The curvilinear trapezoid is bounded by the c | vedclass

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Question

(C) The curve is $y = x^2 + 1$. The point on the curve is $(x_1, x_1^2 + 1)$.The slope of the tangent at $(x_1, x_1^2 + 1)$ is $\frac{dy}{dx} = 2x_1$.

Vedclass · Accepted Answer

$\left( \frac{3}{2}, \frac{13}{4} \right)$

Vedclass · Answer

(C) The curve is $y = x^2 + 1$. The point on the curve is $(x_1, x_1^2 + 1)$.The slope of the tangent at $(x_1, x_1^2 + 1)$ is $\frac{dy}{dx} = 2x_1$.The equation of the tangent is $y - (x_1^2 + 1) = 2x_1(x - x_1)$,which simplifies to $y = 2x_1x - x_1^2 + 1$.Let the tangent intersect the lines $x=1$ and $x=2$ at points $P$ and $Q$ respectively.At $x=1$,$y_P = 2x_1(1) - x_1^2 + 1 = -x_1^2 + 2x_1 + 1$.At $x=2$,$y_Q = 2x_1(2) - x_1^2 + 1 = -x_1^2 + 4x_1 + 1$.The area of the trapezium formed by the lines $x=1, x=2$,the $x$-axis,and the tangent is $A = \frac{y_P + y_Q}{2} 	imes (2 - 1) = \frac{(-x_1^2 + 2x_1 + 1) + (-x_1^2 + 4x_1 + 1)}{2} = \frac{-2x_1^2 + 6x_1 + 2}{2} = -x_1^2 + 3x_1 + 1$.To maximize the area,we find $\frac{dA}{dx_1} = -2x_1 + 3$. Setting $\frac{dA}{dx_1} = 0$ gives $x_1 = \frac{3}{2}$.For $x_1 = \frac{3}{2}$,$y_1 = \left(\frac{3}{2}ight)^2 + 1 = \frac{9}{4} + 1 = \frac{13}{4}$.Thus,the point is $\left( \frac{3}{2}, \frac{13}{4} ight)$.

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