The area of the region bounded by the lines $y=2x+1$,$y=3x+1$ and $x=4$ is

The area (in sq. units) bounded by $x^2=y$,$y=x+2$ and the $X$-axis is

Find the area lying above the $x-$ axis and included between the circle $x^{2}+y^{2}=8x$ and inside the parabola $y^{2}=4x$.

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 $f\left( \frac{\pi}{2} \right)$.

The area bounded by the curve $4y^{2} = x^{2}(4-x)(x-2)$ is equal to ...... .

If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to: