Evaluating $\frac{d}{dx}\int_3^{x^2}e^{t^3}dt$

Question

$\frac{d}{dx}\int_3^{x^2}e^{t^3}dt$

I suppose I don't fully understand the notation used within this problem.

Using the second fundamental theorem of calculus: $\int_a^b f(x)dx = F(x)\bigr|_a^b = F(b)-F(a)$ I am able to evaluate a definite integral, however, I don't understand how $x^2$ can be applied to a function $f(x)=e^{t^3}$.

The function doesn't have any values of x, so far as I can tell.

A further point of confusion, were the function to be: $f(x)=e^{x^3}$, it would evaluate to this.

I don't recall reviewing any material on the $\Gamma$ function, and really have no understanding of it.

I'm not exactly sure what to make of this problem.

Answer 1

Indeed, $\int_a^b f(x)dx = F(x)\bigr|_a^b = F(b)-F(a)$.

Applied to your problem, this reads as $\int_3^{x^2}e^{t^3}dt=F(t)\bigr|_{t=0}^{t=x^2}=F(x^2)-F(3)$, where $F$ is a fairly complicated function such that $\frac {dF(t)}{dt}=f(t)=e^{t^3}$.

Take the derivative, $\frac {d}{dx}(F(x^2)-F(3))=\frac{dF(x^2)}{dx}=\frac{d(x^2)}{dx}.\frac{dF(x^2)}{d(x^2)}=2x.f(x^2)=2x.e^{(x^2)^3}$.