Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

Let $g(x) = \cos(x^2)$,$f(x) = \sqrt{x}$,and $\alpha, \beta$ (where $\alpha < \beta$) be the roots of the quadratic equation $18x^2 - 9\pi x + \pi^2 = 0$. Then,the area (in sq. units) bounded by the curve $y = (g \circ f)(x)$ and the lines $x = \alpha$,$x = \beta$,and $y = 0$ is:

The area of the region bounded by the parabola $y^{2}=8x$ and its latus rectum is

The area bounded by the curve $y = x |\sin(\frac{x}{2})|$ and the $X$-axis between the lines $x = 0$ and $x = 4\pi$ (in square units) is: (in $\pi$)

Find the area under the curve $y=x^{4}$ bounded by the lines $x=1$,$x=5$ and the $x$-axis.

The area of the region bounded by the curve $y=9-x^2$,the $X$-axis,and the lines $x=0$ and $x=3$ is . . . . . . sq. units.