141BExam1Rev - MATH 141B 1 Integration by Parts The formula is Exam 1 Overview u dv = uv We first applied this to the two classes of integrals xn emx dx

# 141BExam1Rev - MATH 141B 1 Integration by Parts The formula...

• Notes
• 8

This preview shows page 1 - 3 out of 8 pages.

MATH 141B Exam 1 Overview 1. Integration by Parts The formula is: integraldisplay udv = uv integraldisplay v du We first applied this to the two classes of integrals: integraldisplay x n e mx dx and integraldisplay x n sin( mx ) dx, In both cases, n is allowed to be any natural number, and m any real number. The latter class of integrals above works equally well with cos( mx ) in place of sin( mx ). In each of these cases, let u = x n and dv = e mx dx or dv = sin mxdx. Integrals from each of these two classes require n applications of the integration by parts formula. We also applied the integration by parts formula to the following class of integrals: integraldisplay x n ln( mx ) dx Both n and m are allowed to be any real number in this case. In this case, let u = ln( mx ) and dv = x n dx . Integrals in this 3rd class only require only one application of the integration by parts formula. The integration by parts formula gave rise to the important integral (worth remembering). integraldisplay ln xdx = x ln x x 2. Improper Integrals There are 3 cases here: I. integraldisplay a f ( x ) dx = lim b →∞ integraldisplay b a f ( x ) dx II. integraldisplay b -∞ f ( x ) dx = lim a →-∞ integraldisplay b a f ( x ) dx III. integraldisplay -∞ f ( x ) dx = integraldisplay c -∞ f ( x ) dx + integraldisplay c f ( x ) dx, for any c 1
If the limit defining these integrals exists, then the improper integral is said to converge . Otherwise, the integral diverges . Case III reduces the integral integraltext -∞ f ( x ) dx to an integral of of type I and an integral of type II. In case III, pick c = 0 unless there’s a discontinuity there. First compute integraltext 0 f ( x ) dx . If it diverges, then there is no need to compute the other integral integraltext 0 -∞ f ( x ) dx . You simply stop declare that the entire integral integraltext -∞ f ( x ) dx diverges. There are 3 ways that an improper integral can diverge. The integral may diverge to + , −∞ or it may be that it diverges, but neither to + nor to −∞ . This happens, for instance, when you end up facing a limit of an oscillating trigonometric function like lim x →∞ sin x , which does not exist (DNE). One important class of limits we discussed in the course of computing improper integrals was: lim n →∞ ln x x n = 0 , for any n > 0. This expresses the fact that the logarithm grows more slowly than any power of x . This fact is obvious for powers n 1 (just look at a graph), but for powers n satisfying 0 < n < 1 , it’s also true. 3. Gaussian Elimination Regarding the solution set of any system of linear equations, there are only 3 possibilies: i) There is exactly one solution. ii) There are no solutions. iii) There are infinitely many solutions. The method of Gaussian elimination uses 3 so-called elementary row operations to reduce a matrix until it satisfies the following conditions: i) All zero rows are grouped at the bottom of the matrix.