Let f(α) denote the area of the region in the first quadrant bounded by x=0, x=1, y^2=x and y=|α x-5|-|1-α x|+α x^2. Then (f(0)+f(1)) is equal to

The value of $\int_{-\pi / 6}^{\pi / 6}\left(\frac{\pi+4 x^{11}}{1-\sin (|x|+\pi / 6)}\right) d x$ is equal to:

The product of all possible values of $\alpha$, for which $\displaystyle\lim_{x \to 0}\left(\dfrac{1 - \cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)}{\sin^2((\alpha+1)x)}\right) = 2$, is:

The value of the integral $\displaystyle\int_0^\infty \dfrac{\log_e(x)}{x^2 + 4}\,dx$ is:

Let $f$ be a differentiable function satisfying $f(x)=1-2 x+\int_{0}^{x} \mathrm{e}^{(x-t)} f(t) \mathrm{dt}, x \in \mathbf{R}$ and let $\mathrm{g}(x)=\int_{0}^{x}(f(\mathrm{t})+2)^{15}(\mathrm{t}-4)^{6}(\mathrm{t}+12)^{17} \mathrm{dt}, x \in \mathbf{R}$. If p and q are respectively the points of local minima and local maxima of g, then the value of $|\mathrm{p}+\mathrm{q}|$ is equal to $\_\_\_\_$.

Let the area of the region bounded by the curve $y=\max \{\sin x, \cos x\}$, lines $x=0, x=\frac{3 \pi}{2}$, and the $x$-axis be A. Then, $\mathrm{A}+\mathrm{A}^{2}$ is equal to $\_\_\_\_$.