Answer
$$\int^{0}_{-\ln2}e^{2w}dw=\frac{3}{8}$$
Work Step by Step
$$A=\int^{0}_{-\ln2}e^{2w}dw$$
We set $u=2w$, which means $$du=2dw$$ $$dw=\frac{1}{2}du$$
For $w=0$, $u=-(-1+1)=0$ and for $w=-\ln2$, $u=-2\ln2=\ln2^{-2}=\ln\frac{1}{4}$
Therefore, $$A=\frac{1}{2}\int^0_{\ln1/4}e^udu$$ $$A=\frac{1}{2}e^u\Big]^0_{\ln1/4}$$ $$A=\frac{1}{2}(e^0-e^{\ln1/4})$$ $$A=\frac{1}{2}(1-\frac{1}{4})$$ $$A=\frac{1}{2}\times\frac{3}{4}=\frac{3}{8}$$
