The area of the bounded region enclosed by the curve $y=3-\left|x-\frac{1}{2}\right|-|x+1|$ and the $x-$axis is

The area of the region bounded by the line $y=3x$ and the curve $y=x^2$ in square units is

If the area bounded by the curve $x^2=4y$,the $X$-axis,and the line $x=4$ is divided into two equal areas by the line $x=\alpha$,then the value of $\alpha$ is ...

The part of the straight line $y = x + 1$ between $x = 2$ and $x = 3$ is revolved about the $x$-axis. The curved surface area of the solid thus generated is

Find the area of the region lying in the first quadrant and bounded by $y=4x^2$,$x=0$,$y=1$,and $y=4$.

Consider a region $R=\{(x, y) \in \mathbb{R}^{2}: x^{2} \leq y \leq 2 x\}$. If a line $y=\alpha$ divides the area of region $R$ into two equal parts,then which of the following is true?