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

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Question

(A) The region is bounded by $x=0$,$y=1+\sqrt{x}$,$x+y=3$,and $y=\frac{x^2}{4}$.From the graph,the intersection points are $(1,2)$ and $(2,1)$.The are

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$\frac{5}{2}$

Vedclass · Answer

(A) The region is bounded by $x=0$,$y=1+\sqrt{x}$,$x+y=3$,and $y=\frac{x^2}{4}$.From the graph,the intersection points are $(1,2)$ and $(2,1)$.The area is given by the integral:$A = \int_{0}^{1} (1+\sqrt{x} - \frac{x^2}{4}) dx + \int_{1}^{2} (3-x - \frac{x^2}{4}) dx$$A = \left[ x + \frac{2}{3}x^{3/2} - \frac{x^3}{12} \right]_{0}^{1} + \left[ 3x - \frac{x^2}{2} - \frac{x^3}{12} \right]_{1}^{2}$$A = (1 + \frac{2}{3} - \frac{1}{12}) + [(6 - 2 - \frac{8}{12}) - (3 - \frac{1}{2} - \frac{1}{12})]$$A = (\frac{12+8-1}{12}) + [(4 - \frac{2}{3}) - (2.5 - \frac{1}{12})]$$A = \frac{19}{12} + (\frac{10}{3} - \frac{29}{12}) = \frac{19}{12} + \frac{40-29}{12} = \frac{19+11}{12} = \frac{30}{12} = \frac{5}{2}$ sq. units.

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

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