Let $F(x)$ be the function defined on $(-\infty,\infty)$ by the formula $$F(x)=\int_{-e^x}^{e^{2x}}|t^2-1|\,{\rm d}t$$ Find $F'(x)$. Show your work.
So I've been trying to figure this out for a while now, and im getting stuck. I've tried splitting the integral into two seperate functions $u$ and $z$ and then applying the fundamental theorum of calculus to find $F'(x)$.
However from observing desmos, and using its ability to graph the derivative of a function, my answer appears to only be correct for $x>0$. I was wondering what I need to do to find the answer for $x<0$. Or if my answer is even right at all. Thanks in advance for the help. The absolute value is the main thing throwing me off I think. Here is desmos, with $F(x$), $F'(x)$ and my answer graphed.
I’ll let $|t^2-1| = g(t)$ for clarity.$$F(x) =\int_{-e^x}^{e^{2x}} g(t) dt$$ By the FTC, we have $$F’(x) = g(e^{2x} ) \ 2e^{2x} - g(-e^x) \ (-e^x) \\ = 2e^{2x}|e^{4x} -1|+e^x|e^{2x} -1| $$ Now, $e^{kx} -1 \lt 0 $ for $x\lt 0$, and positive otherwise. $$\therefore F’(x) = \begin{cases} 2e^{6x} -2e^{2x} +e^{3x} -e^x , \ \text{if} \ x\lt 0 \\ -2e^{6x} +2e^{2x} -e^{3x} +e^x , \ \text{if} \ x\ge 0 \end{cases} $$