Answer
See explanation.
Work Step by Step
Using the rules of integration, we find:
(a) $\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{-a}^{0} f(x) d x=\int_{0}^{a} f(x) d x-\int_{a}^{0} f(-x) d x$
$=\int_{0}^{a} f(x) d x+\int_{a}^{0} f(x) d x=\int_{0}^{a} f(x) d x-\int_{0}^{a} f(x) d x=0$
(b) $\int_{-a}^{a} f(x) d x=\int_{0}^{a} f(x) d x+\int_{-a}^{0} f(x) d x=\int_{0}^{a} f(x) d x-\int_{a}^{0} f(-x) d x$
$=\int_{0}^{a} f(x) d x-\int_{a}^{0} f(x) d x=\int_{0}^{a} f(x) d x+\int_{0}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$
