If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 \{3 - f(x)\} dx = 7$, then find the value of $\int_{-1}^2 f(x) dx$.

$\int_{-\pi}^{\pi} |\pi - |x|| \, dx$ is equal to :

$\int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} d x=$

Read the following mathematical statements carefully:
$I.$ $A$ differentiable function $f$ with maximum at $x = c$ $\implies f''(c) < 0$.
$II.$ Antiderivative of a periodic function is also a periodic function.
$III.$ If $f$ has a period $T$ then for any $a \in R$,$\int\limits_0^T {f(x)\,dx} = \int\limits_0^T {f(x + a)\,dx}$.
$IV.$ If $f(x)$ has a maxima at $x = c$,then $f$ is increasing in $(c - h, c)$ and decreasing in $(c, c + h)$ as $h \to 0$ for $h > 0$. Now indicate the correct alternative.

One of the possible functions $f(x)$ which satisfies $\int_{-2}^{2} f(x) dx = 0$ is

$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=$