The area bounded by the curve y = x^2 + 1 and | vedclass

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(A) Let the point of tangency be $(x_0, y_0) = (x_0, x_0^2 + 1)$.The slope of the tangent at $(x_0, x_0^2 + 1)$ is $\frac{dy}{dx} = 2x_0$.The equation

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$\frac{2}{3}$

Vedclass · Answer

(A) Let the point of tangency be $(x_0, y_0) = (x_0, x_0^2 + 1)$.The slope of the tangent at $(x_0, x_0^2 + 1)$ is $\frac{dy}{dx} = 2x_0$.The equation of the tangent line is $y - (x_0^2 + 1) = 2x_0(x - x_0)$.Since the tangent passes through the origin $(0, 0)$,we have:$0 - (x_0^2 + 1) = 2x_0(0 - x_0)$$-x_0^2 - 1 = -2x_0^2$$x_0^2 = 1 \implies x_0 = \pm 1$.For $x_0 = 1$,the tangent is $y = 2x$. For $x_0 = -1$,the tangent is $y = -2x$.The area $A$ bounded by the curve $y = x^2 + 1$ and the tangents $y = 2x$ and $y = -2x$ is symmetric about the $y$-axis.$A = 2 \int_{0}^{1} ((x^2 + 1) - 2x) dx$$A = 2 \int_{0}^{1} (x - 1)^2 dx$$A = 2 \left[ \frac{(x - 1)^3}{3} \right]_{0}^{1}$$A = 2 \left( 0 - \frac{(-1)^3}{3} \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$.

The area bounded by the curve $y = x^2 + 1$ and the tangents to it drawn from the origin is

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