# Thread: Why Integral 1/x^4 from -1 to 2 is undefined?

1. ## Why Integral 1/x^4 from -1 to 2 is undefined?

$\displaystyle \int \frac {1}{x^4} dx$ from -1 to 2

My teacher said this is undefined

I did a calculation and it actually gave me somthing.

$\displaystyle \int \frac {1}{x^4} dx$
$\displaystyle = \int x^{-4}dx$
$\displaystyle =\frac 1{3} x^{-3}$

$\displaystyle f(x) = F(b) - F(a)$
$\displaystyle = [\frac 1{3} (2)^{-3}] - [\frac 1{3} (-1)^{-3}]$
$\displaystyle = - \frac 8{3} + \frac 1{3}$
$\displaystyle = -\frac 7{3}$

My question is why the answer is undefined and not $\displaystyle -\frac 7{3}$?

2. Originally Posted by cammywhite
$\displaystyle \int \frac {1}{x^4} dx$ from -1 to 2

My teacher said this is undefined

I did a calculation and it actually gave me somthing.

$\displaystyle \int \frac {1}{x^4} dx$
$\displaystyle = \int x^{-4}dx$
$\displaystyle =\frac 1{3} x^{-3}$

$\displaystyle f(x) = F(b) - F(a)$
$\displaystyle = [\frac 1{3} (2)^{-3}] - [\frac 1{3} (-1)^{-3}]$
$\displaystyle = - \frac 8{3} + \frac 1{3}$
$\displaystyle = -\frac 7{3}$

My question is why the answer is undefined and not $\displaystyle -\frac 7{3}$?
I would imagine it is because 1/x^4 is undefined at x=0

3. Originally Posted by e^(i*pi)
I would imagine it is because 1/x^4 is undefined at x=0
Would this be undefined too?

$\displaystyle \int \frac {1}{x^4} dx$ from 1 to 2

It is from 1 to 2 which 0 is not included

4. Originally Posted by e^(i*pi)
I would imagine it is because 1/x^4 is undefined at x=0
A really great example is the following

$\displaystyle \int_{-1}^1 \frac{dx}{x^2} = - \left. \frac{1}{x} \right|_{-1}^1 = -2$. However $\displaystyle \frac{1}{x^2} > 0$ so $\displaystyle \int_{-1}^1 \frac{dx}{x^2} > 0$ but we got -2. Something is going on!

5. Originally Posted by danny arrigo
A really great example is the following

$\displaystyle \int_{-1}^1 \frac{dx}{x^2} = - \left. \frac{1}{x} \right|_{-1}^1 = -2$. However $\displaystyle \frac{1}{x^2} > 0$ so $\displaystyle \int_{-1}^1 \frac{dx}{x^2} > 0$ but we got -2. Something is going on!
So can I think it this way... whenever a= -1 and x is the denominator, the answer is undefined?

6. Originally Posted by cammywhite
So can I think it this way... whenever a= -1 and x is the denominator, the answer is undefined?
No.

Whenever the integrand is undefined for values in the interval of integration then there is going to be problems and special care is required. This is an example of an improper integral.

7. Ok... I will understand better if I can see a pattern...which of these are undefined

1. $\displaystyle \int \frac {1}{x^4} dx$ from -1 to 2 <--- undefined
2. $\displaystyle \int \frac {1}{x^4} dx$ from 1 to 2
3. $\displaystyle \int \frac {1}{x^3} dx$ from -1 to 2
4. $\displaystyle \int \frac {1}{x^3} dx$ from 1 to 2

8. Originally Posted by cammywhite
Ok... I will understand better if I can see a pattern...which of these are undefined

1. $\displaystyle \int \frac {1}{x^4} dx$ from -1 to 2 <--- undefined
2. $\displaystyle \int \frac {1}{x^4} dx$ from 1 to 2
3. $\displaystyle \int \frac {1}{x^3} dx$ from -1 to 2
4. $\displaystyle \int \frac {1}{x^3} dx$ from 1 to 2
Well, which of them do you think has an integrand that is undefined for a value of x in the interval of integration ....?

9. Originally Posted by mr fantastic
Well, which of them do you think has an integrand that is undefined for a value of x in the interval of integration ....?
1. and 2. both end up with the number $\displaystyle -\frac {7}3$

According to #4, since $\displaystyle \frac{1}{x^4}> 0$, so both are undefined.

Is my thinking correct?

10. Originally Posted by cammywhite
1. and 2. both end up with the number $\displaystyle -\frac {7}3$

According to #4, since $\displaystyle \frac{1}{x^4}> 0$, so both are undefined.

Is my thinking correct?
Is 1/x^3 defined for all values of x in [-1, 2]??

Do you understand what I'm saying when I said:

Whenever the integrand is undefined for values in the interval of integration then there is going to be problems and special care is required
and

which of them do you think has an integrand that is undefined for a value of x in the interval of integration ....?

11. Originally Posted by mr fantastic
Is 1/x^3 defined for all values of x in [-1, 2]??

Do you understand what I'm saying when I said:

and
ok, I think I am getting it

1/x^4 is NOT defined for all values of x in [-1, 2], 0 is not defined
1/x^4 is defined for all values of x in [1, 2]

1/x^3 is NOT defined for all values of x in [-1, 2], 0 is not defined
1/x^3 is defined for all values of x in [1, 2]

12. Originally Posted by cammywhite
ok, I think I am getting it

1/x^4 is NOT defined for all values of x in [-1, 2], 0 is not defined
1/x^4 is defined for all values of x in [1, 2]

1/x^3 is NOT defined for all values of x in [-1, 2], 0 is not defined
1/x^3 is defined for all values of x in [1, 2]
You've got the idea I see.

13. Originally Posted by mr fantastic
You've got the idea I see.
If I am not mistaken, I can also say $\displaystyle \int \frac 1{x^2-1}$ from 0 to 2 is undefined because x is undefined when x = 1

14. Originally Posted by cammywhite
If I am not mistaken, I can also say $\displaystyle \int \frac 1{x^2-1}$ from 0 to 2 is undefined because x is undefined when x = 1
Yes, that's right

15. Originally Posted by cammywhite
If I am not mistaken, I can also say $\displaystyle \int \frac 1{x^2-1}$ from 0 to 2 is undefined because x is undefined when x = 1