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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area of the region enclosed by the parabola $x^2=y$ and the line $y=x+2$.

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  • Let $f(x)$ be a continuous function defined on $[a,b]$.Then the area bounded by the curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $x=b$ is given by
  • $A=\int_a^bf(x) dx$ or $\int_a^b ydx.$
Step 1:
Given $x^2=y$ and the line $y=x+2$
Consider the curve $x^2=y$
This is a parabola which is open upwards.
The focus of the parabola is $(4a=1\Rightarrow a=\large\frac{1}{4})$$\Rightarrow (0,\large\frac{1}{4}$)
$y=x+2$ is a straight line.
Step 2:
The points of intersection can be obtained by solving the two equations.
$x^2=x+2\Rightarrow x^2-x-2=0$
On factorising we get,
Hence $x=2$ or $-1$
The points of intersection are $(2,4)(-1,1)$
Step 3:
The required area is the shaded portion shown in the fig.
Area of the shaded portion can be evaluated
Where $y_2=x+2$ and $y_1=x^2$
Hence $A=\int_{-1}^2[(x+2)-x^2]dx.$
Step 4:
On integrating we get,
On applying limits we get,
answered Apr 26, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai
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