Answer
(a) $\vert \int_{a}^{b} f(x)~dx \vert \leq \int_{a}^{b} \vert f(x) \vert~dx$
(b) $\vert \int_{0}^{2\pi} f(x)~sin~2x~dx \vert \leq \int_{0}^{2\pi} \vert f(x)\vert~dx$
Work Step by Step
(a) On the interval $a \leq x \leq b$:
$ -\vert f(x) \vert \leq f(x) \leq \vert f(x) \vert$
Therefore, by Property 7:
$\int_{a}^{b} -\vert f(x) \vert~dx \leq \int_{a}^{b} f(x)~dx \leq \int_{a}^{b} \vert f(x) \vert~dx$
$-\int_{a}^{b} \vert f(x) \vert~dx \leq \int_{a}^{b} f(x)~dx \leq \int_{a}^{b} \vert f(x) \vert~dx$
$\vert \int_{a}^{b} f(x)~dx \vert \leq \int_{a}^{b} \vert f(x) \vert~dx$
(b) $\vert \int_{0}^{2\pi} f(x)~sin~2x~dx \vert$
$\leq \int_{0}^{2\pi} \vert f(x)~sin~2x \vert~dx$
$\leq \int_{0}^{2\pi} \vert f(x) \vert \cdot \vert~sin~2x \vert~dx$
$\leq \int_{0}^{2\pi} \vert f(x)\vert~(1)~dx$
$= \int_{0}^{2\pi} \vert f(x)\vert~dx$
