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?

  • 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

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.

  • $\begingroup$ 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

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