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Math Help - Evaluating definite integral of ln(x)/x from 1 to e

  1. #1
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    Evaluating definite integral of ln(x)/x from 1 to e

    I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

    \int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx

    Let

    u = ln(x)
    du = \frac {1}{x} dx

    Substituting values gives:

    \int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(ln(x))^2}{2} \right ]^e_1 = \frac {(ln(e))^2}{2} - \frac {(ln(1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}
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  2. #2
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    Quote Originally Posted by drew.walker View Post
    I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

    \int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx

    Let

    u = ln(x)
    du = \frac {1}{x} dx

    Substituting values gives:

    \int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(ln(x))^2}{2} \right ]^e_1 = \frac {(ln(e))^2}{2} - \frac {(ln(1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}
    It would be  \int_{0}^{1} u \ du since  u = \ln 1 = 0 and  u = \ln e = 1 .
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  3. #3
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    hmmm, so the interval of the integral changes with the substitution? Is there a name for this or somewhere where I can see more examples that you know of?
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  4. #4
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    Also, if it's from 0 to 1, ln(1) = 0 and ln(0) is undefined, therefore the answer would now be 0. Is that right?
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  5. #5
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    Quote Originally Posted by drew.walker View Post
    I think I'm getting the hang of evaluating integrals, but I just wanted to check my answer. Is someone able to confirm if this working and result are correct?

    \int^e_1 \frac{ln(x)}{x} dx = \int^e_1 \frac{1}{x}ln(x) dx

    Let

    u = ln(x)
    du = \frac {1}{x} dx

    Substituting values gives:

    \int^e_1 u du = \left [ \frac{u^2}{2} \right ]^e_1 = \left [ \frac{(\ln (x))^2}{2} \right ]^e_1 = \frac {(\ln (e))^2}{2} - \frac {(\ln (1))^2}{2} = \frac {1}{2} - 0 = \frac{1}{2}
    You are getting things mixed up and confused. When making a substitution in a definite integral, you can either:

    1. Change the integral terminals and then evaluate the new definite integral: \int_0^1 u \, du, or

    2. Get the anti-derivative in terms of the new variable, re-substitute to get the anti-derivative in terms of the old variable and then evaluate the definite integral in terms of the old variable: \left [ \frac{(\ln (x))^2}{2} \right ]^e_1.

    You do one or the other. You're mixing both together.

    The correct answer is 1/2:

    1. \int_0^1 u \, du = \left[ \frac{u^2}{2} \right]_0^1 = \frac{1}{2}.

    2. \left [ \frac{(\ln (x))^2}{2} \right ]^e_1 = \frac{1}{2}.

    Quote Originally Posted by drew.walker View Post
    Also, if it's from 0 to 1, ln(1) = 0 and ln(0) is undefined, therefore the answer would now be 0. Is that right?
    NO!
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