The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$,is

The area of the region,bounded by the curves $y=\sin ^{-1} x+x(1-x)$ and $y=\sin ^{-1} x-x(1-x)$ in the first quadrant,is

Area of the region enclosed between the curves $y^2=4(x+7)$ and $y^2=5(2-x)$ is

The area bounded by the curve $y = \begin{cases} x^2, & x < 0 \\ x, & x \geq 0 \end{cases}$ and the line $y = 4$ is

The area common to the curves $5x^2 - y = 0$ and $2x^2 - y + 9 = 0$ is equal to

The area (in sq. units) of the part of the circle $x^2+y^2=169$ which is below the line $5x-y=13$ is $\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$ where $\alpha, \beta$ are coprime numbers. Then $\alpha+\beta$ is equal to . . . . . . .