# Fundamental Theorem of Calculus With Function Containing Limit Variable

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?

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