Consider the functions f(x) and g(x), both defined from `R rarrR` and are defined as `f(x)=2x-x^(2) and g(x)=x^(n)` where `n in N`. If the area between f(x) and g(x) is 1/2, then the value of n is

Explanation

`y=2x-x^(2)` is downward parabola intersecting x-axis at x = 0 and x = 2
Solving `f(x)=2x-x^(2) and g(x)=x^(n)`
we have `2x-x^(2)=x^(n)`
`rArr" "x=0 and x=1`
`A=int_(0)^(1)(2x-x^(2)-x^(n))dx=[x^(2)-(x^(3))/(3)-(x^(n+1))/(n+1)]_(0)^(1)`

`1-(1)/(3)-(1)/(n+1)=(2)/(3)-(1)/(n+1)`
hence, `(2)/(3)-(1)/(n+1)=(1)/(2)rArr(2)/(3)-(1)/(2)=(1)/(n+1)`
`rArr" "n=5`