y = Ln [ ∫√(1-t^{3}) dt ]
Limits of integration are from π^{2}/16 to x
I know the first FTOC is involved... but I don't know how to handle the natural log... any help would be appreciated.
Hello,
You didn't tell us what you are trying to do! The fundamental theorem of Calculus has two parts....
My guess is that you were asked to find the derivative $\displaystyle y'$
IF that is the case then you have a function composistion
$\displaystyle u=\int_{\frac{\pi^2}{16}}^{x}\sqrt{1-t^3}dt $
Then by the fundamental theorem you get
$\displaystyle \frac{du}{dx}=\sqrt{1-x^3}$
Then you need to notice that you have
$\displaystyle y=\ln(u)$
so by the chain rule you get
$\displaystyle \frac{dy}{dx}=\left( \frac{d}{du}\ln(u)\right)\frac{du}{dx} =\frac{1}{u}\frac{du}{dx}$
You know both u and its derivative this gives that
$\displaystyle y'=\left(\frac{1}{\int_{\frac{\pi}{16}}^{x}\sqrt{1-t^3}dt }}\right)\sqrt{1-x^3}=\frac{\sqrt{1-x^3}}{\int_{\frac{\pi}{16}}^{x}\sqrt{1-t^3}dt }}$