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

For which of the following values of $m$,the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $\frac{9}{2}$?

The area enclosed between the curves $y=x|x|$ and $y=x-|x|$ is :

The area of the region bounded by $x^2=4y$,$y=1$,$y=4$ and the y-axis lying in the first quadrant is $ . . . . . . $ square units.

If the straight line $x=b$ divides the area enclosed by $y=(1-x)^2$,$y=0$,and $x=0$ into two parts $R_1 (0 \leq x \leq b)$ and $R_2 (b \leq x \leq 1)$ such that $R_1 - R_2 = \frac{1}{4}$,then $b$ equals:

The area of the region bounded by the lines $y=mx$,$x=1$,$x=2$,and the $x$-axis is $6$ sq. units. Then,the value of $m$ is: