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: Parity
$f(-x)\sin(-x) = -f(x)\sin x$, so the product is odd and contributes nothing.

Step 2: Remaining part
$\int_{-2}^{2}|x|dx$ is the area of two triangles, each with base 2 and height 2, so $2 + 2 = 4$.

Step 3: Answer
Option (B).

Final Answer:
Option (B). \[ \boxed{4} \]