The area of the region bounded by y^2=2x+1 and x-y=1 is :

The correct answer is Option (4) → \(\displaystyle\frac{16}{3}\) sq. units

Both curves are easy to write with x in terms of y:

    \( \displaystyle \text{parabola: } x = \frac{y^2-1}{2},\qquad \text{line: } x = y+1 \)

Points of intersection:

    \(y^2 = 2(y+1)+1\)
    \(y^2-2y-3 = 0\)
    \((y-3)(y+1) = 0,\quad y = -1 \text{ or } 3\)

That is, the points (0, −1) and (4, 3). For y between −1 and 3 the line is to the right of the parabola (at y = 0: x = 1 on the line, \(x = -\frac{1}{2}\) on the parabola), so

    \( \displaystyle A = \int_{-1}^{3}\left[(y+1)-\frac{y^2-1}{2}\right]dy \)

    \( \displaystyle = \int_{-1}^{3}\left(y+\frac{3}{2}-\frac{y^2}{2}\right)dy \)

    \( \displaystyle = \left[\frac{y^2}{2}+\frac{3y}{2}-\frac{y^3}{6}\right]_{-1}^{3} \)

    \( \displaystyle = \left(\frac{9}{2}+\frac{9}{2}-\frac{9}{2}\right)-\left(\frac{1}{2}-\frac{3}{2}+\frac{1}{6}\right) \)

    \( \displaystyle = \frac{9}{2}+\frac{5}{6} = \frac{16}{3} \)

So the area is \(\frac{16}{3}\) sq. units.