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

The correct answer is Option (2) → \(\displaystyle\frac{125}{6}\)

The points of intersection come from

    \(x^{2} = x+6\)
    \(x^{2}-x-6 = 0\)
    \((x-3)(x+2) = 0\)

So x = −2 and x = 3. Between them the line is above the parabola (at x = 0 the line gives 6 and the parabola 0), so the area is the integral of (line − parabola):

    \( \displaystyle A = \int_{-2}^{3}\left(x+6-x^{2}\right)dx \)

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

    \( \displaystyle = \left(\frac{9}{2}+18-9\right)-\left(2-12+\frac{8}{3}\right) \)

    \( \displaystyle = \frac{27}{2}+\frac{22}{3} = \frac{125}{6} \)