Does the Integral from 8 to 666 of 1/((x^(1/3))-2) Converge or Diverge?

[SteliosVas]

Homework Statement

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∫1/((x^1/3)-2)dx between b=666 and a=8

Aim is to prove divergence or convergence

Homework Equations

The Attempt at a Solution

Okay so I know I can prove for 4x^1/3 < x when 8 < x but I don't know where to go from here. There is obviously very large terminal values, and i know you could say evalute each terminal values using fundamental theorem of calculus..

Any easier way to do it?

 

[bigplanet401]

The integrand is not continuous over [a, b], so you can't use the fundamental theorem of calculus (directly).

Try the substitution u^3 = x and write the integrand in terms of u. Make sure you change the limits of integration, too. Then try v = u - 2. Then things will look a little more obvious.

 

[HallsofIvy]

The original problem appears to be to integrate from 2 to 666. There is, of course, a problem at x= 2 but since the problem does NOT require integration to infinity, I do not understand Ray Vickson's "(ii) whether the function grows small quickly enough to yield a finite integration as the upper limit goes to infinity."

 

[Ray Vickson]

The original problem appears to be to integrate from 2 to 666. There is, of course, a problem at x= 2 but since the problem does NOT require integration to infinity, I do not understand Ray Vickson's "(ii) whether the function grows small quickly enough to yield a finite integration as the upper limit goes to infinity."

I had mid-read the original problem; there is no issue at the lower end $x = 2$. And, somehow, I thought that the OP was trying to figure out whether the integral $\int_ a^{\infty} f(x) \, dx$ exists, by looking at $\int_a^b f(x) \, dx$ for some large, finite $b = 666$ and trying to get a conclusion or insight from that. However, when I go back and re-read the original, I see that no such statement seems to have been made, so the question is actually a more-or-less pointless exercise except, of course, for actually doing the integral, which is elementary, but tedious. Anyway, I deleted my previous post.

 

[HallsofIvy]

But there is an issue at x= 2 where the integrand is not defined.

 

[Mark44]

I had mid-read the original problem; there is no issue at the lower end $x = 2$.

I agree that there is no problem at x = 2, but the original integral is over the interval [8, 666], and there is a problem at x = 8.

And, somehow, I thought that the OP was trying to figure out whether the integral $\int_ a^{\infty} f(x) \, dx$ exists, by looking at $\int_a^b f(x) \, dx$ for some large, finite $b = 666$ and trying to get a conclusion or insight from that. However, when I go back and re-read the original, I see that no such statement seems to have been made, so the question is actually a more-or-less pointless exercise except, of course, for actually doing the integral, which is elementary, but tedious. Anyway, I deleted my previous post.

The original problem appears to be to integrate from 2 to 666.

No, the original integral is $\int_8^{666}\frac {dx}{\sqrt[3]{x} - 2}$

There is, of course, a problem at x= 2 but since the problem does NOT require integration to infinity, I do not understand Ray Vickson's "(ii) whether the function grows small quickly enough to yield a finite integration as the upper limit goes to infinity."

 

[SteliosVas]

Okay essentially all I want to prove is convergence or divergence of the integral over the [8,666]. I know no problem exists for x=2 but rather the problem exists at x=8

Using substitution as u= x^1/3 and x= x³ i get ∫3*u²/(u-2)

I than reapply v = u-2 and and therefore u = v + 2 and integrate respectively. But my challenge is, think it is asking you to use some sort of comparison rather than evaluating the integral to calculate the convergence/divergence? Would the comparison test work?

 

[SteliosVas]

Bump

 

[HallsofIvy]

Once you have the integral $3\int \frac{u^2}{u- 2} du$ do the division to get $3\int (v+ 2+ \frac{4}{v-2})dx$. The first two terms are easily integrable and you can "compare" the last to integrating $\frac{1}{x}$ in the vicinity of x= 0.