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Math Help - integrals...

  1. #1
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    integrals...

    (1 pt) Consider the function
    Evaluate the definite integral.


    This is a homework problem. I got an answer of (1/2) - ln(9), but that is incorrect. I used the properties of integrals to split up the function. Can you give me a hint? Thanks!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by tennisgirl View Post
    (1 pt) Consider the function

    Evaluate the definite integral.



    This is a homework problem. I got an answer of (1/2) - ln(9), but that is incorrect. I used the properties of integrals to split up the function. Can you give me a hint? Thanks!

    If you look at it from 0<x<1...and from 1<x it is 1/x.....so the area should be
    \int_0^{1}xdx+\int_1^{9}\frac{1}{x}dx=[\frac{1}{2}-0]+[\ln(9)-\ln(1)=\frac{1}{2}+\ln(9)
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  3. #3
    o_O
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    \int_{0}^{9} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{9} f(x)dx
    = \int_{0}^{1} x dx + \int_{1}^{9} x^{-1} dx
    = \frac{1}{2}x^{2} \bigg|_{0}^{1} \: \: + \: \: \ln x \bigg|_{1}^{9}

    Try going through it more carefully

    Ah darn, 2.5 - 2 now eh ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    If you look at it from 0<x<1...and from 1<x it is 1/x.....so the area should be
    \int_0^{1}xdx+\int_1^{9}\frac{1}{x}dx=[\frac{1}{2}-0]+[\ln(9)-\ln(1)=\frac{1}{2}+\ln(9)
    Note! This applies because \lim_{x\to{1^+}}f(x)=\lim_{x\to{1^-}}f(x)
    and now that we know that the limit exists..this is only true because

    \lim_{x\to{1}}f(x)=f(1)

    i.e. it is continous on [0,9]
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    \int_{0}^{9} f(x) dx = \int_{0}^{1} f(x) dx + \int_{1}^{9} f(x)dx
    = \int_{0}^{1} x dx + \int_{1}^{9} x^{-1} dx
    = \frac{1}{2}x^{2} \bigg|_{0}^{1} \: \: + \: \: \ln x \bigg|_{1}^{9}

    Try going through it more carefully

    Ah darn, 2.5 - 2 now eh ?
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