Improper integral

  • Thread starter tony873004
  • Start date
  • #1
tony873004
Science Advisor
Gold Member
1,751
143
evaluate [tex]\int_0^{33} {\left( {x - 1} \right)^{ - 1/5} } \,dx[/tex]
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:
[tex]
\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}
[/tex]
But here's my problem.
If [tex]
\frac{{5\left( { - 1} \right)^{4/5} }}{4}=\frac{5}{4}[/tex] then I'd have the right answer.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited:

Answers and Replies

  • #2
Curious3141
Homework Helper
2,843
87
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 :

[tex]\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[/tex]

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
Gib Z
Homework Helper
3,346
5
Try [tex]\lim_{a\to 0} \left( \int_0^{1-a} (x-1)^{-1/5} dx + \int_{1+a}^{33} (x-1)^{-1/5} dx[/tex].

EDIT: was too slow :(
 
  • #4
tony873004
Science Advisor
Gold Member
1,751
143
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:
[tex]\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}[/tex]
 
  • #5
Gib Z
Homework Helper
3,346
5
In the Reals [itex](-1)^{4/5} = 1^{1/5} = 1[/itex], giving you the correct answer.
 
  • #6
tony873004
Science Advisor
Gold Member
1,751
143
In the Reals [itex](-1)^{4/5} = 1^{1/5} = 1[/itex], giving you the correct answer.
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
Gib Z
Homework Helper
3,346
5
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
tony873004
Science Advisor
Gold Member
1,751
143
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
tony873004
Science Advisor
Gold Member
1,751
143
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
Gib Z
Homework Helper
3,346
5
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
279
0
In the Reals [itex](-1)^{4/5} = 1^{1/5} = 1[/itex], giving you the correct answer.
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
tony873004
Science Advisor
Gold Member
1,751
143
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

Thanks everyone for your helpful replies.
 
  • #13
Gib Z
Homework Helper
3,346
5
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 [itex](-1)^{4/5}[/itex] 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.
 

Related Threads on Improper integral

  • Last Post
Replies
7
Views
801
  • Last Post
Replies
7
Views
764
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
935
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
887
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
576
Top