The area (in sq. units) in the first quadrant | vedclass

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(B) The equation of the parabola is $y = x^2 + 1$.To find the tangent at $(2, 5)$,we find the derivative: $\frac{dy}{dx} = 2x$.At $x = 2$,the slope $m

Vedclass · Accepted Answer

$\frac{37}{24}$

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(B) The equation of the parabola is $y = x^2 + 1$.To find the tangent at $(2, 5)$,we find the derivative: $\frac{dy}{dx} = 2x$.At $x = 2$,the slope $m = 2(2) = 4$.The equation of the tangent is $y - 5 = 4(x - 2)$,which simplifies to $y = 4x - 3$.The tangent intersects the $x$-axis at $y = 0$,so $4x - 3 = 0$,which gives $x = \frac{3}{4}$.The required area is the area under the parabola from $x = 0$ to $x = 2$ minus the area of the triangle formed by the tangent line,the $x$-axis,and the vertical line $x = 2$.Area $= \int_{0}^{2} (x^2 + 1) dx - 	ext{Area of triangle}$.Area of triangle $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2 - \frac{3}{4}) 	imes 5 = \frac{1}{2} 	imes \frac{5}{4} 	imes 5 = \frac{25}{8}$.Integral part $= [\frac{x^3}{3} + x]_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}$.Required Area $= \frac{14}{3} - \frac{25}{8} = \frac{112 - 75}{24} = \frac{37}{24}$ sq. units.

The area (in sq. units) in the first quadrant bounded by the parabola $y = x^2 + 1$,the tangent to it at the point $(2, 5)$,and the coordinate axes is

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