The area bounded by the curve $y = x(1 - \ln x)$,the line $x = e^{-1}$,and the positive $X$-axis between $x = e^{-1}$ and $x = e$ is:

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$ and $x = 2a$ has the least value,is

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \frac{3\pi}{2}$ is:

If the line $x=\alpha$ divides the area of region $R=\{(x, y) \in \mathbb{R}^2: x^3 \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts,then which of the following is true?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$

Let $f(x) = x^3 - 3x^2 + 3x + 1$ and $g$ be the inverse of $f$. The area bounded by the curve $y = g(x)$ with the $x$-axis between $x = 1$ and $x = 2$ is (in square units):

If a curve $y=a \sqrt{x}+b x$ passes through the point $(1,2)$ and the area bounded by this curve,line $x=4$ and the $X$-axis is $8$ sq.units,then the value of $a-b$ is