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(B) The point of intersection of $y^{2}=x$ and $x+y=2$ is found by substituting $y=2-x$ into the parabola equation:$(2-x)^{2}=x$$4-4x+x^{2}=x$$x^{2}-5

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$\frac{7}{6}$ sq. units

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(B) The point of intersection of $y^{2}=x$ and $x+y=2$ is found by substituting $y=2-x$ into the parabola equation:$(2-x)^{2}=x$$4-4x+x^{2}=x$$x^{2}-5x+4=0$$(x-4)(x-1)=0$$x=1$ or $x=4$.Since we are looking for the area in the first quadrant,we take $x=1$. Substituting $x=1$ into $y^{2}=x$,we get $y=1$ (as $y>0$ in the first quadrant).Thus,the intersection point is $(1,1)$.The line $x+y=2$ intersects the $X$-axis at $(2,0)$.The required area is the sum of the area under the parabola from $x=0$ to $x=1$ and the area under the line from $x=1$ to $x=2$:Area $= \int_{0}^{1} \sqrt{x} \, dx + \int_{1}^{2} (2-x) \, dx$$= \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} + \left[ 2x - \frac{x^{2}}{2} \right]_{1}^{2}$$= \frac{2}{3}(1) + \left( (4-2) - (2-0.5) \right)$$= \frac{2}{3} + (2 - 1.5) = \frac{2}{3} + 0.5 = \frac{2}{3} + \frac{1}{2} = \frac{4+3}{6} = \frac{7}{6}$ sq. units.

The area of the region included between the parabola $y^{2}=x$ and the line $x+y=2$ in the first quadrant is

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