The area bounded by the curve y = 4+3x-x^2 and x-axis is equal to

The correct answer is Option (1) → \(\displaystyle\frac{125}{6}\) Sq. units

The curve meets the x-axis where y = 0:

    \(x^2 - 3x - 4 = 0\)
    \((x - 4)(x + 1) = 0,\quad x = -1 \text{ or } 4\)

The parabola opens downwards, so y ≥ 0 between these roots. The area is the definite integral of y from −1 to 4:

    \(\displaystyle \text{Area} = \int_{-1}^{4} (4 + 3x - x^2)\,dx\)

    \(\displaystyle = \left[4x + \frac{3x^2}{2} - \frac{x^3}{3}\right]_{-1}^{4}\)

    \(\displaystyle = \left(16 + 24 - \frac{64}{3}\right) - \left(-4 + \frac{3}{2} + \frac{1}{3}\right)\)

    \(\displaystyle = \frac{56}{3} + \frac{13}{6} = \frac{125}{6}\) sq. units

Shortcut: The area between a parabola \(y = -a(x - \alpha)(x - \beta)\) and the x-axis is \(\displaystyle\frac{a(\beta - \alpha)^3}{6}\). Here a = 1 and β − α = 5, so the area is \(\displaystyle\frac{125}{6}\).