If A is the area in the first quadrant enclos | vedclass

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(A) The curve is $y = 2x^2 + 1$. The tangent at $(1, 3)$ is found by differentiating: $\frac{dy}{dx} = 4x$. At $x = 1$,the slope is $4$. The equation

Vedclass · Accepted Answer

$16$

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(A) The curve is $y = 2x^2 + 1$. The tangent at $(1, 3)$ is found by differentiating: $\frac{dy}{dx} = 4x$. At $x = 1$,the slope is $4$. The equation of the tangent is $y - 3 = 4(x - 1)$,which simplifies to $y = 4x - 1$.The intersection of the tangent $y = 4x - 1$ and the line $x + y = 1$ is found by substituting $y$: $x + (4x - 1) = 1 \implies 5x = 2 \implies x = 2/5$. Then $y = 3/5$. So the intersection point $S$ is $(2/5, 3/5)$.The area $A$ is bounded by the curve $y = 2x^2 + 1$,the tangent $y = 4x - 1$,and the line $y = 1 - x$. The region is bounded between $x = 0$ and $x = 1$. $A = \int_{0}^{1} (2x^2 + 1) dx - 	ext{Area of triangle formed by } (0, 1), (1, 0), (2/5, 3/5) 	ext{ and the tangent line segment.}$Calculating the integral: $\int_{0}^{1} (2x^2 + 1) dx = [\frac{2}{3}x^3 + x]_{0}^{1} = \frac{2}{3} + 1 = \frac{5}{3}$.The area of the region bounded by the curve,the tangent,and the line is calculated as: $A = \int_{0}^{1} (2x^2 + 1) dx - 	ext{Area}(	riangle TQS) - 	ext{Area}(	ext{trapezoid under tangent})$.Alternatively,using the integral of the upper curve minus the lower boundaries: $A = \int_{0}^{2/5} (2x^2 + 1 - (1 - x)) dx + \int_{2/5}^{1} (2x^2 + 1 - (4x - 1)) dx = \int_{0}^{2/5} (2x^2 + x) dx + \int_{2/5}^{1} (2x^2 - 4x + 2) dx$.$= [\frac{2}{3}x^3 + \frac{x^2}{2}]_{0}^{2/5} + [\frac{2}{3}x^3 - 2x^2 + 2x]_{2/5}^{1} = (\frac{2}{3} \cdot \frac{8}{125} + \frac{1}{2} \cdot \frac{4}{25}) + ((\frac{2}{3} - 2 + 2) - (\frac{2}{3} \cdot \frac{8}{125} - 2 \cdot \frac{4}{25} + 2 \cdot \frac{2}{5})) = \frac{16}{375} + \frac{2}{25} + \frac{2}{3} - \frac{16}{375} + \frac{8}{25} - \frac{4}{5} = \frac{10}{25} + \frac{2}{3} - \frac{4}{5} = \frac{2}{5} + \frac{2}{3} - \frac{4}{5} = \frac{6 + 10 - 12}{15} = \frac{4}{15}$.$60A = 60 	imes \frac{4}{15} = 16$.

If $A$ is the area in the first quadrant enclosed by the curve $C: 2x^2 - y + 1 = 0$,the tangent to $C$ at the point $(1, 3)$ and the line $x + y = 1$,then the value of $60A$ is

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