Answer
$(b) -\frac{\sqrt{2}}{2}+2$
(a) $\frac{5}{2}$
Work Step by Step
Using the rules of integrals:
(a) $\int_{-1}^{1}|-1+2 x| d x=\int_{-1}^{1 / 2}|-1+2 x| d x+\int_{1 / 2}^{1}|-1+2 x| d x$
$=-\int_{-1}^{1 / 2}(-1+2 x) d x+\int_{1 / 2}^{1}(-1+2 x) d x$
$=-\left.\left(-x+x^{2}\right)\right|_{-1} ^{1 / 2}+\left.\left(-x+x^{2}\right)\right|_{1 / 2} ^{1}$
$=-\left(-2-\frac{1}{4}\right)+\left(0+\frac{1}{4}\right)$
$=\frac{5}{2}$
(b) $\int_{0}^{3 \pi / 4}|\cos x| d x=\int_{\pi / 2}^{3 \pi / 4}|\cos x| d x+\int_{0}^{\pi / 2}|\cos x| d x$
$=-\int_{\pi / 2}^{3 \pi / 4} \cos x +\int_{0}^{\pi / 2} \cos x d xd x$
$=-\left.(\sin x)\right|_{\pi / 2} ^{3 \pi / 4}+\left.(\sin x)\right|_{0} ^{\pi / 2}$
$=-\left(\frac{\sqrt{2}}{2}-1\right)+(1-0)$
$=-\frac{\sqrt{2}}{2}+2$
