Question

use symmetry to evaluate the following integral.
\\(\int_{-\pi}^{\pi} 5 \sin x \\, dx\\)

\\(\int_{-\pi}^{\pi} 5 \sin x \\, dx = \square\\) (simplify your answer.)

Explanation:

Step1: Identify odd function property

A function $f(x)$ is odd if $f(-x) = -f(x)$. For $f(x) = \sin x$, $\sin(-x) = -\sin x$, so $\sin x$ is odd. Scaling by a constant 5 gives $5\sin x$, which is also odd.

Step2: Apply odd integral rule

For an odd function $f(x)$, $\int_{-a}^{a} f(x) dx = 0$. Here $a = \pi$, so:
$$\int_{-\pi}^{\pi} 5\sin x dx = 0$$

Answer:

0