calculus
Find the area of the region that is bounded by the given curve and lies in the specified sector. $r = e^{−θ/8}$$\ ,\ π/2 ≤ θ ≤ π$
I know the formula is $\int_a^b \frac{1}{2} r^2 d\theta$
$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/8})^2 d\theta$
From this point I cannot integrate. Help!
My answer was $2e^{\frac{-\pi}{4}}(e^{\frac{\pi}{8}}-1)$
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.
$$ \begin{align} \int_{-1}^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x &=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)\mathrm{d}x\\ &=4\int_0^{\pi/4}\cos^2(\theta)\,\mathrm{d}\theta-\frac23\\ &=2\int_0^{\pi/4}(\cos(2\theta)+1)\,\mathrm{d}\theta-\frac23\\[3pt] &=\frac\pi2+\frac13\\ \end{align} $$
Best Answer
$$\int_{\pi/2}^{\pi} \frac{1}{2} (e^{- \theta/8})^2 d\theta$$ $$=\frac{1}{2}\int_{\pi/2}^{\pi} e^{- \theta/4} d\theta$$
Set $u=-\theta/4$, $\frac{\text{d}u}{\text{d}\theta}=\frac{-1}{4}$
$$=\frac{-4}{2}\int_{-\pi/8}^{-\pi/4} e^{u} du$$
$$=-2e^u|_{-\pi/8}^{-\pi/4} = 2e^{\frac{-\pi}{4}}(e^{\frac{\pi}{8}}-1)$$
You appear to have done it correctly.