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Math Help - Why Integral 1/x^4 from -1 to 2 is undefined?

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

    \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.

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

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

    My question is why the answer is undefined and not  -\frac 7{3}?
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  2. #2
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    Quote Originally Posted by cammywhite View Post
    \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.

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

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

    My question is why the answer is undefined and not  -\frac 7{3}?
    I would imagine it is because 1/x^4 is undefined at x=0
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    I would imagine it is because 1/x^4 is undefined at x=0
    Would this be undefined too?

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

    It is from 1 to 2 which 0 is not included
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    I would imagine it is because 1/x^4 is undefined at x=0
    A really great example is the following

     \int_{-1}^1 \frac{dx}{x^2} = - \left. \frac{1}{x} \right|_{-1}^1 = -2. However \frac{1}{x^2} > 0 so \int_{-1}^1 \frac{dx}{x^2} > 0 but we got -2. Something is going on!
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    Quote Originally Posted by danny arrigo View Post
    A really great example is the following

     \int_{-1}^1 \frac{dx}{x^2} = - \left. \frac{1}{x} \right|_{-1}^1 = -2. However \frac{1}{x^2} > 0 so \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?
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    Quote Originally Posted by cammywhite View Post
    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.
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    Ok... I will understand better if I can see a pattern...which of these are undefined

    1. \int \frac {1}{x^4} dx from -1 to 2 <--- undefined
    2. \int \frac {1}{x^4} dx from 1 to 2
    3. \int \frac {1}{x^3} dx from -1 to 2
    4. \int \frac {1}{x^3} dx from 1 to 2
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  8. #8
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    Quote Originally Posted by cammywhite View Post
    Ok... I will understand better if I can see a pattern...which of these are undefined

    1. \int \frac {1}{x^4} dx from -1 to 2 <--- undefined
    2. \int \frac {1}{x^4} dx from 1 to 2
    3. \int \frac {1}{x^3} dx from -1 to 2
    4. \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 ....?
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    Quote Originally Posted by mr fantastic View Post
    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 -\frac {7}3

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

    Is my thinking correct?
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    Quote Originally Posted by cammywhite View Post
    1. and 2. both end up with the number -\frac {7}3

    According to #4, since \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 ....?
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    Quote Originally Posted by mr fantastic View Post
    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]
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    Quote Originally Posted by cammywhite View Post
    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.
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    Quote Originally Posted by mr fantastic View Post
    You've got the idea I see.
    If I am not mistaken, I can also say \int \frac 1{x^2-1} from 0 to 2 is undefined because x is undefined when x = 1
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  14. #14
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    Quote Originally Posted by cammywhite View Post
    If I am not mistaken, I can also say \int \frac 1{x^2-1} from 0 to 2 is undefined because x is undefined when x = 1
    Yes, that's right
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  15. #15
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    Quote Originally Posted by cammywhite View Post
    If I am not mistaken, I can also say \int \frac 1{x^2-1} from 0 to 2 is undefined because x is undefined when x = 1
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