The slope of the tangent to a curve y = f(x) | vedclass

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

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$\frac{5}{6}$

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(A) The slope of the tangent is given by $\frac{dy}{dx} = 2x + 1$.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 $y = 2$ to find $C$:$2 = (1)^2 + 1 + C \Rightarrow 2 = 2 + C \Rightarrow C = 0$.Thus,the equation of the curve is $y = x^2 + x$.The region is bounded by the curve $y = x^2 + x$,the $x$-axis $(y = 0)$,and the line $x = 1$. The curve intersects the $x$-axis at $x^2 + x = 0$,which gives $x(x + 1) = 0$,so $x = 0$ and $x = -1$. The region bounded by the curve,the $x$-axis,and the line $x = 1$ is in the interval $[0, 1]$.Required Area $= \int_{0}^{1} (x^2 + x) dx$$= \left[ \frac{x^3}{3} + \frac{x^2}{2} \right]_{0}^{1}$$= \left( \frac{1}{3} + \frac{1}{2} \right) - (0 + 0) = \frac{2 + 3}{6} = \frac{5}{6}$.Therefore,the correct option is $(A)$.

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 of the region bounded by the curve,the $x$-axis,and the line $x = 1$ is:

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