JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Area Under Curves

The parabola $(y-2{)}^2=x-1$ can be written as $x=g(y)=(y-2{)}^2+1$, a single-valued rightward curve with vertex $(1,2)$, meeting the x-axis at $(5,0)$.

Solving with the line $x=2y-4$: $(y-2{)}^2=2y-5\Rightarrow y^2-6y+9=0\Rightarrow (y-3{)}^2=0\Rightarrow y=3$, so the line touches the parabola at $(2,3)$. The line meets the y-axis at $(0,2)$.

For $0\le y\le 2$, the line's x-value $2y-4$ is negative, so in the first quadrant the left boundary here is simply the y-axis; for $2\le y\le 3$, the line's x-value lies between $0$ and $2$, which is less than $g(y)$, so the line becomes the left boundary there. The region is bounded on the right by the parabola throughout.

$\text{Area}={\int }_0^{2}g(y)dy+{\int }_2^3(g(y)-(2y-4))dy={\int }_0^2(y^2-4y+5)dy+{\int }_2^3(y^2-6y+9)dy$.

${\int }_0^2(y^2-4y+5)dy={\left[\frac{y^3}{3}-2y^2+5y\right]}_0^2=\frac{8}{3}-8+10=\frac{14}{3}$.

${\int }_2^3(y^2-6y+9)dy={\left[\frac{y^3}{3}-3y^2+9y\right]}_2^3=(9-27+27)-\left(\frac{8}{3}-12+18\right)=9-\frac{26}{3}=\frac{1}{3}$.

Total area $=\frac{14}{3}+\frac{1}{3}=5$.