Derivative of Integrals FTC

[Alphonse]

The problem I'm trying to figure out is 8.4. I understand what's going on but the part I get stuck at is how do I write the function of the integral of e^{t^2} from x to 1 as a function of (sin(u)) over (u). I have to leave it as an integral because when plugging it in, it's not the derivative as apposed to applying chain rule.



Thanks so much for the help!

 

[HallsofIvy]

The "Fundamental Theorem of Calculus", FTC, that you cite, says that $\displaystyle \frac{d}{dx}\int_a^x f(t)dt= f(x)$. "Leibniz's formula" extends that:
$\displaystyle \frac{d}{dx}\int_{u(x)}^{v(x)} f(t,x) dt= \frac{dv}{dx}f(v(x),x)- \frac{du}{dx}f(u(x),x)+ \int_{u(x)}^{v(x)}\frac{\partial f(t,x)}{\partial x} dt$.

Here, $\displaystyle u(x)= \int_1^x e^{t^2}dt$, $\displaystyle v(x)= x^2$ and the integrand is not a function of x so that third term is 0. Yes, you will need to write $\displaystyle f(u(x), x)$ as $\displaystyle \frac{sin\left(\int_1^x e^{t^2}dt\right)}{\int_1^x e^{t^2}dt}$ which cannot be simplified.

 

[Alphonse]

Wow thanks man, I suppose it's simple enough. I was scared I'd might have to learn IBP .