7.8 Improper Integrals

Transcription

1 CHAPTER 7. TECHNIQUES OF INTEGRATION Improper Integrals Eample. Find Solution. Z e d. Z e d = = e e!! e = (e ) = Z Eample. Find d. Solution. We do this prolem twice: once the WRONG way, and once the correct way. WRONG WAY Z d = = = ( + /) = 3/ Can you see how we can tell that this answer must e wrong? Here is the graph of this function: graphics/area_under over squared-eps-converted-to.pdf

2 CHAPTER 7. TECHNIQUES OF INTEGRATION 68 This graph is always positive, so the area under the curve has to e positive, and so there s no way the integral should e negative. If this integral is defined, it has to e positive, ut we ll see in a moment that it s not defined. Here s the right way to do this. CORRECT WAY Z d = Z =! = Z d + d! a! + + = ( + ) ( )) a! + a = = + Since the integral comes out to e infinite, we say that the integral diverges. Challenge. Can you come up with an eample, like the previous one ut where the integral, when done correctly, converges? Eample 3. Find the volume of the shape known as Gariel s Horn. This is defined y rotating f() = around the -ais, from =to =. Solution. We start y picturing the original region, and the volume generated y rotating it. graphics/_over for_rotating-eps-converted-to.pdf

3 CHAPTER 7. TECHNIQUES OF INTEGRATION 69 graphics/gariels_horn_rotated-eps-converted-to.pdf Ihopethepicturesstimulateyoutothink volumeydisks. Weslicethe volume into disks, moving along the -ais. Thus, the volume is given y Z V = d. Using the techniques for improper integrals that we have learned, this integral is fairly straight forward, Z Z d = d =! = (! = ( +) = ( + ) = ) This eample shows a physical solid, that is infinitely long, ut has a finite volume. Here s a nice way to think aout this: if you turned this horn vertically, you could fill it with liquid, using a finite amount of liquid. Eample Z 4. Figure out which of the following converge: (a) d Z () d.5 Z (c) d.7

4 CHAPTER 7. TECHNIQUES OF INTEGRATION 7 Solution. (a) Z So in this case, the integral is divergent. () Z d = ln()! =ln() ln()! = = d =.5 Z.5 d.5 =!.5 =.5.5! =.5!.5 =.5 = ( ).5 =.5 So in this case, the integral is convergent. (c) Z d =.7 Z.7 d.3 =!.3 =.3 = So in this case the integral is divergent.!.3 ( ).3 = Z Eample 5. For which values of p does d converge? p Solution. We saw in the previous eample that the integral diverges for p =.

5 CHAPTER 7. TECHNIQUES OF INTEGRATION 7 For p>welookatcase()andseethesamecalculationswouldworkin general: the anti-derivative will come out to e,withp >, and this p function has a horizontal asymptote. So the it will eist, and the integral converges. For p<welookatcase(c)andseethatthesamecalculationswillwork in general: the anti-derivative will come out to e p with p>, and this function does not have an horizontal asymptote. So the it will not eist, and the integral diverges. In conclusion: Z ( d = convergent if p> p divergent if p apple Eample Z 6. Figure out which of the following converge. (a) d Z () d.3 Z (c) d.6 Solution. (a) Z So in this case, the integral is divergent. () Z d = a! a =ln() ln(a) a! = ( ) = d =.3 Z.3 d.3 = a!.3 a = a.3.3 a! =.3 a! a.3 =.3 = ( ).3

6 CHAPTER 7. TECHNIQUES OF INTEGRATION 7 = ± So in this case, the integral is divergent. (c) Z d =.6 Z.6 d.4 = a!.4 =.4 a a.4 a! = ( ).4 =.4 So in this case the integral is convergent. Eample 7. Figure out which values of p make the following integral converge. Z p d Solution. We saw in the previous eample that d is divergent. For p>welookatcase()andseethesamecalculationswouldworkin general: the anti-derivative will come out to e,withp >, and this p function has a vertical asymptote. So the it will not eist, and the integral diverges. For p<welookatcase(c)andseethatthesamecalculationswillwork in general: the anti-derivative will come out to e p with p>, and this function can e evaluated at =. Sotheitwilleistandtheintegral converges. In conclusion: Z ( d = convergent if p> p divergent if p Z Eample 8. Find the following, or show that it diverges Z 3 ( ) /3 d Solution. Note: you have to spot what the prolem is with this eample, it s not made eplicit. You have to rememer that the negative power means that you have,andthenyou have to see that this could give division y ( ) /3

7 CHAPTER 7. TECHNIQUES OF INTEGRATION 73 whenyouplugin =. Thismeansthatthereisaverticalasymptoteat =,andsowehavetosplittheintegralupthere: Z 3 d = ( ) /3 Z d + ( ) /3 Now we do a very simple u-sustitution, u = = = 3 ( )/3 3 Z 3 d ( ) /3 andget ( )/3 + /3 /3 (! )/3 ( ) /3 +(3 ) /3 ( ) /3! = 3 ( + ) = Note, this integral comes out to e zero since ( ) /3 is odd, and the area on the left, elow the -ais cancels the area on the right, aove the -ais.