Question

If \(f(x)\) is an even function, then \(\int _{-2}^2(|x|+f(x)sinx)\,\text{d}x\) is

• A. \(2\)
• B. \(4\)
• C. \(6\)
• D. \(8\)

Hint:

f even and sin x odd make f(x) sin x an odd function, so its integral over [-2, 2] is zero.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For an odd function g, \(\int_{-a}^{a}g(x)\,dx = 0\). For an even function, \(\int_{-a}^{a}g(x)\,dx = 2\int_0^{a}g(x)\,dx\).

Step 2: Split
\[ \int_{-2}^{2}\big(|x| + f(x)\sin x\big)dx = \int_{-2}^{2}|x|\,dx + \int_{-2}^{2}f(x)\sin x\,dx \]
Since f is even and \(\sin x\) is odd, \(f(x)\sin x\) is odd, so the second integral is 0.

Step 3: Evaluate
\[ \int_{-2}^{2}|x|\,dx = 2\int_0^2 x\,dx = 2\cdot\frac{4}{2} = 4 \]
So the value is 4, option (B).

Final Answer:
The value is 4, option (B). \[ \boxed{4} \]