If $\int_{-4}^4 f(x)(\sin x +1)\, dx = 8, \; \int_{-2}^4 f(x)\, dx = 4$ where $f(x)$ is an even function, what is $\int_{-2}^0 f(x)\mathrm dx\ ?$

Question

If $$\int_{-4}^4 f(x)(\sin x +1)\mathrm dx = 8, \quad \int_{-2}^4 f(x)\mathrm d x = 4$$ where $f(x)$ is an even function, then what is the value of $$\int_{-2}^0 f(x)\mathrm dx\ ?$$ The answer is $0$.

Answer 1

\begin{align*} &\int_{-4}^{4} f(x)(\sin(x) + 1)\,dx = 8\\[4pt] \implies\;&\int_{-4}^{4} f(x)\sin(x)\,dx + \int_{-4}^{4} f(x)\,dx = 8\\[4pt] \implies\;&\int_{-4}^{4} f(x)\,dx = 8\qquad\text{[since$\;f(x)\sin(x)\;$is odd]}\\[4pt] \implies\;&\int_{0}^{4} f(x)\,dx = 4\qquad\text{[since$\;f(x)\;$is even]}\\[4pt] \end{align*} hence $$\int_{-2}^{0} f(x)\,dx = \int_{-2}^{4} f(x)\,dx - \int_0^{4} f(x)\,dx = 4 - 4 = 0$$

Answer 2

Note that $$ \int _{-4} ^4 f(x) (\sin x +1) dx = \int _{-4} ^4 f(x)\sin x dx +\int _{-4} ^4 f(x) dx =8$$

Given that $f(x) \sin x $ is odd, we have $$ \int _{-4} ^4 f(x)\sin x dx =0$$

Thus $$ \int _{-4} ^4 f(x) dx =8 \implies \int _{-4} ^0 f(x) dx =4 $$

Given $$ \int _{-4} ^{-2} f(x) dx =4 $$

We conclude that $$ \int _{-2} ^{0} f(x) dx =0$$