f(x) = 1/(x^0.6) , 1

To calculate the area, we'll have to evaluate the improper
integral of the function f(x) = 1/(x^0.6) = 1/x^(6/10) =
1/x^(3/5)

y
`oo`

`int`
dx/x^(3/5) = lim     `int` dx/x^(3/5)

1                    
y->`oo` 1

We'll evaluate the definte integral using
Leibniz Newton
formula:

y

`int` dx/x^(3/5) =
F(y) - F(1)

1

We'll calculate
the indefinite integral of the function:

`int` x^(-3/5)dx =
x^(-3/5 + 1)/(-3/5 + 1) + c

`int` x^(-3/5)dx = 5x^(2/5)/2 +
c

F(y) - F(1) = 5y^(2/5)/2 -
5/2

Now, we'll evaluate the
limit:

lim [F(y) - F(1)] = lim F(y) - lim
F(1)

y->`oo` y->`oo`
y->`oo`

lim [F(y) - F(1)] = `oo` -
5/2

y->`oo`

lim [F(y) -
F(1)] =
`oo`

y->`oo`

Since
the limit is infinite, the integral is divergent and the area is not
finite.