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}$.

Column-$I$	Column-$II$
$(A)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $2(a^2-b^2)=c^2$ and $\lambda=\frac{\sin(X-Y)}{\sin Z}$,then possible values of $n$ for which $\cos(n\pi\lambda)=0$ is (are)	$(P)$ $1$
$(B)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $1+\cos 2X-2\cos 2Y=2\sin X\sin Y$,then possible value$(s)$ of $\frac{a}{b}$ is (are)	$(Q)$ $2$
$(C)$ In $\mathbb{R}^2$,let $\sqrt{3}\hat{i}+\hat{j}$,$\hat{i}+\sqrt{3}\hat{j}$ and $\beta\hat{i}+(1-\beta)\hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$,respectively. If the distance of $Z$ from the bisector of the acute angle of $\overline{OX}$ with $\overline{OY}$ is $\frac{3}{\sqrt{2}}$,then possible value$(s)$ of $\|\beta\|$ is (are)	$(R)$ $3$
$(D)$ Suppose that $F(\alpha)$ denotes the area of the region bounded by $x=0, x=2, y^2=4x$ and $y=\|\alpha x-1\|+\|\alpha x-2\|+\alpha x$,where $\alpha \in \{0, 1\}$. Then the value$(s)$ of $F(\alpha)+\frac{8}{3}\sqrt{2}$,when $\alpha=0$ and $\alpha=1$,is (are)	$(S)$ $5$
	$(T)$ $6$

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

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