Let `f(x)=M a xi mu m{x^2,(1-x)^2,2x(1-x)},` where `0lt=xlt=1.` Determine the area of the region bounded by the curves `y=f(x),x-a xi s ,x=0,` and `x=1.`

Answer

`(17)/(27)` sq unit

Explanation

We can draw the graph of `y=x^(2), y=(1-x^(2)) and y=2x(1-x)` in following figure

Now, to get the point of intersection of `y=x^(2) and y=2x(1-x),` we get
`x^(2)=2x(1-x)`
`rArr 3x^(2)=2x`
`rArr x(3x-2)=0`
`rArr x=0, 2//3`
Similarly, we can find the coordinate of the points of intersection of
`y=(1-x^(2)) and y=2x(1-x) " are " x=1//3 and x=1`
From the figure , it is clear that,
`f(x)={((1-x)^(2)",",if 0le x le 1 le 1//3), (2x(1-x)",", if 1//3 le x le 2//3),(x^(2)",",if 2//3 le x le 1):}`
`therefore` The required area
`A=int_(0)^(1//3) f(x)dx`
`=int_(0)^(1//3)(1-x)^(2)dx+int_(1//3)^(2//3) 2x(1-x)dx+int_(2//3)^(1)x^(2)dx`
`=[-(1)/(3)(1-x)^(3)]_(0)^(1//3)+[x^(2)-(2x^(3))/(3)]_(1//3)^(2//3)+[(1)/(3) x^(3)]_(2//3)^(1)`
`=[-(1)/(3)((2)/(3))^(3)+(1)/(3)]+[((2)/(3))^(2)-(2)/(3)((2)/(3))^(3)-((1)/(3))^(2)+(2)/(3)((1)/(3))^(3)]+[(1)/(3)(1)-(1)/(3)((2)/(3))^(3)]`
`=(19)/(81)+(13)/(81)+(19)/(81)=(17)/(27)` sq unit