L8 - Math 1B UCB Spring 2013 M Christ Summary of Lecture 8 Friday Improper Integrals Todays lecture is based on 7.8 of our text Please study this

# L8 - Math 1B UCB Spring 2013 M Christ Summary of Lecture 8...

• Notes
• 5

This preview shows page 1 - 3 out of 5 pages.

Math 1B — UCB, Spring 2013 — M. Christ Summary of Lecture 8, Friday 2/8/2013 Improper Integrals Today’s lecture is based on § 7.8 of our text. Please study this section if you have not already done so. We may return to numerical integration, § 7.7, briefly next week, after you have had a chance to work some problems and to absorb what has already been discussed. Improper integrals are those which either involve integration over unbounded intervals, or involve integration of unbounded functions. Examples: R 1 x - 1 dx , R 1 1 1+ x 2 dx . Definition. For improper integrals of the form R a f ( x ) dx : Z a f ( x ) dx = lim t →∞ Z t a f ( x ) dx, if this limit exists , as a finite number . In this case, the improper integral is convergent . (Equivalently, we say that the integral converges.) If the limit fails to exist, including the case where lim t →∞ R t a f ( x ) dx = ±∞ , the improper integral is divergent . There are corresponding definitions for improper integrals of the related form R b -∞ f ( x ) dx ; see text. Example: R 1 x - 1 dx diverges. Solution. First consider any t 1 and evaluate Z t 1 x - 1 dx = ln( x ) t 1 = ln( t ) - ln(1) = ln( t ) . Now take the limit as t → ∞ : lim t →∞ Z t 1 x - 1 dx = lim t →∞ ln( t ) = . The limit is + , so this improper integral diverges. (It is common to write R 1 x - 1 dx = + .) Example: R 0 1 1+ x 2 dx converges. Solution. For t 0, Z t 0 1 1 + x 2 dx = arctan( x ) t 0 = arctan( t ) - arctan(0) = arctan( t ) . 1
A quick review of the graph of y = arctan( x ) reminds us that arctan( t ) π 2 as t + . Therefore lim t →∞ Z t 0 1 1 + x 2 dx = π 2 . Therefore this improper integral converges, and Z 0 1 1 + x 2 dx = π 2 . Example: R 25 1 1+ x 2 dx : Same analysis as preceding example gives lim t →∞ Z t 25 1 1 + x 2 dx = π 2 - arctan(25) . This improper integral converges, and R 25 1 1+ x 2 dx = π 2 - arctan(25). Changing the lower limit of integration from 0 to 25 ( Affects the numerical value of the convergent integral, But is irrelevant to deciding whether or not the integral converges.