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

The area bounded by the curve $y = x^3$,the $x$-axis,and the ordinates $x = 1$ and $x = 4$ is:

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is

The area bounded by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$ and $y = |\ln |x||$ is ......... $sq. \,unit$.

If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

The area of the region bounded by the curve $y = \tan x$,$x = 0$,$x = \frac{\pi}{4}$,and the $X$-axis is . . . . . . sq. units.