The area (in sq. units) bounded by the parabo | vedclass

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(A) The equation of the parabola is $y = x^2 - 1$. To find the tangent at $(2, 3)$,we find the derivative: $\frac{dy}{dx} = 2x$. At $x = 2$,the slope

Vedclass · Accepted Answer

$\frac{8}{3}$

Vedclass · Answer

(A) The equation of the parabola is $y = x^2 - 1$. To find the tangent at $(2, 3)$,we find the derivative: $\frac{dy}{dx} = 2x$. At $x = 2$,the slope $m = 2(2) = 4$. The equation of the tangent line is $y - 3 = 4(x - 2)$,which simplifies to $y = 4x - 5$ or $x = \frac{y + 5}{4}$. The parabola can be written as $x = \sqrt{y + 1}$ (for $x > 0$). The tangent intersects the $y$-axis at $x = 0$,where $y = -5$. The area bounded by the parabola,the tangent,and the $y$-axis is given by the integral with respect to $y$ from $y = -1$ to $y = 3$: Area $= \int_{-1}^{3} (x_{	ext{tangent}} - x_{	ext{parabola}}) dy = \int_{-1}^{3} \left( \frac{y + 5}{4} - \sqrt{y + 1} ight) dy$. $= \left[ \frac{y^2}{8} + \frac{5y}{4} ight]_{-1}^{3} - \left[ \frac{2}{3}(y + 1)^{3/2} ight]_{-1}^{3}$. $= \left( (\frac{9}{8} + \frac{15}{4}) - (\frac{1}{8} - \frac{5}{4}) ight) - \left( \frac{2}{3}(4)^{3/2} - 0 ight)$. $= (\frac{39}{8} - (-\frac{9}{8})) - \frac{2}{3}(8) = \frac{48}{8} - \frac{16}{3} = 6 - \frac{16}{3} = \frac{18 - 16}{3} = \frac{2}{3}$. Wait,re-evaluating the integral bounds and region: The area is $\int_{-1}^{3} (x_{	ext{tangent}} - x_{	ext{parabola}}) dy = \frac{2}{3}$. Since the provided options do not match,we re-check the calculation. The area is $\int_{-1}^{3} (\frac{y+5}{4} - \sqrt{y+1}) dy = [\frac{y^2}{8} + \frac{5y}{4}]_{-1}^{3} - [\frac{2}{3}(y+1)^{3/2}]_{-1}^{3} = (\frac{9}{8} + \frac{15}{4} - \frac{1}{8} + \frac{5}{4}) - \frac{16}{3} = (1 + 5) - \frac{16}{3} = 6 - \frac{16}{3} = \frac{2}{3}$. Given the options,there might be a typo in the question or options. Assuming the standard interpretation,the result is $\frac{2}{3}$.

The area (in sq. units) bounded by the parabola $y = x^2 - 1$,the tangent at the point $(2, 3)$ to it,and the $y$-axis is

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