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

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$\frac{5}{6}$

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(C) Given that the slope of the tangent to the curve $y=f(x)$ 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 the point $(1, 2)$,we substitute $x=1$ and $y=2$ into the equation:$2 = (1)^2 + 1 + c$$2 = 1 + 1 + c$$2 = 2 + c \Rightarrow c = 0$Thus,the equation of the curve is $y = x^2 + x$.The area bounded by the curve,the $X$-axis,and the line $x=1$ is the integral from $x=0$ to $x=1$ (since the curve intersects the $X$-axis at $x^2+x=0$,i.e.,$x(x+1)=0$,giving $x=0$ and $x=-1$):Area $= \int_0^1 (x^2+x) dx$$= \left[ \frac{x^3}{3} + \frac{x^2}{2} ight]_0^1$$= \left( \frac{1^3}{3} + \frac{1^2}{2} ight) - (0)$$= \frac{1}{3} + \frac{1}{2} = \frac{2+3}{6} = \frac{5}{6} 	ext{ sq. units}$.

The slope of the tangent to a curve $y=f(x)$ at $(x, f(x))$ is $2x+1$. If the curve passes through the point $(1,2)$,then the area (in sq. units),bounded by the curve,the $X$-axis and the line $x=1$,is

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