Answer
$ \frac{8}{3} $
Work Step by Step
Step 1 of 2:
Consider the function $ y=x^{2},y=4x-x^{2}. $
The region enclosed by the curves $ y=x^{2},y=4x-x^{2} $
is shown below:
Step 2 of 2:
From the above graph the point of intersection of the curves $ y=x^{2},y=4x-x^{2} $
are $ \left( 0,0 \right) and $
$ \left( 2,4 \right) $
.
Consider
$ y=4x-x^{2}=f \left( x \right) $
$ y=x^{2}=g \left( x \right) $
The area between the curves $ y=f \left( X \right) ,y=g \left( x \right)$ and between x=a and x=b is
$ A= \int _{0}^{2} \vert f \left( x \right) -g \left( x \right) \vert dx $
.
Here the area of the region bounded by the curves $ y=x^{2},y=4x-x^{2} $ and between $ x=0 $ and $ x=2 $ is
$ A= \int _{0}^{2} \vert 4x-x^{2}-x^{2} \vert dx $
$ = \vert 2x^{2}-\frac{2x^{3}}{3} \vert _{0}^{2} $
$ = \vert \left( 2 \left( 2 \right) ^{2}-\frac{2 \left( 2 \right) ^{3}}{3} \right) - \left( 2 \left( 0 \right) ^{2}-\frac{2 \left( 0 \right) ^{3}}{3} \right) \vert $
$ =\frac{8}{3} $
Hence the required area of the given region is $ \frac{8}{3} $.
