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Math Help - diverging or converging?

  1. #1
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    diverging or converging?

    determine whether the integral diverges or converges
    if it converges then evaluate the integral

    integral from 0 to infinity: xe^-(x^2) dx

    integral from 0 to 1: xln(x) dx

    integral from 1 to infinity: (ln t) / (t^2) dt

    integral from 0 to 2: 1 / (t-1) dt


    i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?
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  2. #2
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    Quote Originally Posted by razorfever View Post
    determine whether the integral diverges or converges
    if it converges then evaluate the integral

    integral from 0 to infinity: xe^-(x^2) dx

    integral from 0 to 1: xln(x) dx

    integral from 1 to infinity: (ln t) / (t^2) dt

    integral from 0 to 2: 1 / (t-1) dt


    i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?
    The integral converges if it's value is finite. If it's value is infinite, it is divergent.

     \int_0^1 {x\ln(x) dx} = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x^2}{2} \times \frac{1}{x} dx}

     = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x}{2} dx}


     = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \bigg[{\frac{x^2}{4}}\bigg]^1_0

    This integral is divergent because when you plug in the limits in the first expression, you end up with the term  \ln(0) . This is not defined, and hence the integral has no finite value, it does not converge.

    Try the other 2 and let us know what results you obtain.
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  3. #3
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    Quote Originally Posted by Mush View Post

    The integral converges if it's value is finite. If it's value is infinite, it is divergent.

     \int_0^1 {x\ln(x) dx} = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x^2}{2} \times \frac{1}{x} dx}

     = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \int_0^1{\frac{x}{2} dx}


     = \bigg[\frac{x^2}{2}\ln(x)\bigg]_0^1 - \bigg[{\frac{x^2}{4}}\bigg]^1_0

    This integral is divergent
    Since \underset{x\to 0}{\mathop{\lim }}\,x\ln x=0, then x\ln x is continuous on (0,1], hence integrable on [0,1].
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  4. #4
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    so does that mean its diverging or converging, now im totally confused??
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  5. #5
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    Convergent.
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  6. #6
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    hah so i was right ... thanks for clearing that up
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  7. #7
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    Quote Originally Posted by razorfever View Post

    determine whether the integral diverges or converges
    if it converges then evaluate the integral

    integral from 0 to infinity: xe^-(x^2) dx

    i have solved the first integral and i got the answer as 1/2 but how do i show if it is diverging or converging?
    \int_{0}^{\infty }{x{{e}^{-{{x}^{2}}}}\,dx}=\int_{0}^{1}{x{{e}^{-{{x}^{2}}}}\,dx}+\int_{1}^{\infty }{x{{e}^{-{{x}^{2}}}}\,dx}. Let's just worry about the second piece since the first one is integrable on [0,1] for being a continuous function on that interval. Now, for x\ge1 we have \frac{x}{{{e}^{{{x}^{2}}}}}\le \frac{x}{{{e}^{x}}}=x{{e}^{-x}}, hence by direct comparison test your integral converges because \int_{1}^{\infty }{x{{e}^{-x}}\,dx}, does.

    Second and third question were already answered, and for your last question, you just need to split that integral into two ones. I mean, \int_{0}^{2}{\frac{dt}{t-1}}=\int_{0}^{1}{\frac{dt}{t-1}}+\int_{1}^{2}{\frac{dt}{t-1}}.

    And there's no much here, these are clearly divergent integrals, so your integral diverges.
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