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Math Help - u subs and x bounds on integrals

  1. #1
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    u subs and x bounds on integrals

    Could someone please explain how to use u subs when taking an integral? We learned this awhile ago and I never completely understood. Here is an example of a problem where, I think, I should be using a u sub:
    ack! I can't figure out how to type this in........ sooooo....

    The integral of x squared minus 1 over x between the bounds of e and 1

    Hope that makes sense.
    Thank you

    Also, I know I'm only supposed to ask one question, but how would the answer change if the upper bound were x? Would I just plug in x to the final solution, or is there a special rule?
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  2. #2
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    Talking

    If I understand you correctly, you're asking what sort of substitution might be helpful in the following:

    . . . . . \int_1^e\, \frac{x^2\, -\, 1}{x}\, dx

    It should be noted that there are oftentimes many ways of doing an integral. No one is more "right" than another. Since this can be simplified, you could work with the following:

    . . . . . \int_1^e\, x\, -\, \frac{1}{x}\, dx

    In this way, no substitution would be necessary. But you could fiddle around with substitutions, and notice that:

    . . . . . \mbox{for }\, \ln(x)\, =\, u,\, \mbox{ we have }\, \frac{1}{x}\, dx\, =\, du

    Then:

    . . . . . x^2\, -\, 1\, =\, e^{2u}\, -\, 1

    Does that help at all...?
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  3. #3
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    Quote Originally Posted by squid View Post
    Could someone please explain how to use u subs when taking an integral? We learned this awhile ago and I never completely understood. Here is an example of a problem where, I think, I should be using a u sub:
    ack! I can't figure out how to type this in........ sooooo....

    The integral of x squared minus 1 over x between the bounds of e and 1

    Hope that makes sense.
    Thank you

    Also, I know I'm only supposed to ask one question, but how would the answer change if the upper bound were x? Would I just plug in x to the final solution, or is there a special rule?
    I'm assuming 1 is the lower limit of integration and e is the upper.

    now ... is it

    \int_1^e \frac{x^2-1}{x} dx

    or

    \int_1^e x^2 - \frac{1}{x} \, dx

    ???

    fyi, no substitution required in either case.

    \int_1^e x - \frac{1}{x} \, dx

    \left[\frac{x^2}{2} - \ln{x}\right]_1^e = \left(\frac{e^2}{2} - 1\right) - \left(\frac{1}{2} - 0 \right) = \frac{e^2-3}{2}

    or ...

    \int_1^e x^2 - \frac{1}{x} \, dx

    \left[\frac{x^3}{3} - \ln{x}\right]_1^e = \left(\frac{e^3}{3} - 1\right) - \left(\frac{1}{3} - 0\right) = \frac{e^3-4}{3}
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    haha I see now, thank you both. And I meant the equation the way staple wrote it, or skeeter's first one.

    Would you also mind answering if there is a special way to solve an integral with an upper bound of x? For example, if the equation I used above had an upper bound of x rather than e, would I just plug in x at the end, or do I do something different?
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  5. #5
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    Quote Originally Posted by squid View Post
    haha I see now, thank you both. And I meant the equation the way staple wrote it, or skeeter's first one.

    Would you also mind answering if there is a special way to solve an integral with an upper bound of x? For example, if the equation I used above had an upper bound of x rather than e, would I just plug in x at the end, or do I do something different?
    \int_1^x \frac{t^2 - 1}{t} \, dt

    t is a dummy variable, just so that it's different from x.

    the value of the definite integral is now a function of x.
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  6. #6
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    so would that mean that my final answer would be (x squared over 2) minus lnx minus 1/2?
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  7. #7
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    Quote Originally Posted by squid View Post
    so would that mean that my final answer would be (x squared over 2) minus lnx minus 1/2?
    yes, and that would be a function.
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