Question

solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt

Answer 1

derivative of the integral is the integrand, plus chain rule
replace \(t\) by \(x^2\) and multiply by the derivative of \(x^2\)
get
\[\sin(x^2)\sqrt{x^2}\times 2x\] for the top part
bottom part is the same, but make it negative

Answer 2

I think I am suppose to break it up. integral from sqrt x to 0 plus 0 to x^2. But i have no idea why this is the way to do it

Answer 3

yes but you can break up anyway you like

Answer 4

the top part gives what i wrote above, the bottom part is analogous, just make it negative

Answer 5

the bottom can be any number??

Answer 6

I know I'm suppose to use the second fundamental theorem

Answer 7

yes
\[\int_{\sqrt{x}}^{x^2}f(t)dt=\int_{\sqrt{x}}^af(t)dt+\int_a^{x^2}f(t)dt\]

Answer 8

My teacher chose 0 though. Is there a reason for that? should I just choose 0 everytime?

Answer 9

take the derivative of the second part, get \(f(x^2)\times 2x\)

Answer 10

the derivative of the first part is \(-f(\sqrt{x})\times \frac{1}{2\sqrt{x}}\)

Answer 11

if you like to pick zero, go ahead a pick zero
makes no difference

Answer 12

Ok I understand it now. Thanks!!!! :)