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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
 one year ago
 one year 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
 one year ago
 one year ago

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AustralopithecusBest ResponseYou've already chosen the best response.0
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.
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
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
 one year ago

colorfulBest ResponseYou've already chosen the best response.1
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
 one year ago
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