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Australopithecus

  • 3 years ago

Apply the Comparison Test to determine if the integral converges or diverges. The Integral is \[\int\limits_{0}^{1} e^{2x}dx/x^{3} \] So far I have that \[\int\limits_{0}^{1} e^{2x}dx/x^{3} = \lim_{t \rightarrow 0^{+}} \int\limits_{t}^{1} e^{2x}dx/x^{3}\] Since \[1/0^{+} = \infty\] I assume this integral is Divergent Therefore, I use f(x) < g(x) where f(x) = e^(2x)/x^(3) \[e^{2x}/x^{3} \le 1/x^{3}\] Since \[\lim_{t \rightarrow 0^{+}} \int\limits_{t}^{1} dx/x^{3} \] is Divergent by the C.T. \[\int\limits_{0}^{1} e^{2x}dx/x^{3} \] is Divergent

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  1. Australopithecus
    • 3 years ago
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    did I do this correctly, I'm a little lost when trying to acquire the new function, g(x) and how to do it correctly from my understanding you simply remove the smallest element until it is simplified.

  2. colorful
    • 3 years ago
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    this looks right to me... as far as tips as to what to choose for g(x), just try to thing of a function that is easy to improperly integrate, and compare it to the original function on that interval

  3. colorful
    • 3 years ago
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    i.e. since \[\int_{0}^{1}\frac1{x^3}dx\]is easy to show as divergent, we choose it as g(x) and compare it to the original integrand like I said, looks good to me

  4. Australopithecus
    • 3 years ago
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    Thanks :)

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