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Math Help - Integration problem

  1. #1
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    Integration problem

    I am having trouble with an integration

    The problem is looking for e raised to the integrated function \int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx

    The answer to the problem states that the result = 1 +.1x

    Of course if the .1 was not in the numerator it would be e to the natural log of (1+.1x) divided by e raised to 'ln (1)' and that of course would be (1+.1x) divided by one and we'd have the answer.

    When I pull the numerator (.1) outside of the integrand and then integrate I get .1 * (1+.1x), and that is not the stated answer. Help please.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by tedmetro View Post
    I am having trouble with an integration

    The problem is looking for e raised to the integrated function \int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx

    The answer to the problem states that the result = 1 +.1x

    Of course if the .1 was not in the numerator it would be e to the natural log of (1+.1x) divided by e raised to 'ln (1)' and that of course would be (1+.1x) divided by one and we'd have the answer.

    When I pull the numerator (.1) outside of the integrand and then integrate I get .1 * (1+.1x), and that is not the stated answer. Help please.
    If u=1+.1x, then du=.1dx. So \displaystyle\int_0^{x}\frac{.1dx}{1+.1x}=\int_{1}  ^{1+.1x}\frac{du}{u}=\ldots

    Taking \exp\left(\int_{1}^{1+.1x}\frac{du}{u}\right) will then yield the desired result.

    Does this make sense?
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  3. #3
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    Quote Originally Posted by tedmetro View Post
    I am having trouble with an integration

    The problem is looking for e raised to the integrated function \int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx

    The answer to the problem states that the result = 1 +.1x

    Of course if the .1 was not in the numerator it would be e to the natural log of (1+.1x) divided by e raised to 'ln (1)' and that of course would be (1+.1x) divided by one and we'd have the answer.

    When I pull the numerator (.1) outside of the integrand and then integrate I get .1 * (1+.1x), and that is not the stated answer. Help please.
    It's not clear what you're trying to do here. Is the function you've been given e^{\int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx}, ? What are you asked to do with it?
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  4. #4
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    Quote Originally Posted by adkinsjr View Post
    It's not clear what you're trying to do here. Is the function you've been given e^{\int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx}, ? What are you asked to do with it?
    Yes - that's the function. The manual is just simplifying and states that e^{\int\limits_{0}^{x}\frac{.1}{(1+.1x)}\,dx} = 1 + .01x

    They are losing me there because I'm pulling the constant in the numerator out before
    I integrate and so I don't see where they are getting rid of that .1.
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  5. #5
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    To be perfectly accurate (I apologize) it's really e^{\int\limits_{0}^{t}\frac{.1}{(1+.1r)}\,dr} = 1 + .01t

    The variable being integrated is r from 0 to t.
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    If u=1+.1x, then du=.1dx. So \displaystyle\int_0^{x}\frac{.1dx}{1+.1x}=\int_{1}  ^{1+.1x}\frac{du}{u}=\ldots

    Taking \exp\left(\int_{1}^{1+.1x}\frac{du}{u}\right) will then yield the desired result.

    Does this make sense?
    I went to Wolfram Alpha and saw their integration steps, which used this substitution. I basically get it, but in their steps I am not sure I understand why the .1 moves outside of the integrand to become the reciprocal.

    Wolfram Alpha
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  7. #7
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    Chris L T521 showed you how to do this in his first post. Use substitution.

    Let u= 1+ .1x so that du= .1dx or dx= du/.1. Because when you substitute a new variable for an expression, you have to substitute for the differential as well. Since u is only "one tenth x", integrating with respect to u rather than x will give you a measure only one tenth as large as it should be. To fix that you have to multiply by 10 (or divide by .1). That's why the ".1 moves outside of the integrand to become the reciprocal". And, of course, that will cancel the ".1" multiplying dx in the original integral. Another way of looking at it is the ".1 dx" in the original integral is simply replaced by du so that there is no constant to take out. That's why the ".1 moves outside of the integrand to become the reciprocal". And, of course, that will cancel the ".1" multiplying dx in the original integral. Another way of looking at it is the ".1 dx" in the original integral is simply replaced by du so that there is no constant to take out. Since this is a definite integral, with x= 0, u= 1 and with x= x, u= 1+ .1x.

    As Chris L T521 said, the integral should be \int_1^{1+ .1x}\frac{du}{u}= ln(u)\right|_1^{1+ .1x}= ln(1+ .1x)- ln(1)= ln(1+ .1x) since ln(1)= 0.

    Then e^{\int_1^x}\frac{.1dx}{1+ .1x}= e^{ln(1+ .1x}}= 1+ .1x
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