The area of the region bounded by $y=\cos x$,$x=0$,$x=\pi$,and the $x$-axis is ... sq. units.

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

The area of the region bounded by the curve $y=4x-x^{2}$ and the $x$-axis is

The area bounded by the curve $y=x|x|$,the $x$-axis,and the ordinates $x=-1$ and $x=1$ is given by [Hint: $y=x^{2}$ if $x>0$ and $y=-x^{2}$ if $x <  0$].

If the ordinate $x = a$ divides the area bounded by the curve $y = \left( 1 + \frac{8}{x^2} \right)$,the $x$-axis,and the ordinates $x = 2$ and $x = 4$ into two equal parts,then $a = $