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Q)

Draw a rough sketch of the curve $y^2+1=x,x\leq 2$.Find the area enclosed by the curve and the line $x=2$.

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

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A)
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  • Area of the region bounded between a curve and a line is given by \[A=\int_a^by\;dx=\int_a^bf(x)\;dx\]
  • where a and b are the point of intersection of the line and the curve.
  • The point of intersection can be found by solving the two equations.
The area of the required region is the shaded part of the curve below $y^2 + 1 = x$, intercepted by the line $x=2$
This is the shaded portion shown in the fig.
clearly the point of intersection is (1,0) with the x-axis and (2,1) and (2,-1)
Area of the region bounded between a curve and a line is given by $A=\large \int_a^b$$y\;dx=\large \int_1^2 $$y \; dx$
$A = 2 \times \large \int_1^2$$ \sqrt (x^2-1)\;dx$
$A = 2 \Large \frac{[(x-1)^{3/2}]_1^2}{3/2}$
Applying Limits, $A = 2 \times \large \frac{2}{3}$$ \times (1)^{3/2}$
$\Rightarrow A = \large\frac{4}{3}$ sq units
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