A curve y = f(x) is such that `f(x)ge 0 and f(0)=0` and bounds a curvilinear triangle with the base [0,x] whose area is proportional to `(n+1)^(th)` power of `f(x)cdot" If "f(1)=1` then find f(x).

Answer

`x^(1//n)`

Explanation

According to the question `overset(x)underset(0)intf(x)dx=lamda{f(x)}^(n+1)`
Where `lamda` is constant of proportionality
Differentiating both sides w.r.t x,
`f(x)=lamda(n+1)(f(x))^(n)f'(x)`
`"or "(f(x)^(n-1))f'(x)=(1)/(lamda(n+1))`
Integrating both sides w.r.t x, `((f(x))^(n))/(n)=(x)/(lamda(n+1))+C`
`f(0)=0, therefore C=0`
`(f(x))^(n)=(nx)/(lamda(n+1))`
`f(1)=1`
`therefore" "(n)/(lamda(n+1))=1`
`therefore" "f(x)=x^(1//n)`