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 bounded by the curves \(y = x^2 + 2, y = x, \: x = 0\) and \(x = 3.\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If we are given two curves represented by y=f(x);y=g(x),where $f(x)\geq g(x)$ in [a,b],the points of intersection of two curves are given by x=a and x=b,by taking common values of y from the equation of the two curves.
Given the equation of the curves are:
$y=x^2+2$
$y=x$
$x=0\;and\;x=3.$
Here clearly $y=x^2+2$ represents a parabola with vertex at (0,2) and axis along positive direction of y-axis.The parabola is open upwards.
y=x is a line passing through the origin .
x=0 is y-axis and x=3 is a line parallel to y-axis at a distance of 3units from it.
Area of the required region is bounded by these curves and it is the shaded region as shown in the fig.
Required area is $A=\int_0^3(y_2-y_1)dx$
$A=\int_0^3[(x^2+2)-x]dx$
on integrating we get,
$A=\begin{bmatrix}\frac{x^3}{3}+2x-\frac{x^2}{2}\end{bmatrix}_0^3$
on applying the limits we get,
$A=\begin{bmatrix}9+6-\frac{9}{2}\end{bmatrix}=\frac{21}{2}$sq.units.
Hence the required area is $\frac{21}{2}$ sq.units.
answered Dec 21, 2013 by yamini.v
 

Related questions

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
...