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Home  >>  CBSE XII  >>  Math  >>  Application of Integrals
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Find the area between the curves \(y = x\) and \(y = x^2.\)

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2 Answers

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Suppose we are given two curves represented by y=f(x),y=g(x) where $f(x)\geq g(x)$ in [a,b].
Here the points of intersection of these two curves are given by x=a and y=b obtained by taking common values of y from the given equation of two curves.
Hence dA=[f(x)-g(x)]dx and the total area A can be taken as $A=\int_a^b[f(x)-g(x)]dx$.
Given two curves f(x)=x and g(x)=$x^2$.
To find the limits,let us find the points of interaction of the curves.
since y=x and y=x^2
on equating we get x=$x^2$
$(x^2-x)=0$
x(x-1)=0
x=0 or x=1
Hence the points of intersection are (0,0) and(1,1)
Hence,the limits can be taken as x=0 and x=1
Now the required area can be found by using the information in the tool box:
$A=\int_a^b[f(x)-g(x)dx]$
Here a=0,b=1 f(x)=x and $g(x)=x^2$
Therefore \begin{array}{1 1}A=\int_0^1[x-x^2]dx \\ =\begin{bmatrix}\frac{x^2}{2}-\frac{x^3}{3}\end{bmatrix}_0^1\\=\frac{1}{6}sq.units\end{array}
Hence the required area is $\frac{1}{6}$sq.units.

 

answered Jan 21, 2013 by sreemathi.v
 
0 votes
Given two curves f(x)=x and g(x)=$x^2$.
To find the limits,let us find the points of interaction of the curves.
since y=x and y=x^2
on equating we get x=$x^2$
$(x^2-x)=0$
x(x-1)=0
x=0 or x=1
Hence the points of intersection are (0,0) and(1,1)
Hence,the limits can be taken as x=0 and x=1
Now the required area can be found by using the information in the tool box:
$A=\int_a^b[f(x)-g(x)dx]$
Here a=0,b=1 f(x)=x and $g(x)=x^2$
Therefore \begin{array}{1 1}A=\int_0^1[x-x^2]dx \\ =\begin{bmatrix}\frac{x^2}{2}-\frac{x^3}{3}\end{bmatrix}_0^1\\=\frac{1}{6}sq.units\end{array}
Hence the required area is $\frac{1}{6}$sq.units.
answered Dec 21, 2013 by yamini.v
 

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