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Math Help - understanding this improper integral

  1. #1
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    understanding this improper integral

    My teacher worked this problem one way and I think I have found an easier way. Also, I don't understand how he came to his conclusion.

    Here is the problem: \int_{0}^{e} ln {x}^2~dx This is how he worked the problem: \lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx=2x~ln x -\int_{0}^{e} 2~dx=(2e-0)-2x (from~0~to~e)=2e-2e=0. He also included \lim_{x\to 0}\frac{2~lnx}{1/x}=\lim_{x\to 0^+}\frac{2/x}{-1/x^2}=\lim_{x\to 0^+}-2x=0. This is what I did: \lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}

    Is my method correct? Any idea how he got what he did?
    Thanks!
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  2. #2
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    Your answer is correct. The answer is 0.
    Attached Thumbnails Attached Thumbnails understanding this improper integral-june28.gif  
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  3. #3
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    Quote Originally Posted by Possible actuary View Post
    My teacher worked this problem one way and I think I have found an easier way. Also, I don't understand how he came to his conclusion.

    Here is the problem: \int_{0}^{e} ln {x}^2~dx This is how he worked the problem: \lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx=2x~ln x -\int_{0}^{e} 2~dx=(2e-0)-2x (from~0~to~e)=2e-2e=0. He also included \lim_{x\to 0}\frac{2~lnx}{1/x}=\lim_{x\to 0^+}\frac{2/x}{-1/x^2}=\lim_{x\to 0^+}-2x=0. This is what I did: \lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}

    Is my method correct? Any idea how he got what he did?
    Thanks!
    It looks to me like he integrated by parts to go from
    \lim_{b\to 0^+}\int_{b}^{e} 2~ln {x}~dx
    to
    2x~ln x -\int_{0}^{e} 2~dx

    In your work we have the problem that
    \lim_{b\to 0^+}\frac{2}{b} \to \infty

    -Dan
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  4. #4
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    Quote Originally Posted by Possible actuary View Post
    \lim_{b\to 0^+}2\int_{b}^{e} lnx~dx=\lim_{b\to 0^+}\frac{2}{x}~(from~b~to~e)=\frac{2}{e}-\frac{2}{b}=\frac{2}{e}
    You might confuse the integral of lnx (which is x(lnx-1)) with the derivative of lnx (which is 1/x).
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