applying first part fundamental theorem of calc problem I was advised to use this site for help. I am a freshman who has an exam in two days and I am flipping out because I have no idea how to do the problem below. If someone could help me out and show me how to do this problem, you would make my day. Thank you.

 A: This is the Fundamental Theorem of Calculus at work. You could integrate, plug in your upper and lower bounds, and then differentiate. But notice if you integrate, you get have $F(5/x)-F(\sqrt{x})$, where $F$ is the antiderivative. So you're then taking the derivative of a function of $x$. Not only that, you're essentially undoing the integral! 
So the idea is this: why even do the integral? But you have to be a bit careful because undoing the integral isn't quite the same thing as taking the derivative after integrating because you're missing the chain rule you would have had to do if you did the problem as written. So we can just plug in the upper and lower bounds of the integral into the integrand, but we also need to take account of the chain rule. We do this by multiplying by the derivative of the upper and lower limit, respectively. So let 
$$
f(x)=\frac{\sin t}{\sqrt{t+\sqrt{1-t}}}
$$
(simply so I can write less). Then the problem reduces down to 
$$
f(5/x)\bigg(\frac{5}{x}\bigg)'-f(\sqrt{x})\bigg(\sqrt{x}\bigg)'
$$
Where $(stuff)'$ means to take the derivative. So we have
$$
f(5/x)\bigg(\frac{-5}{x^2}\bigg)-f(\sqrt{x})\bigg(\frac{1}{2}x^{-1/2}\bigg)
$$
I'll leave it to you to plut $5/x$ and $\sqrt{x}$ into the function. Notice this is how we do these in general. So if we have 
$$
\frac{d}{dx} \int_{2}^{2x^2} \frac{1}{t-1}dt
$$
then the solution is 
$$
\frac{1}{2x^2-1}\cdot(4x)-\frac{1}{2-1}\cdot (0)
$$
because the derivative of $2$ is $0$. So we really have 
$$
\frac{1}{2x^2-1}\cdot(4x)=\frac{4x}{2x^2-1}
$$
I encourage you to see we get the same thing by actually doing the integral then taking the derivative to see why this works!
Good luck with your exam!
