Answer
(b) $\sqrt{3}-1-\frac{\pi}{12}$
(a) $(-4 \sqrt{2}+3 \sqrt{3}+8)\frac{2}{3}$
Work Step by Step
Using the rules of integrals:
(a) $\int_{-1}^{2} \sqrt{|x|+2} d x=\int_{-1}^{0} \sqrt{-x+2} d x+\int_{0}^{2} \sqrt{x+2} d x$
$=\left.\left(-\frac{2}{3}(-x+2)^{3 / 2}\right)\right|_{-1} ^{0}+\left.\left(\frac{2}{3}(2+x)^{3 / 2}\right)\right|_{0} ^{2}$
$=\left(8-2^{3 / 2}\right)\frac{2}{3}-\frac{2}{3}\left(2^{3 / 2}-3^{3 / 2}\right)$
$=\frac{2}{3}(-4 \sqrt{2}+3 \sqrt{3}+8)$
(b) $\int_{0}^{\pi / 2}\left|-\cos x+\frac{1}{2}\right| d x=-\int_{0}^{\pi / 3}\left(-\cos x+\frac{1}{2}\right) d x+\int_{\pi / 3}^{\pi / 2}\left(-\cos x+\frac{1}{2}\right) d x$
$=-\left.\left(-\sin x+\frac{1}{2} x\right)\right|_{0} ^{\pi / 3}+\left.\left(-\sin x+\frac{1}{2} x\right)\right|_{\pi / 3} ^{\pi / 2}$
$=-\left(\frac{\pi}{6}-(\frac{\sqrt{3}}{2} \right)+\left(\frac{\pi}{4}-\frac{\pi}{6}-1+\frac{\sqrt{3}}{2}\right)$
$=-\frac{\pi}{12}-1+\sqrt{3}$
