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

Given inequalities are
\(\begin{aligned}
& x \geq 0 \\
& x+y \leq 3 \\
& x^2 \leq 4 y, \\
& y \leq 1+\sqrt{x}
\end{aligned}\)
\(\therefore \quad\) The equalities are
\(\begin{aligned}
& x+y=3 ...(i)\\
& x^2=4 y ...(ii)\\
& y=1+\sqrt{x}...(iii)
\end{aligned}\)
from (i) and (iii), we get
\(\begin{array}{ll}
& 3-x=1+\sqrt{x} \\
\therefore \quad & x+\sqrt{x}-2=0
\end{array}\)
\(\begin{aligned}
\therefore \quad & (\sqrt{x}+2)(\sqrt{x}-1)=0 \\
& \Rightarrow \sqrt{x}=1 \quad \ldots[\because \sqrt{x} \text { cannot be negative }] \\
& \Rightarrow x=1 \text { and } y=2
\end{aligned}\)
From (i) and (ii), we get
\(\begin{array}{ll}
& x+\frac{x^2}{4}=3 \\
\therefore \quad & x^2+4 x-12=0 \\
\therefore \quad & (x+6)(x-2)=0 \\
& \Rightarrow x=2 \\
& \Rightarrow y=1
\end{array}\)
\(\ldots[\because x \geq 0]\)
\(\therefore \quad\) Required area
\(\begin{aligned}
& =\int_0^1\left(1+\sqrt{x}-\frac{x^2}{4}\right) \mathrm{d} x+\int_1^2\left(3-x-\frac{x^2}{4}\right) \mathrm{d} x \\
& =\int_0^1(1+\sqrt{x}) \mathrm{d} x+\int_1^2(3-x)-\frac{1}{4} \int_0^2 x^2 \mathrm{~d} x \\
& =[x]_0^1+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_0^1+3[x]_1^2-\frac{1}{2}\left[x^2\right]_1^2-\frac{1}{12}\left[x^3\right]_0^2 \\
& =1+\frac{2}{3}+3-\frac{3}{2}-\frac{2}{3} \\
& =\frac{5}{2}
\end{aligned}\)