Integral Question- Help would be appreciated! $\int_{1}^{x^2} t^2 \cos(\pi t)$ Just started talking about integration in calculus class.
I'm thrown off by this problem:
Let $$f(x) = \int_{1}^{x^2} t^2 \cos(\pi t)$$
Evaluate $F'(x)$ and $F'(\sqrt{3})$.
By the way, we just started integration so the only technique I know is u-sub.
What throws me off is there being two variables (why is $x^2$ in the upper boundary?)

$$f(x)=\int_{1}^{x^2} t^2 \cos(\pi t)\,\mathrm dt$$
Using Leibniz Integral Rule
 We have

$$\frac{\partial}{\partial x}\int_{a(x)}^{b(x)}p(t,x)\,\mathrm dt=\int_{a(x)}^{b(x)}\frac{\partial p}{\partial x}\,\mathrm dt+p(b(x),x)\frac{\partial b}{\partial x}-p(a(x),x)\frac{\partial a}{\partial x}$$

Now put $$p(t,x)=t^2 \cos(\pi t)$$
So,
$$\begin{align}f'(x)&=\int_{1}^{x^2} 0\,\mathrm dt + (x^2)^2 cos(\pi x^2)\cdot 2x- 1^2\cos(\pi\cdot 1)\cdot0\\
&=[0]_{1}^{x^2}+2x\cdot(x^2)^2 \cos(\pi x^2)-0\\
&=0+2x\cdot(x^2)^2 \cos(\pi x^2)\\
&=2x\cdot(x^2)^2 \cos(\pi x^2)\\
f'(x)&=2x^5\cos(\pi x^2)\\
\end{align}$$
and then

$$f'(\sqrt3)=-18\sqrt3$$

