The area of the region enclosed by the curve $y=x^3$ and its tangent at the point $(-1,-1)$ is

The area bounded by the circle $x^2 + y^2 = 4$,the line $x = \sqrt{3}y$,and the $x$-axis lying in the first quadrant 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 $f(x) = \sin(\pi x)$ and the $X$-axis for $x \in [1, 3]$ is . . . . . . sq. units.

Find the area of the region bounded by the curve $y^{2}=x$,the lines $x=1$,$x=4$,and the $x$-axis in the first quadrant.

Let $f(x)$ be a non-negative continuous function such that the area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = \frac{\pi}{4}$ and $x = \beta > \frac{\pi}{4}$ is given by $\left( \beta \sin \beta + \frac{\pi}{4} \cos \beta + \sqrt{2} \beta \right)$. Then,find the value of $f\left( \frac{\pi}{2} \right)$.