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

The area bounded by the curve $y = 4 + 3x - x^2$ and the x-axis is found by determining where the curve intersects the x-axis and integrating between those points.

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The curve touches the x-axis when $y = 0$:

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

$x^2 - 3x - 4 = 0$

$(x - 4)(x + 1) = 0$

Therefore $x = 4$ or $x = -1$

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The curve is above the x-axis between $x = -1$ and $x = 4$, so the area is:

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

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

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At $x = 4$:

$4(4) + \frac{3(4)^2}{2} - \frac{(4)^3}{3}$

$= 16 + \frac{48}{2} - \frac{64}{3}$

$= 16 + 24 - \frac{64}{3}$

$= 40 - \frac{64}{3}$

$= \frac{120}{3} - \frac{64}{3}$

$= \frac{56}{3}$

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At $x = -1$:

$4(-1) + \frac{3(-1)^2}{2} - \frac{(-1)^3}{3}$

$= -4 + \frac{3}{2} + \frac{1}{3}$

$= \frac{-24}{6} + \frac{9}{6} + \frac{2}{6}$

$= \frac{-13}{6}$

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$\text{Area} = \frac{56}{3} - \left(\frac{-13}{6}\right)$

$= \frac{56}{3} + \frac{13}{6}$

$= \frac{112}{6} + \frac{13}{6}$

$= \frac{125}{6}$

Therefore, the area bounded by the curve and the x-axis is $\dfrac{125}{6}$ square units.