# Improper integral

1. Oct 20, 2007

### tony873004

evaluate $$\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx$$
The back of the book gives 75/4 as the answer. I get 80/4. It has an asymptote at 1 since 0^(-1/5) DNE, and 0.00000001^(1/5) equals a number that grows larger as I add more 0's.

But any time I try to raise a negative number ^(1/5) I get an error. It just doesn't seem like 0-1 is in the Domain of this function. Even when I graph it, I see it shoot down from the x=1 asymptote, and then make a hard turn right as it approaches the x-axis. I don't see any Range in this function below x=1.

But somehow, the integral from 0 to 1 must equal -5/4 for my answer to match the back of the book's.

Here's my effort:
$$\begin{array}{l} \int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx \\ \\ \,\,\,\,\,\,\,\,\,\,u = x - 1,\,\,\frac{{du}}{{dx}} = 1,\,\,dx = du \\ \\ \int_*^* {u^{ - 1/5} } \,du = \mathop {\lim }\limits_{t \to 0^ + } \int_*^* {u^{ - 1/5} } \,du \\ \\ \mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \,\, = \,\,\mathop {\lim }\limits_{t \to 0^ + } \left. {\frac{{u^{4/5} }}{{4/5}}} \right|_*^* \left. { = \mathop {\lim }\limits_{t \to 0^ + } \frac{{5\left( {x - 1} \right)^{4/5} }}{4}} \right|_t^{33} = \mathop {\lim }\limits_{t \to 0^ + } \left[ {\left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {t - 1} \right)^{4/5} }}{4}} \right)} \right] = \\ \\ \left( {\frac{{5\left( {33 - 1} \right)^{4/5} }}{4}} \right) - \left( {\frac{{5\left( {0^ + - 1} \right)^{4/5} }}{4}} \right) = \frac{{5 \cdot 16}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} = \frac{{80}}{4} - \frac{{5\left( { - 1} \right)^{4/5} }}{4} \\ \end{array}$$
But here's my problem.
If $$\frac{{5\left( { - 1} \right)^{4/5} }}{4}=\frac{5}{4}$$ then I'd have the right answer.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 20, 2007
2. Oct 20, 2007

### Curious3141

If you restrict the range of the function to reals, then f(x) = x^(4/5) will give f(-1) = 1. Negative numbers can have real odd roots.

But I still can't follow your working. In the original integral, your asymptote occurs at x = 1. You should be splitting up your integral like so :

$$\int_{0}^{33}(x-1)^{-\frac{1}{5}}dx = \lim_{t \rightarrow 1}\int_{0}^{t}(x-1)^{-\frac{1}{5}}dx + \lim_{t \rightarrow 1}\int_{t}^{33}(x-1)^{-\frac{1}{5}}dx$$

then working out the individual integrals.

You decided to make the substitution, which changes the bounds and the value of the asymptote. In this case, the bounds go from u = -1 to u = 32, and the value of t becomes 0. You still get the same answer.

3. Oct 20, 2007

### Gib Z

Try $$\lim_{a\to 0} \left( \int_0^{1-a} (x-1)^{-1/5} dx + \int_{1+a}^{33} (x-1)^{-1/5} dx$$.

EDIT: was too slow :(

4. Oct 20, 2007

### tony873004

Thanks for the replies :)

I know that substitution changes the bounds. In my examples, I just put astericks on the bounds after converting to du, so I don't have to compute them. I switch back to dx before solving which is why I don't bother to compute the new bounds.

I tried splitting it up from 0 to 1 and 1 to 33, but I ran into the exact same problem. Here's my effort. This is only the 0-1 part. The 1 to 33 part I get 80/4:
$$\begin{array}{l} \int_0^1 {\left( {x - 1} \right)^{ - 1/5} \,dx} = \mathop {\lim }\limits_{t \to 1^ - } \int_0^1 {\left( {x - 1} \right)^{ - 1/5} \,dx} = \mathop {\lim }\limits_{t \to 1^ - } \left( {\frac{5}{4}\left( {\left( {t - 1} \right)^{4/5} - \left( { - 1} \right)^{4/5} } \right)} \right) = \\ \\ \frac{5}{4}\left( {0 - \left( { - 1} \right)^{4/5} } \right) = \frac{5}{4}\left( { - \left( { - 1} \right)^{4/5} } \right) = \frac{{ - 5\left( { - 1} \right)^{4/5} }}{4} \\ \end{array}$$

5. Oct 20, 2007

### Gib Z

In the Reals $(-1)^{4/5} = 1^{1/5} = 1$, giving you the correct answer.

6. Oct 21, 2007

### tony873004

Thanks. I'll take your word for it, but how do you get that? I just looked in a pre-calc textbook about imaginary numbers and didn't find anything that allowed me to draw that conclusion.

7. Oct 21, 2007

### Gib Z

haha well you did look at a pre calc textbook about imaginary numbers and this integral is talking completely about the reals. Your integral is over the real interval [0,33], no complex numbers.

8. Oct 21, 2007

### tony873004

ok, but if I type (-1)^(4/5) into Google, it tells me (-1)^(4 / 5) = -0.809016994 + 0.587785252 i . I guess that's where I got confused. My TI-84 tells me (-1)^(4/5)=1 like you say. My computer eval calculator gives me an error. As long as my TI-84 says it's so, I guess I don't have to justify that step. But I'm still curious: how do you get 1^(1/5) from (-1)^(4/5)? And why does Google calculator want to give me the answer in imaginary form?

9. Oct 21, 2007

### tony873004

and also the online graphing calculator refuses to plot anything less than x=1. How can there be an area when there's no curve?

10. Oct 21, 2007

### Gib Z

Are you checking all quadrants? Because graphamatica has no problem graphing it...and the inbuilt numerical integrator confirms the 75/4 answer to 3 sig figs.

As a side exercise, note that the (x-1) arguement rather than x just shifts the graph 1 unit to the right. So graph y=(x-1)^(-1/5). It is the same curve as xy^5 =1.

What that tells me is that its a similar shape to a rectangular hyperbola, but the y axis becomes as asmytote much faster than the x axis because of the higher power of y. To see what i mean, graph it on some program.

11. Oct 21, 2007

### coomast

Is it correct to say that you used the following steps:

(-1)^(4/5)=[(-1)^4]^(1/5)=[1]^(1/5)=1

This is tricky. I would start with (-1)^(1/5). This presents 5 different numbers. These are approximately (0.809+i0.588); (0.809-i0.588); (-0.309+i0.951); (-0.309-i0.951) and -1.
So, only one real number. Raising these to the power 4 gives approx. (-0.809+i0.588); (-0.809-i0.588); (0.309-i0.951); (0.309+i0.951) and 1. The outcome is the same, only
one real number. If this is used in the integral you get the correct answer 75/4. This way of working gives a detailed view on what is happening, albeit a way bit more work.
Also the reason for the different results with the different methods.

Integral solved, case closed.

12. Oct 22, 2007

### tony873004

I asked in office hours. The teacher pointed out that if I treat -0.809016994 + 0.587785252 i, with the 1st number on the x-axis, and the 2nd number on the y-axis, and then use pothagorean theorum, I'd get 1. She was right:
sqrt(0.809016994^2 + 0.587785252^2)=1

That is true but you shouldn't have needed to go there at all! It is a completely real integral, and the only solution in the reals for $(-1)^{4/5}$ is 1 !! This kind of thinking might harm you when doing functions that are one to one in the reals, but not so in the complex numbers such as the exponential function.