The area bounded by the parabola y^2=x and th | vedclass

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(A) To find the area bounded by the parabola $y^2=x$ and the line $x+y=2$ in the first quadrant,we first find the points of intersection.Substituting

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

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(A) To find the area bounded by the parabola $y^2=x$ and the line $x+y=2$ in the first quadrant,we first find the points of intersection.Substituting $y = 2-x$ into $y^2=x$,we get $(2-x)^2 = x$,which simplifies to $x^2 - 4x + 4 = x$,or $x^2 - 5x + 4 = 0$.Factoring gives $(x-4)(x-1) = 0$,so $x=1$ or $x=4$.In the first quadrant,the intersection point is $(1, 1)$.The line $x+y=2$ intersects the $x$-axis at $(2, 0)$.The required area is the integral of the parabola from $x=0$ to $x=1$ plus the integral of the line from $x=1$ to $x=2$.Area $= \int_0^1 \sqrt{x} \, dx + \int_1^2 (2-x) \, dx$$= \left[ \frac{2}{3} x^{3/2} ight]_0^1 + \left[ 2x - \frac{x^2}{2} ight]_1^2$$= \left( \frac{2}{3} - 0 ight) + \left( (4 - 2) - (2 - \frac{1}{2}) ight)$$= \frac{2}{3} + (2 - \frac{3}{2}) = \frac{2}{3} + \frac{1}{2} = \frac{4+3}{6} = \frac{7}{6} 	ext{ sq. units}$.

The area bounded by the parabola $y^2=x$ and the line $x+y=2$ in the first quadrant is

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