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

If the tangent to the curve $y = 1 - x^2$ at $x = \alpha,$ where $0 < \alpha < 1,$ meets the axes at $P$ and $Q.$ Also $\alpha$ varies,the minimum value of the area of the triangle $OPQ$ is $k$ times the area bounded by the axes and the part of the curve for which $0 < x < 1,$ then $k$ is equal to

The ratio of the areas bounded by the curves $y = \cos x$ and $y = \cos 2x$ between $x = 0$ and $x = \frac{\pi}{3}$ with the $X$-axis is:

The area of the region bounded by the curve $y = \cos x$,$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ and the $X$-axis is . . . . . . sq. units.

Area inside the parabola ${y^2} = 4ax$ between the lines $x = a$ and $x = 4a$ is equal to

The area bounded by the curve $y=|x-2|$,$x=1$,$x=3$ and $X$-axis is