Answer:
The area of the searched region is [tex]A= a+b+ \frac{2n}{n+1}- \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1}-2[/tex]
Step-by-step 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)
[tex]A(a<x<1)=\int\limits^1_a {\int\limits^{x^{\frac{1}{n}} }_{x} } {} \, dy } \, dx = \int\limits^1_a {x^{\frac{1}{n} } -x \, dx = a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A(1<x<b)=\int\limits^b_1 {\int\limits^{x}_{x^{\frac{1}{n} } } {} \, dy } \, dx = \int\limits^b_1 {x-x^{\frac{1}{n} } \, dx =b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1}[/tex]
[tex]A=a-1 +\frac{n}{n+1} - \frac{na^{\frac{n+1}{n} } }{n+1} + b-1 + \frac{n}{n+1} - \frac{nb^{\frac{n+1}{n} } }{n+1} = a+b+ \frac{2n}{n+1} - \frac{n(a^{\frac{n+1}{n} }+b^{\frac{n+1}{n} }) }{n+1} -2[/tex]
