The area (in sq. units) of the region in the first quadrant enclosed…

The correct answer is Option (1) → \( \displaystyle\frac{7}{6} \)

The line meets the parabola where \( x=y^2 \) satisfies \( x+y=2 \):

    \( \displaystyle y^2+y-2=0 \)

    \( \displaystyle (y+2)(y-1)=0 \)

    \( \displaystyle y=1\ (\text{first quadrant}),\ x=1 \)

The region lies between y = 0 and y = 1, with the parabola \( x=y^2 \) on the left and the line \( x=2-y \) on the right. Integrating with respect to y,

    \( \displaystyle \text{Area}=\int_{0}^{1}\left[(2-y)-y^2\right]dy \)

    \( \displaystyle =\left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{0}^{1} \)

    \( \displaystyle =2-\frac{1}{2}-\frac{1}{3} \)

    \( \displaystyle =\frac{7}{6} \)

Shortcut: Integrating in x splits the region at x = 1: under the parabola from 0 to 1, then a triangle under the line from 1 to 2.

    \( \displaystyle \int_{0}^{1}\sqrt{x}\,dx=\frac{2}{3} \)

    \( \displaystyle \text{Triangle}=\frac{1}{2}\times 1\times 1=\frac{1}{2} \)

    \( \displaystyle \text{Area}=\frac{2}{3}+\frac{1}{2}=\frac{7}{6} \)

Why the other options are incorrect

• Option (2) \( \displaystyle\frac{4}{3} \) — Incorrect. The area is \( \frac{7}{6} \), not \( \frac{4}{3} \).

• Option (3) \( \displaystyle\frac{2}{3} \) — Incorrect. \( \frac{2}{3} \) is only the part under the parabola from x = 0 to x = 1; the triangle under the line from x = 1 to x = 2 has been left out.

• Option (4) \( \displaystyle\frac{7}{3} \) — Incorrect. The area is \( \frac{7}{6} \), not \( \frac{7}{3} \).