The area (in sq. units) of the region {(x, y) | vedclass

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(D) Given inequalities are:$x \geq 0$$x+y \leq 3$$x^2 \leq 4y$$y \leq 1+\sqrt{x}$The boundary curves are:$x+y=3 \quad (i)$$x^2=4y \quad (ii)$$y=1+\sqr

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$\frac{5}{2}$

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(D) Given inequalities are:$x \geq 0$$x+y \leq 3$$x^2 \leq 4y$$y \leq 1+\sqrt{x}$The boundary curves are:$x+y=3 \quad (i)$$x^2=4y \quad (ii)$$y=1+\sqrt{x} \quad (iii)$From $(i)$ and $(iii)$:$3-x = 1+\sqrt{x}$$x+\sqrt{x}-2=0$$(\sqrt{x}+2)(\sqrt{x}-1)=0$Since $\sqrt{x} \geq 0$,we have $\sqrt{x}=1 \Rightarrow x=1, y=2$.From $(i)$ and $(ii)$:$x + \frac{x^2}{4} = 3$$x^2+4x-12=0$$(x+6)(x-2)=0$Since $x \geq 0$,we have $x=2, y=1$.The required area is:$A = \int_0^1 (1+\sqrt{x} - \frac{x^2}{4}) dx + \int_1^2 (3-x - \frac{x^2}{4}) dx$$A = [x + \frac{2}{3}x^{3/2} - \frac{x^3}{12}]_0^1 + [3x - \frac{x^2}{2} - \frac{x^3}{12}]_1^2$$A = (1 + \frac{2}{3} - \frac{1}{12}) + ((6 - 2 - \frac{8}{12}) - (3 - \frac{1}{2} - \frac{1}{12}))$$A = (\frac{12+8-1}{12}) + (4 - \frac{2}{3} - 3 + \frac{1}{2} + \frac{1}{12})$$A = \frac{19}{12} + (1 - \frac{8}{12} + \frac{6}{12} + \frac{1}{12}) = \frac{19}{12} + (1 - \frac{1}{12}) = \frac{19}{12} + \frac{11}{12} = \frac{30}{12} = \frac{5}{2}$ sq. units.

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

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