practice How would you find: e x + e x e 2x e x 1 dx 1 e today: improper integrals

Transcription

1 prctice How would you find: dx e x + e x e 2x e x 1 dx e 2x 1 e x dx 1. Let u=e^x. Then dx=du/u. Ans = rctn ( e^x ) + C 2. Let u=e^x. Becomes u du / (u-1), divide to get u/(u-1)=1+1/(u-1) Ans = e^x + ln e^x C 3. integrtion by prts, u=e^x dv=e^x (1-e^x)^(-1) dx Ans = -sqrt(1-e^x) * ( 4/3 + 2e^x / 3 ) + C (This is Leithold ) tody: improper integrls fridy: lst drop dy webwork 5 11:55 pm mslc webwork 5 workshop in SEL 12:30, 1:30, 2:30, 3:30, 4:30 mondy: webwork extr credit ii help session in EA 5:30 tuesdy: rc length thursdy, 12 november: surfce re quiz iv: 4.4, 7.8 homework 6 due (4.4.28, , , , , ) mondy, 16 november: webwork extr credit ii help session in EA 5:30 mondy, 23 november: webwork extr credit ii 6:00 m Extr Credit II due dte hs been moved to Mondy, 23 November (this gives you n extr week thn before.) The extr week is the week of midterm iii.

2 review of l Hôpitl ln(1 2x) lim x 1/2 tn(π x) This is - /, so use l Hôpitl. Answer=0. (Leithold ) Ask if there re ny other l Hôpitl questions. review of l Hôpitl lim x x2 e x Answer=0. This is Stewrt s

3 review of l Hôpitl ( 1 lim x 1 ln x 1 ) x 1 Get common denomintor. l Hôpitl. multiply top nd bottom by x. l Hôpitl. Answer=1/2 This is Stewrt review of l Hôpitl lim (1 2x)1/x x 0 Let y=(1-2x)^(1/x), find limit of ln(y) using l Hôpitl. Limit of ln(y) is -2, so the nswer is exp(-2). This is Stewrt s

4 improper integrls Here is number. In prticulr, is not negtive infinity. If f is continuous for ll x, we define the improper integrl with infinite upper limit f(x)dx := lim f(x)dx b provided this limit exists. Improper integrls re integrls where infinity comes up in some wy. This presenttion follows Leithold 7.9 nd If the limit does not exist, the integrl is divergent. If it does, then convergent. improper integrls If f is continuous for ll x b, we define the improper integrl with infinite lower limit f(x)dx := lim f(x)dx provided this limit exists. b is number. b is NOT positive infinity. Integrls tht re improper becuse of hving bounds of or - re clled improper integrls of type 1.

5 exmple 0 xe x2 dx Answer = 1/2 exmple 0 sin x dx This is divergent integrl. No limit exists.

6 exmple 1 dx x This is divergent integrl. The limit is. Fct: If insted the denomintor ws x^p, would be convergent if p>1, divergent if p 1. exmple Wht is the volume of the figure obtined by revolving the region under f(x)=1/x, x 1 bout the x-xis? We just sw tht the re ws unbounded, but the volume is finite. Answer = π

7 improper integrls If f is continuous for ll x nd c is ny rel number, we define the improper integrl with both lower nd upper infinite limits f(x)dx := c lim f(x)dx + lim f(x)dx b c provided this limit exists. It is left s n exercise for the reder to show tht the nswer is the sme regrdless of the choice of c. In prticulr, we do NOT just tke one limit. We must split it up, so tht we lwys hve t lest one finite bound. exmple 4 x (x 2 + 1) 3 dx Split t 0 (sy). Integrl from 0 to is 1, from - to 0 is -1, so whole integrl is 0.

8 exmple sin x dx We split this t 0, but we ve lredy found the integrl from 0 to infinity is divergent, so this integrl is divergent. improper integrls If f is continuous for ll x on (, b] nd lim f(x) = we define the improper x + integrl with infinite discontinuity t its lower limit f(x)dx := lim f(x)dx t + t provided this limit exists. These remining types of improper integrls re sometimes clled type 2 improper integrls.

9 improper integrls If f is continuous for ll x on [, b) nd lim f(x) = we define the improper x b integrl with infinite discontinuity t its upper limit t f(x)dx := lim f(x)dx t b provided this limit exists. exmple 4 0 dx 4 x Antiderivtive is -2 sqrt(4-x), tking limit nd evluting nswer is 4.

10 exmple 1 0 ln(x)dx This is improper since ln is undefined t 0. Antiderivtive is x ln x - x + C Must use l Hôpitl to limit, nswer=-1. This is Stewrt s exmple improper integrls If f is continuous for ll x on [, b] except c (<b<c) nd if lim f(x) =, we define the x c improper integrl with infinite discontinuity in the interior t f(x)dx := lim f(x)dx + lim f(x)dx t c s c + s provided this limit exists.

11 exmple 33 0 (x 1) 1/5 dx This is improper since the function is undefined t 1, so we must integrte from 0 to 1 nd from 1 to 33. Answer=75/4. This is Stewrt s comprison theorem Suppose f nd g re continuous nd suppose f(x) g(x) 0 for x. If f(x)dx is convergent, then so is g(x)dx If g(x)dx is divergent, then so is f(x)dx Smller thn convergent is convergent; bigger thn divergent is divergent. Note: The converse does not hold! Functions tht re bigger thn convergent functions my or my not converge.

12 exmple is the following integrl convergent or divergent? 1 cos 2 (x) 1+x 2 dx Convergent since smller thn 1/(1+x^2) which converges. This is Stewrt s coming soon red 8.1 webwork 5 due fridy homework 6 due next thursdy strt extr credit project 2, due 23 6:00 m Agin, this mens tht you now hve n extr week to do extr credit project 2...