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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area enclosed by the curve $y^2+1=x,x\leq 2$ and the line $x=2$.

$\begin {array} {1 1} (A)\;\large\frac{2}{3}\: sq\: units & \quad (B)\;\large\frac{5}{3}\: sq\: units \\ (C)\;\large\frac{3}{4}\: sq\: units & \quad (D)\;\large\frac{4}{3}\: sq\: units \end {array}$

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Toolbox:
  • 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
answered Dec 22, 2013 by balaji.thirumalai
 

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