Radical integral question calculus I have a question in calculus.
Let $F(x)=\int_1^\sqrt{x} t^2\cos( \pi t)dt$ Find $F'(4)$
I know $F'(X$)  $=\int_1^\sqrt{x} x^2\cos( \pi x)dx$ 
So I made $u=x^{\frac{1}{2}}$
and I got 
$u^2\cos\pi(u)$
which is  
$F'(x)=x^{1/4}\cdot \left(\frac{1}{2\sqrt{x}}\right)\cdot \cos\pi\sqrt{x}$
when I plugged in four I got 
$F'(4)=\sqrt{2}\frac{1}{\sqrt{2}}$
but did I do this correctly?
 A: Your post is very confusing: but I think I finally follow:
Note we have substituted $u = \sqrt x$ and $du = \dfrac 1{2\sqrt x  }dx$
giving us $u^2\cos\pi(u)\,du$
So your expression for  $F'(x)$ is off and needs to be, no in terms of $x$: $$F'(x)=(\sqrt x)^2\cdot  \cos\pi\sqrt{x}\cdot \left(\frac{1}{2\sqrt{x}}\right)$$
Now evalate $F'(4)$:
The procedure is a little mixed up and the exposition was a little confusing.
A: The integral bound is from 1 to 2 because, if you have gotten to u-substitution, u becomes sqrt(x) and you evaluate u(4)=sqrt(4)=2 and then use integration by parts. Here is a link to integration by parts: 
http://tutorial.math.lamar.edu/Classes/CalcII/IntegrationByParts.aspx
And the derivative should be x*cos(pi*sqrt(x)) because of Leibniz's rule.
A: This is an indirect application of the Fundamental Theorem of Calculus which is quickly passed through in many introductory texts.  You have an integral function  $F(x) = \int^{x}_{1} f(t) dt$; since $t^2 \cos(\pi t )$ is continuous everywhere, we can safely say that  $F'(x) = f(x)$ .  
When the upper limit is a function of $x$, the FTC will let us write $\int^{u}_{1} f(t) dt  =  F(u) - F(1)$.  When we differentiate this with respect to $x$, the Chain Rule gives us  $$\frac{d}{dx}\int^{u}_{1} f(t) dt  =  \frac{d}{dx} [F(u) - F(1)]  =  \frac{dF}{du} \cdot \frac{du}{dx}  = f(u(x)) \cdot \frac{du}{dx}$$
So for this problem, 
$$\frac{d}{dx}\int^{\sqrt{x}}_{1} t^2 \cos(\pi t ) dt    = (\sqrt{x})^2 \cos(\pi \sqrt{x} )) \cdot \frac{d}{dx}(\sqrt{x}).$$
To answer your question, you would complete the differentiation and evaluate the result at  x = 4 .
(I'll mention, incidentally, that a problem of this type is a favorite final exam question.)
