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

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Aicharya · Accepted Answer

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 $\begin{aligned} & =\int_0^1\left(1+\sqrt{x}-\frac{x^2}{4}\right) \mathrm{d} x+\int_1^2\left(3-x-\frac{x^2}{4}\right) \mathrm{d} x \\ & =\int_0^1(1+\sqrt{x}) \mathrm{d} x+\int_1^2(3-x)-\frac{1}{4} \int_0^2 x^2 \mathrm{~d} x \\ & =[x]_0^1+\frac{2}{3}\left[x^{\frac{3}{2}}\right]_0^1+3[x]_1^2-\frac{1}{2}\left[x^2\right]_1^2-\frac{1}{12}\left[x^3\right]_0^2 \\ & =1+\frac{2}{3}+3-\frac{3}{2}-\frac{2}{3} \\ & =\frac{5}{2} \end{aligned}$

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…

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