Class 18 - Comparison of improper integrals (10/13/2017, Friday)

2nd Gateway

You must pass the second Gateway by today. Otherwise, your grade will be damaged a lot!

Webwork

• 8.3 due 10/18 Wed
• 7.6 due 10/19 Thurs

Comparison of improper integrals

We recall that

\[\int_{1}^{\infty}\frac{1}{\sqrt{x^{3} + x + 1}} = \lim_{r \rightarrow \infty}\int_{1}^{r}\frac{1}{\sqrt{x^{3} + x + 1}}.\]

The integral above is very difficult to compute. Can we tell whether the above integral converges without computing?

The answer is yes. For any $x \geq 1,$ we have 

\[\frac{1}{\sqrt{x^{3} + x + 1}} \leq \frac{1}{\sqrt{x^{3} + x^{3} + x^{3}}} = \frac{1}{\sqrt{3} x^{3/2}}.\]

This implies that

\[\int_{1}^{\infty}\frac{1}{\sqrt{x^{3} + x + 1}} dx \leq \int_{1}^{\infty}\frac{1}{\sqrt{3}} \frac{1}{x^{3/2}} dx < \infty,\]

by $p$-test with $p = 3/2 > 1$, so we see that the original integral must converge to a finite quantity as well.

Exercise. Determine whether the following improper integral is convergent:

\[\int_{1}^{\infty}\frac{2\cos(x) + 3}{\sqrt{x}}dx.\]

Exercise. Determine whether the following improper integral is convergent:

\[\int_{1}^{\infty}e^{-x^{2}/3}dx.\]

Exercise*. Solve #5 of Exam 2, Winter 2017.

Exercise*. Solve #9 of Exam 2, Winter 2017.