Question

question 28
the net area of the \\(\int_{-5}^{5} \frac{\sin x}{\sqrt{1 + x^2}} dx\\) is equivalent to which of the following, and why?
\\(\circ\\) 0, because the integrand is an odd function.
\\(\circ\\) \\(2 \cdot \int_{0}^{5} \frac{\sin x}{\sqrt{1 + x^2}} dx\\), because the integrand is an odd function.
\\(\circ\\) 0, because the integrand is an even function.
\\(\circ\\) \\(2 \cdot \int_{0}^{5} \frac{\sin x}{\sqrt{1 + x^2}} dx\\), because the integrand is an even function.
\\(\circ\\) no correct answer choice is given.

Explanation:

Step1: Recall Odd Function Property

A function \( f(x) \) is odd if \( f(-x) = -f(x) \). For an odd function, the integral over a symmetric interval \([-a, a]\) is \( \int_{-a}^{a} f(x) dx = 0 \).

Step2: Check if Integrand is Odd

Let \( f(x) = \frac{\sin x}{\sqrt{1 + x^2}} \). Compute \( f(-x) \):
\( f(-x) = \frac{\sin(-x)}{\sqrt{1 + (-x)^2}} = \frac{-\sin x}{\sqrt{1 + x^2}} = -f(x) \).
So, \( f(x) \) is odd.

Step3: Apply Integral Property

The interval \([-5, 5]\) is symmetric about 0. For an odd function \( f(x) \), \( \int_{-5}^{5} f(x) dx = 0 \).

Answer:

0, because the integrand is an odd function.