Answer
$$\int^1_02^{-\theta}d\theta=\frac{1}{2\ln2}$$
Work Step by Step
$$A=\int^1_02^{-\theta}d\theta$$
We set $-\theta=u$, which means $$-d\theta=du$$ $$d\theta=-du$$
- For $\theta=1$, we have $u=-1$ and for $\theta=0$, we have $u=0$
Therefore, $$A=-\int^{-1}_02^udu$$ $$A=-\frac{2^u}{\ln2}\Big]^{-1}_0$$ $$A=-\frac{1}{\ln2}(2^{-1}-2^0)$$ $$A=-\frac{\frac{1}{2}-1}{\ln2}$$ $$A=\frac{\frac{1}{2}}{\ln2}=\frac{1}{2\ln2}$$
