$r = e^{- \theta/4}$ $\pi /2 \leq \theta \leq \pi$
I know the formula is
$\int_a^b \frac{1}{2} r^2 d\theta$
$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/4})^2 d\theta$
From here I cannot figure out an easy way to integrate this.
$r = e^{- \theta/4}$ $\pi /2 \leq \theta \leq \pi$
I know the formula is
$\int_a^b \frac{1}{2} r^2 d\theta$
$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/4})^2 d\theta$
From here I cannot figure out an easy way to integrate this.
Note that $(e^{-\theta/4})^2=e^{-\theta/2}$. It is a special case of the general exponential rule $(a^b)^c=a^{bc}$.
Hint: $(x^a)^b=x^{ab}$. After that, it falls right out, either through FTC or a substitution.
The integral you compute should is:
$\frac{1}{2} \int_{\pi/2}^{\pi}\left(e^{-\theta/4}\right)^2\, d\theta=\frac{1}{2} \int_{\pi/2}^{\pi}e^{-\theta/2}\, d\theta$
Using the exponent rule $(a^b)^c=a^{bc}$. Letting $u=-\theta/2$, $\theta=-2u$, $d\theta=-2\, du$
$\frac{1}{2}\int_{\pi/2}^{\pi}e^{-\theta/2}\, d\theta=-\int_{-\pi/4}^{-\pi/2}e^u\, du=-\left[e^u\right]^{-\pi/2}_{-\pi/4}=e^{-\pi/2}(e^{\pi/4}-1)$