# f(x) = 1/(x^0.6) , 1 < or = x < infinity Find the area between this curve and the x-axis, if it is finite. Find the volume of the solid obtained by the revolution of this curve about the...

f(x) = 1/(x^0.6) , 1 < or = x < infinity

Find the area between this curve and the x-axis, if it is finite.

Find the volume of the solid obtained by the revolution of this curve about the x-axis, if it is finite

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### 1 Answer

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.**