logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
0 votes

Find the area of the region enclosed by the parabola $x^2=y$ and the line $y=x+2$.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • 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})$$\Rightarrow (0,\large\frac{1}{4}$)
$y=x+2$ is a straight line.
Step 2:
The points of intersection can be obtained by solving the two equations.
$x^2=x+2\Rightarrow x^2-x-2=0$
On factorising we get,
$(x-2)(x+1)=0$
Hence $x=2$ or $-1$
The points of intersection are $(2,4)(-1,1)$
Step 3:
The required area is the shaded portion shown in the fig.
Area of the shaded portion can be evaluated
$A=\int_{-1}^2(y_2-y_1)dx.$
Where $y_2=x+2$ and $y_1=x^2$
Hence $A=\int_{-1}^2[(x+2)-x^2]dx.$
Step 4:
On integrating we get,
$A=\begin{bmatrix}(\large\frac{x^2}{2}\normalsize+2x)-\large\frac{x^3}{3}\end{bmatrix}_{-1}^2$
On applying limits we get,
$A=[(2+4)-\large\frac{1}{2}$$-2)]-[\large\frac{8}{3}+\frac{1}{3}]$$=6+\large\frac{3}{2}$$-3$
$\;\;=3+\large\frac{3}{2}=\frac{9}{2}$sq.units.
answered Apr 26, 2013 by sreemathi.v
edited Dec 22, 2013 by balaji.thirumalai
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...