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Thread: Help with finding an expression for an integral...

  1. #1
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    Help with finding an expression for an integral...

    Find an expression for $\displaystyle \int_1^s \frac {1}{x^m} dx$ in terms of m and s, where m is a positive rational number, m != -1 and s > 1. Show that the infinte integral $\displaystyle \int_1^\infty \frac {1} {x^m} dx$ has a meaning if m > 1, and state its value in terms of m.

    Is it asking me to integrate and come up with a simplified expression? I did that, but it's not matching the answer on multiple attempts.
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    What did you come up with?
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  3. #3
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    Quote Originally Posted by struck View Post
    Find an expression for $\displaystyle \int_1^s \frac {1}{x^m} dx$ in terms of m and s, where m is a positive rational number, m != -1 and s > 1. Show that the infinte integral $\displaystyle \int_1^\infty \frac {1} {x^m} dx$ has a meaning if m > 1, and state its value in terms of m.

    Is it asking me to integrate and come up with a simplified expression? I did that, but it's not matching the answer on multiple attempts.
    1. $\displaystyle \int_1^s{ \frac {1}{x^m}\, dx} = \int_1^s{ x^{-m}\, dx}$

    $\displaystyle = \left[\frac{1}{1-m}x^{1 - m}\right]_1^s$

    $\displaystyle = \left[\frac{1}{1-m}s^{1 - m}\right] - \left[-\frac{1}{m}\right]$

    $\displaystyle = \frac{1}{m} + \frac{1}{1-m}s^{1 - m}$.


    2. The question is exactly the same, but this time you have to take the limit as s approaches infinity.
    Last edited by Prove It; Dec 26th 2008 at 05:15 PM.
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  4. #4
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    Integral

    Hello struck
    Quote Originally Posted by struck View Post
    Find an expression for $\displaystyle \int_1^s \frac {1}{x^m} dx$ in terms of m and s, where m is a positive rational number, m != -1 and s > 1. Show that the infinte integral $\displaystyle \int_1^\infty \frac {1} {x^m} dx$ has a meaning if m > 1, and state its value in terms of m.

    Is it asking me to integrate and come up with a simplified expression? I did that, but it's not matching the answer on multiple attempts.
    I'm assuming that m != -1 means $\displaystyle m \ne 1$, because if $\displaystyle m=1$ the answer is quite different (it's $\displaystyle \text{ln }x$)

    So:

    Re-write $\displaystyle \frac{1}{x^m}$ as $\displaystyle x^{-m}$, and then use the usual integration rule: Increase the power by 1, and then divide by the number you just thought of.

    When you increase $\displaystyle -m$ by $\displaystyle 1$ you get $\displaystyle -m+1$, so let's see what it looks like:

    $\displaystyle \int_0^s x^{-m}= \left[\frac{x^{-m+1}}{-m+1}\right]$

    (Note that this is where we need $\displaystyle m \ne 1$, because if $\displaystyle m=1$, we'd be violating the first commandment of arithmetic: Thou shalt not divide by zero.)

    Now what you have to do:

    • Return $\displaystyle x$ to the denominator of the fraction, by changing the sign of its power.
    • Put in the limits $\displaystyle s$ and $\displaystyle 1$ being careful with the minus signs.

    This should give you the answer to the first part:

    $\displaystyle \frac{1}{m-1}\left(1-\frac{1}{s^{m-1}}\right)$

    Now you need to look at what happens when $\displaystyle m>1$ and $\displaystyle m<1$. So, think about the sign of $\displaystyle m-1$, and what happens to $\displaystyle \frac{1}{s^{m-1}}$ as $\displaystyle s \to \infty$ if $\displaystyle m-1$ is positive, and if $\displaystyle m-1$ is negative.

    Can you see what to do now?

    Grandad


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