{d} s} F(s) = e^{s^3} $ however the solution suggests calculating $ e^{s^3} + \int^s_0 \frac{\partial }{\partial s} e^{st^2}$.

What is the intuition here?

Improve this question
$\endgroup$
  • 2
    $\begingroup$ There is no $x$ in your integral so you probably don't mean $\frac {\mathrm d}{\mathrm dx}$ $\endgroup$ – GFauxPas May 15 '15 at 14:42
  • $\begingroup$ I took the liberty of interpreting it as $\frac{d}{ds}$. $\endgroup$ – Demosthene May 15 '15 at 14:51
  • $\begingroup$ You're right, my mistake. It should be $ \frac{\mathrm{d} }{\mathrm{d} s} $. $\endgroup$ – kw3rti May 15 '15 at 20:38
2
$\begingroup$

The rule for differentiation under the integral sign is: $$\dfrac{d}{dx}\left(\int_{a(x)}^{b(x)}f(x,t)dt\right)=f(x,b(x))\cdot b'(x)-f(x,a(x))\cdot a'(x)+\int_{a(x)}^{b(x)}f_x(x,t)dt$$ Thus: $$\dfrac{d}{ds}\left(\int_0^se^{st^2}dt\right)=e^{s\cdot s^2}\cdot 1-e^{s\cdot 0^2}\cdot 0+\int_0^s\dfrac{\partial}{\partial s}e^{st^2}dt$$ which leads to the given solution.

Improve this answer
$\endgroup$
  • $\begingroup$ Could you expand on why this is the case? $\endgroup$ – kw3rti May 15 '15 at 20:35
  • $\begingroup$ @kw3rti The Wikipedia link contains a very nice proof of this rule, derived directly from the fundamental theorem of calculus. $\endgroup$ – Demosthene May 16 '15 at 9:06

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.