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Solve the following.
Evaluate ∫₋₂² x f(x) dx given, f(x) = x + g(x) where g(x) is an even function.
Evaluate $\int_{-2}^{2} x f(x)\,dx$ given, $f(x) = x + g(x)$ where $g(x)$ is an even function.
Answer
Answer
Integral: $\int_{-2}^{2} \left(x^2\right) dx$
$\int_{-2}^{2} x f(x)\,dx = \int_{-2}^{2} x^2\,dx + \int_{-2}^{2} x\,g(x)\,dx$.
$x\,g(x)$ is an odd function (odd times even), so its integral over $[-2, 2]$ is $0$.
$\int_{-2}^{2} x^2\,dx = 2\int_0^2 x^2\,dx = 2\cdot\dfrac{8}{3} = \dfrac{16}{3}$.
Final answer: $\dfrac{16}{3}$