Question

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

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

Thanks :)