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Math 104: Improper Integrals (With Solutions)
Ryan Blair
University of Pennsylvania
Tuesday March 12, 2013
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 1 / 15Outline
1Improper Integrals
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 2 / 15Improper Integrals
Improper integrals
Definite integrals?
b a f(x)dxwere required to have finite domain of integration [a,b] finite integrandf(x)<±∞ Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 3 / 15Improper Integrals
Improper integrals
Definite integrals?
b a f(x)dxwere required to have finite domain of integration [a,b] finite integrandf(x)<±∞Improper integrals
1Infinite limits of integration
2Integrals with vertical asymptotes i.e. with infinite discontinuity
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 3 / 15Improper Integrals
Infinite limits of integration
Definition
Improper integrals are said to be
convergentif the limit is finite and that limit is the value of the improper integral. divergentif the limit does not exist. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 4 / 15Improper Integrals
Infinite limits of integration
Definition
Improper integrals are said to be
convergentif the limit is finite and that limit is the value of the improper integral. divergentif the limit does not exist. Each integral on the previous page is defined as a limit. If the limit is finite we say the integralconverges, while if the limit is infinite or does not exist, we say the integraldiverges. Convergence is good (means we can do the integral); divergence is bad (means we can"t do the integral). Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 4 / 15Improper Integrals
Example 1
Find?∞
0 e-xdx. (if it even converges)Solution:
0 e-xdx = limb→∞? b 0 e-xdx= limb→∞? -e-x?b0 = lim b→∞-e-b+e0= 0 + 1 = 1.So the integral converges and equals 1.
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 5 / 15Improper Integrals
Example 2
Find?∞
-∞11 +x2dx.
(if it even converges)Solution: By definition,
-∞11 +x2dx=?
c -∞11 +x2dx+? c11 +x2dx, where we get to pick whatevercwe want. Let"s pickc= 0. ?0 -∞11 +x2dx= limb→-∞?
arctan(x)? 0 b= limb→-∞[arctan(0)-arctan(b)] = 0-limb→-∞arctan(b) =π2 Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 6 / 15Improper Integrals
Example 2, continued
Similarly,?∞
011 +x2dx=π2.
Therefore,
-∞11 +x2dx=π2+π2=π.
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 7 / 15Improper Integrals
Example 3, thep-test
The integral?∞
11 xpdx1Convergesifp>1;
For example:
11 x3/2dx= limb→∞-?2x1/2? b 1= 2, while 11 x1/2dx= limb→∞?2⎷x?
b1= limb→∞2⎷b-2=∞,
and 11 xdx= limb→∞? ln(x)? b1= limb→∞ln(b)-0=∞.
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 8 / 15Improper Integrals
Convergence vs. Divergence
In each case, if the limit exists (or if both limits exist, in case 3!), we say the improper integralconverges. If the limit fails to exist or is infinite, the integraldiverges. In case 3, if eitherlimit fails to exist or is infinite, the integral diverges. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 9 / 15Improper Integrals
Example 4
Find?2
02x x2-4dx. (if it converges)Solution: The denominator of2x
x2-4is 0 whenx= 2, so the function is not even defined whenx= 2. So ?2 02x x2-4dx= limc→2-? c02xx2-4dx= limc→2-?
ln|x2-4|? c 0 = lim c→2-ln|x2-4| -ln(4) so the integral diverges. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 10 / 15Improper Integrals
Example 5
Find? 301(x-1)2/3dx,if it converges.
Solution:
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 11 / 15Improper Integrals
Example 5
Find? 301(x-1)2/3dx,if it converges.
Solution: We might think just to do
3 01 (x-1)2/3dx=?3(x-1)1/3?30,
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 11 / 15Improper Integrals
Example 5
Find? 301(x-1)2/3dx,if it converges.
Solution: We might think just to do
3 01 (x-1)2/3dx=?3(x-1)1/3?30,
but this is not okay: The functionf(x) =1 (x-1)2/3isundefined when x= 1 , so we need to split the problem into two integrals. 3 01 (x-1)2/3dx=? 101(x-1)2/3dx+?
311(x-1)2/3dx.
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 11 / 15Improper Integrals
Example 5
Find? 301(x-1)2/3dx,if it converges.
Solution: We might think just to do
3 01 (x-1)2/3dx=?3(x-1)1/3?30,
but this is not okay: The functionf(x) =1 (x-1)2/3isundefined when x= 1 , so we need to split the problem into two integrals. 3 01 (x-1)2/3dx=? 101(x-1)2/3dx+?
311(x-1)2/3dx.
The two integrals on the right hand side both converge and addup to3[1 + 21/3], so?3
01 (x-1)2/3dx= 3[1 + 21/3]. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 11 / 15Improper Integrals
Tests for convergence and divergence
The gist:
1If you"re smaller than something that converges, then youconverge.
2If you"re bigger than something that diverges, then you diverge.
Theorem
x≥a. Then1?∞
af(x)dx converges if?∞ ag(x)dx converges.2?∞
ag(x)dx diverges if?∞ af(x)dx diverges. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 12 / 15Improper Integrals
Example 6
Which of the following integrals converge?
(a)? 1 e-x2dx,(b)? 1sin 2(x) x2dx.Solution: Both integrals converge.
see?∞1e-xdx=1
e, so?∞1e-x2dxconverges.
for allx≥1. Since?∞ 11 x2dxconverges (byp-test), so does?∞ 1sin 2(x) x2dx. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 13 / 15Improper Integrals
Limit Comparison Test
Theorem
If positive functions f and g are continuous on[a,∞)and lim x→∞f(x) g(x)=L,0Example 7
: Letf(x) =1⎷x+1; consider?∞ 11 ⎷x+ 1dx.Does the integral converge or diverge?
Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 14 / 15Improper Integrals
Example 7, continued
Solution: We note thatflooks a lot likeg(x) =1⎷x, and?∞1g(x)dxdiverges by thep-test. Furthermore,
lim x→∞f(x) g(x)=⎷ x⎷x+ 1= 1, so the LCT says 11 ⎷x+1dxdiverges. Ryan Blair (U Penn)Math 104: Improper IntegralsTuesday March 12, 2013 15 / 15quotesdbs_dbs30.pdfusesText_36[PDF] convergence et divergence définition
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