Question

Find the area of the following regions, expressing your results in terms of the positive integer n ≥ 2. The region bounded by f(x)=x and g(x)=x^1/n, for x≥0

Answer 1

The area of the searched region is
`A= a+b+ (2n)/(n+1)- \frac{n(a^{(n+1)/(n) }+b^{(n+1)/(n) }) }{n+1}-2`

Explanation:

If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.

In this case, for 0<x<1, f(x)<g(x)

while for 1<x, g(x)<f(x).

Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:

A=A(a<x<1)+A(1<x<b)

`A(a<x<1)=\int\limits^1_a {\int\limits^{x^{(1)/(n)} }_(x) } {} \, dy } \, dx = \int\limits^1_a {x^{(1)/(n) } -x \, dx = a-1 +(n)/(n+1) - \frac{na^{(n+1)/(n) } }{n+1}`

`A(1<x<b)=\int\limits^b_1 {\int\limits^(x)_{x^{(1)/(n) } } {} \, dy } \, dx = \int\limits^b_1 {x-x^{(1)/(n) } \, dx =b-1 + (n)/(n+1) - \frac{nb^{(n+1)/(n) } }{n+1}`

`A=a-1 +(n)/(n+1) - \frac{na^{(n+1)/(n) } }{n+1} +  b-1 + (n)/(n+1) - \frac{nb^{(n+1)/(n) } }{n+1} = a+b+ (2n)/(n+1)  - \frac{n(a^{(n+1)/(n) }+b^{(n+1)/(n) }) }{n+1}   -2`