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 2 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
 2 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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Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0did 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.

colorful
 2 years ago
Best ResponseYou've already chosen the best response.1this 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

colorful
 2 years ago
Best ResponseYou've already chosen the best response.1i.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
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