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

$\frac{125}{6}$ Sq. units

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

Given curve: $y = 4 + 3x - x^2$

To find area bounded by the curve and x-axis, find limits where $y = 0$:

$4 + 3x - x^2 = 0 \Rightarrow -x^2 + 3x + 4 = 0 \Rightarrow x^2 - 3x - 4 = 0$

$\Rightarrow x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}$

$\Rightarrow x = 4,\, -1$

Now integrate between $x = -1$ and $x = 4$:

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

Integrate term by term:

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

At $x = 4$: $4(4) + \frac{3(16)}{2} - \frac{64}{3} = 16 + 24 - \frac{64}{3} = 40 - \frac{64}{3} = \frac{120 - 64}{3} = \frac{56}{3}$

At $x = -1$: $4(-1) + \frac{3(1)}{2} - \frac{(-1)^3}{3} = -4 + \frac{3}{2} + \frac{1}{3} = \frac{-24 + 9 + 2}{6} = \frac{-13}{6}$

Total area = $\frac{56}{3} - \left( -\frac{13}{6} \right) = \frac{56}{3} + \frac{13}{6} = \frac{112 + 13}{6} = \frac{125}{6}$