The area (in sq. units) of the region enclosed by the parabola y^2 =…

The correct answer is Option (2) → 9

Writing the line as \(x = \frac{y+4}{2}\) and the parabola as \(x = \frac{y^{2}}{4}\), the points of intersection come from

    \( \displaystyle \frac{y^{2}}{4} = \frac{y+4}{2} \)

    \(y^{2}-2y-8 = 0\)
    \((y-4)(y+2) = 0\)

So the curves meet at y = −2 and y = 4, that is at (1, −2) and (4, 4). For each y between them the line is to the right of the parabola (at y = 0 the line gives x = 2 and the parabola x = 0), so, integrating with respect to y,

    \( \displaystyle A = \int_{-2}^{4}\left(\frac{y+4}{2}-\frac{y^{2}}{4}\right)dy \)

    \( \displaystyle = \left[\frac{y^{2}}{4}+2y-\frac{y^{3}}{12}\right]_{-2}^{4} \)

    \( \displaystyle = \left(4+8-\frac{16}{3}\right)-\left(1-4+\frac{2}{3}\right) \)

    \( \displaystyle = \frac{20}{3}+\frac{7}{3} = 9 \)

Why the other options are incorrect

• Option (1) 18 — Incorrect. The integral gives \(\frac{20}{3}+\frac{7}{3} = 9\), and 18 is twice this.

• Option (3) 6 — Incorrect. This is \(\int_{-2}^{4}\frac{y^{2}}{4}\,dy\), the area between the parabola and the y-axis, which is only the part subtracted.

• Option (4) 15 — Incorrect. This is \(\int_{-2}^{4}\frac{y+4}{2}\,dy\), the area between the line and the y-axis; the part under the parabola is not subtracted.