AP Calculus Fundamental Theorem of Calculus Please help me go over this problem; I am a bit confused.
Find ${\displaystyle \frac{\mathrm d}{\mathrm dt} \int_2^{x^2}e^{x^3}\mathrm dx}$.
 A: The Fundamental Theorem of Calculus states that if
$$g(x) = \int_{a}^{f(x)} h(t)~{\rm d}t$$
where $a$ is any constant, then
$$g'(x) = h(f(x)) \cdot f'(x)$$
Using this with the integral, $g(x) = g(x)$, $f(x) = x^2$, and $h(x) = e^{x^3}$.
So, $$\frac{\mathrm d}{\mathrm dx} \int_2^{x^2}e^{x^3}\mathrm dx=(2x)e^{x^6}$$
A: First off, I'm assuming that you want to evaluate this:
$$
\frac{d}{dx} \int_2^{x^2}e^{t^3}dt
$$
You want to find the derivative of the integral, so we replace the $x$ inside the integral with a dummy variable to avoid confusion. Now, let's set $f(x)=e^{x^3}$ and say that $F(x)$ is the antiderivative, so $F'(x)=f(x)=e^{x^3}$. We don't know what $F(x)$ is yet, but as it turns out it doesn't actually matter. By the second part of the Fundamental Theorem of Calculus,
$$
\int_a^bf(x)dx = F(b)-F(a)
$$
and in our case $a=2, b=x^2, and f(x)=e^{x^3}$. Now we want to find the derivative of this, so we differentiate the right hand side:
$$
\frac{d}{dx}[F(b)-F(a)]
$$
Applying the chain rule, this becomes:
$$
F'(b)\cdot b'-F'(a)\cdot a'
$$
But wait! We already defined $F'(x)=f(x)$! So this simplifies down to:
$$
f(b)\cdot b' -f(a)\cdot a'
$$
Plugging in our values, and noting that the derivative of 2 is zero, we get:
$$
\frac{d}{dx} \int_2^{x^2}e^{t^3}dt=e^{(x^2)^3}\cdot 2x=2xe^{x^6}
$$
