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(C) Given the slope of the function is $\frac{dy}{dx} = 2x + 1$. Integrating both sides with respect to $x$,we get: $y = \int (2x + 1) dx = x^2 + x +

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$1/6 \, 	ext{sq. unit}$

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(C) Given the slope of the function is $\frac{dy}{dx} = 2x + 1$. Integrating both sides with respect to $x$,we get: $y = \int (2x + 1) dx = x^2 + x + c$. Since the curve passes through $(1, 2)$,we substitute $x = 1$ and $y = 2$: $2 = (1)^2 + 1 + c \implies 2 = 2 + c \implies c = 0$. Thus,the equation of the curve is $y = x^2 + x$. To find the area bounded by the curve and the $x$-axis,we find the points where the curve intersects the $x$-axis by setting $y = 0$: $x^2 + x = 0 \implies x(x + 1) = 0 \implies x = 0, x = -1$. The area is given by the integral: $	ext{Area} = \left| \int_{-1}^{0} (x^2 + x) dx ight| = \left| \left[ \frac{x^3}{3} + \frac{x^2}{2} ight]_{-1}^{0} ight|$. Evaluating the definite integral: $	ext{Area} = \left| (0) - \left( \frac{(-1)^3}{3} + \frac{(-1)^2}{2} ight) ight| = \left| - \left( -\frac{1}{3} + \frac{1}{2} ight) ight| = \left| - \left( \frac{1}{6} ight) ight| = \frac{1}{6} \, 	ext{sq. unit}$.

Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x$-axis is

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