Question

The area (in sq. units) of the region $\left\{\right(x,y):x\ge 0,x+y\le 3,x^2\le 4y$ and $y\le 1+\sqrt{x}\}$ is.

• A. $\frac{59}{12}$
• B. $\frac{3}{2}$
• C. $\frac{7}{3}$
• D. $\frac{5}{2}$

Answer 1

The correct option is D $\frac{5}{2}$

$C_1:x\ge 0$

$C_2:x+y\le 3$

$C_3:x^2\le 4y$

$C_4:y\le 1+\sqrt{x}$

Solving these curves, we get their intersection points as shown in the figure.

Now, we have to find the area of the shaded region, which is formed using the constraint.

$Area={\int }_0^1(1+\sqrt{x})dx+{\int }_1^2(3-x)dx-{\int }_0^2\left(x^2\right)dx$

$\Rightarrow Area={[x+\frac{2}{3}{x}^{3/2}]}_0^1+[3x-x^2{]}_1^3-{\left[\frac{x^3}{3}\right]}_0^2=\frac{5}{2}$