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by: Kennith Herman


Kennith Herman
GPA 3.54


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This 3 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 114 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/228154/ma-114-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15
1 Lecture 17 Improper integrals 1 Consider improper integrals at in nity 2 Find areas as improper integrals 3 The comparison theorem A question Can an region have in nite perimeter and nite area If so7 we can color it in7 but we cannot draw its outline Example Find the area bounded by z 17 y 0 and y e w Solution This is a new problemeuntil now7 we have found the area of regions which were of nite length in every direction We attempt to solve this problem7 by rst nding the area bounded by z 17 y0ye w andzN Thisis New 1 7N e dx77e 1 e If we take the limit as N goes to in nity7 we expect to obtain a reasonable value for the in nite area 1 N 1 area lim fee 7 Naoo e 6 De nition of improper integral De nition If a is a real number and f is a function which is Riemann integrable on each interval 1 N for N gt 17 then we de ne the improper integral as 00 N ft dt lim ft dt a Naoo a If this limit exists and is nite7 we say the improper integral is convergent If the limit is in nite or does not exist7 we say the improper integral is divergent Epereise Write out the corresponding de nition for improper integrals at 700 Epereise Re read the de nition of the Riemann integral ft dt and explain what dif culties would arise if we let b oo in this de nition Eat ample How should we treat the integral zdz Solution If we break the integral up as follows 00 0 co xdx xdx xdx oo 700 0 then each of the integrals on the right is divergent I Eat ample Evaluate the following irnproper integrals7 or state ifthe integral diverges 0 1 ezmdx 0 2 e zmdx 0 1 3 7dr 1 1x2 Eat ample Deterrnine for which p7 the integral 0 1 ids 1 x17 converges Solution If p 31 17 then 00 1 717M mN N ids lirn z pdx lirn 1 17 Naoo 1 Naoo 7p1 This limit will exist and be nite ifthe exponent of N is negative That is the integral converges ifp gt 1 The limit will be in nite and the integral diverges if p lt 1 The case p 1 is treated separately because it involves the logarithm lfp 17 then the integral is divergent I N Ptl 1 1 0 1 1 1 lirn F1 Naoo The comparison test Theorem 1 ff and g are continuous functions on 100 and 0 S f S gz then we have 00 00 u do 3 g dx If u dx diverges then g dx diverges If g dx converges then u dx converges Notice that these results do not give us the value only whether the integral converges or diverges Eat ample Show that 1 sin xe w dx e w2 dx 0 0 converges Solution We have 1 sinxe w S 2e w and we know the integral of f5 e w con verges and thus the comparison test tells us that the integral of e m2 converges We only need to worry about 1 e m2du and in this region we have x 2 z 7z2 3 in and nally e m2 S e w Now the comparison test tells us the integral 5 e m2 is convergent I 2 Example Determine if the following integrals converge or diverge using the compar ison test 7m 2 e 2 sinu 1dx 6 dx dm 0 1 u 1 u March 1 2004


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