The area of the region bounded by the curve $y = x^2 + 2$ and the lines $y = x$,$x = 0$,and $x = 3$ is

If the area bounded by the curves $y=ax^2$ and $x=ay^2$ $(a>0)$ is $3$ sq units, then the value of $a$ is

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$, and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

If the area of the region ${(x, y) : -2x + 1 \le y \le 4 - x^2, x \ge 0, y \ge 0}$ is $\frac{\alpha}{\beta}$, where $\alpha, \beta \in N$ and $\gcd(\alpha, \beta) = 1$, then the value of $(\alpha + \beta)$ is:

The area of the region bounded by the curves $x(1+y^2)=1$ and $y^2=2x$ is :

The area of the region $R=\{(x,y):xy \le 8, 1 \le y \le x^{2}, x \ge 0\}$ is