Given: f(x) = the integral from 0 to sin(x) of sqrt.(1 + t^2) dt
and g(y) = the integral from 3 to y of f(x) dx.
Find g''(pi/6)
I'm not sure how to put it all together since there are two different integrals involved.
Thanks!
Given: f(x) = the integral from 0 to sin(x) of sqrt.(1 + t^2) dt
and g(y) = the integral from 3 to y of f(x) dx.
Find g''(pi/6)
I'm not sure how to put it all together since there are two different integrals involved.
Thanks!
By the fundamental theorem of calculus:
$\displaystyle g'(y)=f(y)$
Then, $\displaystyle g''(y)=f'(y)$. You just have to take the derivative of $\displaystyle f(x)$. Use the Leibniz integral rule.
Okay, so f'(x) = sqrt (1 + sin^2 (x) ), right?
But I just don't quite understand how you get f'(y) when you only have f'(x). (I did try reading the Leibniz integral page, but I wasn't sure whether I was supposed to use it when differentiating f(x) or what.)
Sorry to ask you to spell out every little step; this doesn't even seem like it should be such a hard problem, I'm just getting hung up with the switching between x and y.
You shouldn't worry about the switching between x and y, that is just the name of the variable. The real question you have to answer is the derivative of f(x). Using the Leibniz integral rule should look someting like this:
$\displaystyle \begin{matrix}\frac{df}{dx}&=\frac{d}{dx}\int_0^{\ sin x}\sqrt{1+t^2}dt=\frac{d(\sin x)}{dx}.\sqrt{1+\sin^2 x}=\sqrt{1-\sin^4 x}\end{matrix}$