Practice Problems: Improper Integrals
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1 Pracice Problem: Improper Inegral Wrien by Vicoria Kala December 6, Solion o he pracice problem poed on November 3. For each of he folloing problem: a Eplain hy he inegral are improper. b Decide if he inegral i convergen or divergen. converge o.. Solion: + d a Improper becae i i an infinie inegral called a Type I. b Rerie: d + So he inegral diverge. + d 3 + 3/ 3 = If i i convergen, find hich vale i + / d 3 + 3/. d Solion: Thi qeion a on my bjec GRE. a Improper becae he fncion i diconino a = called a Type II. b There are o ay o do hi problem, o I ill po boh olion. One ay: Spli p he inegral a = : d = d + d d + + d Boh of he i diverge o he inegral diverge. Anoher ay: i an even fncion, o i i ymmeric abo = : d = d d + = = So he inegral diverge.
2 3. r dr Solion: a Improper becae i i an infinie inegral called a Type I. b Rerie: r dr Convergen! r r dr = ln ln ln = ln = ln. Solion: y 3 3y dy a Improper becae i i an infinie inegral called a Type I. b Need o pli i p, ry abo y = : y 3 3y dy = y 3 3y dy+ = y 3 3y dy + 3 Boh of hee i diverge, o he inegral diverge. y 3 3y dy y y 3 3y dy y y3 + y Solion: co π d a Improper becae i i an infinie inegral called a Type I. b Need o pli i p, ry abo = : co π d = co π d + co π d = in π π + r in π r π = Boh of hee i diverge, o he inegral diverge. co π d + r in π π r co π d + in πr r π
3 6. ln d Solion: a Improper becae ln i ndefined a = called a Type II. b Try a -biion fir. Le =, d = d d = d. When =, = and hen =, = : ln ln d = d = ln d = ln d Thi i ill improper becae ln i ndefined a =. Rerie ih a i: ln d + ln d Ue inegraion by par e did ln d in cla once pon a ime...: ln + = + ln = ln = ln + = ln + + The righ i i ha e call indeerminae becae if e ake he i e ge omehing ha look like, hich i no beno. So e need o e L Hôpial Rle Secion., pg 3 in yor ebook: ln ln + + ln = + Thi ho ha or inegral i convergen, and i converge o + ln = =. 7. e e + 3 d Solion: a Improper becae i i an infinie inegral called a Type I. b Le do a -biion fir. Le = e, hen d = e d. When =, = and hen, : Convergen! e e + 3 d = e e + 3 d = + 3 d + 3 d 3 3 an 3 an an = 3 π 3 π 6 = 3 π π 6 = π 3 3 3
4 8. 5 d Solion: a Improper becae he fncion i diconino a = called a Type II. b Try a -biion fir. Le =, hen = +, d = d. When =, =, and hen = 5, = 3: d = + d = + Thi i ill a Type II inegral ince fncion + i diconino a =. Need o pli p he inegral: 3 + d = + 3 d + + d + d d d + ln + + ln ln + ln ln 3 + ln + Boh of he i diverge, o he inegral diverge. Ue he Comparion Theorem o decide if he folloing inegral are convergen or divergen e d Solion: a Improper becae i i an infinie inegral called a Type I. b Le ge ha hi inegral i divergen. Tha mean e need o find a fncion maller han +e ha i divergen. To make i maller, e can make he op maller or he boom bigger. Le make he op maller: Then ake he inegral: d So he inegral diverge. Since + e d ln ln = d diverge, hen +e d diverge.. π in d Solion:
5 a Improper becae he fncion in i ndefined a = called a Type II. b Le ge ha hi inegral i convergen. Tha mean e need o find a fncion bigger han in ha i convergen. To make i bigger, e can make he op bigger or he boom maller. Le make he op bigger: in Then ake he inegral: π d + π d + π = π + = π = π So he inegral converge. Since π d converge, hen π in d converge. 5
MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.
![MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.](/thumbs/86/93221008.jpg "MATH 31B: MIDTERM 2 REVIEW. x 2 e x2 2x dx = 1. ue u du 2. x 2 e x2 e x2] + C 2. dx = x ln(x) 2 2. ln x dx = x ln x x + C. 2, or dx = 2u du.")
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![1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.](/thumbs/71/65799522.jpg "1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.")
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