Sean Fitzpatrick |
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University of Lethbridge |
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Trig substitution:
Partial fractions:
For which values of \(p\) does the integral
Which of the following integrals would be considered improper?
An integral that uses the same fork for its main course and dessert.
All of the above.
Although we can evaluate many improper integrals, the main question is not “What is the value?”, but “Does it converge?”
Improper integrals are potentially infinite. An improper integral converges if it has a finite value.
For a function \(f(x)\) continuous on \([a,\infty)\text{,}\) we set
If \(f(x)\) is unbounded at \(x=c\text{,}\) we set
\(\di \int_1^\infty \frac{1}{x^4}\,dx\)
\(\di \int_3^\infty \frac{x}{x^2-4}\,dx\)
\(\di \int_{-\infty}^\infty \frac{1}{x^2+16}\,dx\)
\(\di \int_0^\infty e^{-5x}\,dx\)
\(\di \int_5^\infty\frac{1}{x-4}\,dx\)
The improper integral \(\di \int_1^\infty\frac{1}{x^p}\,dx\) converges if \(p\gt 1\text{,}\) and diverges if \(p\leq 1\text{.}\)
\(\di \int_0^1\ln(x)\,dx\)
\(\di \int_0^1 \frac{1}{\sqrt[3]{x}}\,dx\)
\(\di \int_0^1\frac{1}{x^p}\,dx\) (For what \(p\) does it converge?)
Premise is simple:
If \(f(x)\leq g(x)\) and \(\int_a^\infty g(x)\,dx\) converges, so does \(\int_a^\infty f(x)\,dx\text{.}\)
If \(f(x)\geq g(x)\) and \(\int_a^\infty g(x)\,dx\) diverges, so does \(\int_a^\infty f(x)\,dx\text{.}\)
The hard part is setting up the comparison.
Use comparison to decide on convergence or divergence of the improper integral:
\(\di \int_2^\infty\frac{1}{x^2+x}\,dx\)
\(\di \int_1^\infty\frac{1}{\sqrt{x^2-1}}\,dx\)
\(\di \int_2^\infty\frac{1}{\sqrt{x^2+1}}\,dx\)
What about \(\di \int_3^\infty\frac{1}{x\ln(x)}\,dx\text{?}\)
Useful when direct comparison is hard to set up.
Suppose \(\di \lim_{x\to\infty}\frac{f(x)}{g(x)}=L\text{,}\) where \(0\lt L\lt \infty\text{.}\) Then the integrals
Determine the convergence of
What can we say about \(\di \int_4^\infty\frac{x^2}{e^x}\,dx\text{?}\)