AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

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1 AP Clulus BC Chpter 8: Integrtion Tehniques, L Hopitl s Rule nd Improper Integrls

2 8. Bsi Integrtion Rules In this setion we will review vrious integrtion strtegies. Strtegies: I. Seprte the integrnd into two frtions II. Rewrite the integrte in n equivlent form. + e (Hint: Add nd sutrt e in the numertor nd then rewrite s the sum of two frtions. OR Multiply the frtion y e e. )

3 III. Complete the squre. (Hint: When you omplete the squre in the denomintor, you my end up with n integrl for r sin, r tn, or r se.) IV. Divide Rtionl Funtion. + (Hint: If the degree of the numertor is greter thn or equl to the degree of the denomintor, it my e helpful to divide the numertor y the denomintor.)

4 V. Epnd the epression. ( + e ) VI. Use Trigonometri Identities. ot (Hint: Use trig identity to rewrite the integrnd. In this se, use Pythgoren Identity.)

5 VII. Multiply nd Divide y the Conjugte. + sin (Hint: Multiply numertor nd denomintor y ( sin ) )

6 8. Integrtion y Prts Question: How do we integrte epressions suh s ln or e or e sin??? We will use new method, integrtion y prts, to integrte these omplited looking integrls. The produt rule is the sis for deriving the formul for integrtion y prts. Given u() nd v(), the derivtive of the produt u( ) v( ) is given y: Integrting oth sides of the eqution gives: d u ( ) v ( ) = u ( ) v '( ) + v ( ) u '( ) d [ u() v() ] = [ u() v'() + v() u'() ] u() v() = u() v'() + v() u'() u() v() = u() dv + du v() u() v() = u() dv + v() du Rerrnge the eqution to get the formul for Integrtion y Prts: u()dv = u() v() v()du Using simpler nottion we get: u dv = uv v du The new formul epresses the originl integrl in terms of nother integrl tht should e esier to evlute. To pply the Integrtion y Prts formul:. Determine the pproprite sustitution for u nd dv.. Integrte dv to solve for v, nd differentite u to solve for du. 3. Sustitute the vlues for u, v, du, nd dv into the formul. 4. Solve the new integrl.

7 Emple sin Courtesy of Emple e Emple 3 e os

8 Emple 4 ln The Tulr Method This method is useful when you must pply integrtion y prts multiple times. For instne, when integrting ny of the forms elow, the tulr method is helpful. Emple 5 n sin or n os or n e sin 4 Let u = nd dv = sin 4 The sign pplies to Differentite until you get to 0. Integrte. u nd u'. Alternting Sign +/- u nd du dv nd v The solution is otined y dding the signed produt of the digonl entries.

9 8.3 Trigonometri Integrls Some simple trig integrls tht you lredy hve done Use U-su.. sin 3 os. tn 4 se 3. se 3 se tn Now let s mke it little more hllenging In this setion we will e solving integrls like: sin m osn The following trig identities will prove quite useful in this setion. Pythgoren Identity sin + os = Power Redution Formuls sin os = os + os = Notie tht the Power Redution Formuls re just nother rrngement of the Doule Angle Identities from trig: os = sin nd os = os

10 . When the power of either sin or os is odd nd positive, keep one ftor of sin nd onvert ll other ftors to os, OR keep one ftor of os nd onvert ll other ftors to sin. Use sin or os s your du for u-sustitution. Emple: sin 3 os 4. When the powers of oth sine nd/or osine re even (you n't use the first strtegy) mke repeted use of the power reduing identities ove. Emple: os 4

11 8.3 Trigonometri Integrls Dy In this setion we will e solving integrls of the form: se m tn n Strtegies:. If the power of sent is even nd positive, sve sent squred ftor for the du nd onvert the remining ftors to tngent. Use the identity: tn + = se Emple: se 4 tn 3. If the power of tngent is odd nd positive, sve sent-tngent ftor for the du, nd onvert the remining ftors to sents. Emple: tn 3 se

12 3. If there re no sent ftors nd the power of tngent is even nd positive, onvert tngent-squred ftor to sent-squred ftor. Let the sent-squred ftor e your du. Emple: tn 4 4. If there re no tngent ftors nd the power of sent is odd nd positive, use Integrtion y prts. Emple: se 3 5. If none of the 4 strtegies ove pply, try onverting to sines nd osines! se Emple: tn

13 8.5 Prtil Frtions The deomposition of rtionl funtion into simpler rtionl frtions, otherwise known s prtil frtions n e helpful in the integrtion of diffiult rtionl funtions.. Divide n improper frtion (degree of numertor is greter thn or equl to the degree of the denomintor).. Deompose rtionl funtions into prtil frtions. (see ll ses elow) 3. Integrte the resulting terms. ) 4 ) 3 6

14 3) 5 3

15 8.7 Indeterminte Forms nd L'Hopitl's Rule Emples of some indeterminte forms: 0 0 0,,,, 0,,, 0, 0 In the proess of determining limit we osionlly rrive t one of the indeterminte forms listed ove. We n only speulte out the mening of epressions suh s these. Previously, we might hve used lgeri tehniques or grphing lultor to solve suh dilemm, ut now we will lern to use L'Hopitl's rule to determine the vlue of n indeterminte form when it is the solution to given limit prolem. First, we'll review tehniques with whih you re lredy fmilir. Algeri Method Emple : lim = 0 y diret sustitution Rewrite the frtion using lger nd try evluting the limit gin. 3 lim ( )( + ) = lim 3( + ) = 6 ( ) Emple : 5 lim + = y diret sustitution 6 7 Rewrite the frtion using lger nd try evluting the limit gin. 5 5 lim lim = = 5 6 Grphing Clultor Method e 0 Emple 3: lim = 0 0 y diret sustitution From the grph nd the tle of vlues in the neighorhood of = 0, you n pproimte the limit. e lim 0 =

16 L'Hopitl's Rule Let f nd g e funtions tht re differentile on n open intervl (, ) ontining, eept possily t itself. Assume tht g'() is not equl to 0 for ll in (, ), eept possily t itself. If the limit f ( ) s pprohes produes the indeterminte form g( ) 0/0, then lim f () g() = lim f '() g'() provided the limit on the right eists or is infinite. This result lso pplies if the limit of f ( ) g( ) s pprohes produes ny one of the indeterminte forms,,,. We n lso pply L'Hopitl's Rule to limits pprohing. f lim ( ) f lim '( = ) g( ) g'( ) It is importnt to note tht L'Hopitl's rule pplies only to limits tht yield the indeterminte forms 0 / 0 or ± / ±. If limit yields one of the other indeterminte forms listed ove nd n e rewritten to yield the form 0 / 0 or ± / ±, then you n pply L'Hopitl's rule to the new form. Note lso, there re other forms known to e "determinte" forms suh s: The lst determinte form is perhps the most diffiult to ept. It is not intuitively ovious. A proof of the lst determinte form follows t the end of the notes for the prtiulrly inquisitive Clulus student. Let's use L'Hopitl's Rule to determine the limit in Emple 3 from pge. Emple 3: (gin) e lim 0 =

17 Emple 4: lim sin 0 = Emple 5: lim os 0 = Emple 6: lim = Emple 7: sin 8 lim = 0 Emple 8: lim e = Emple 9: ln lim Emple 0: lim sin

18 Emple : lim e Emple : lim ( sin ) HINT: Set the limit equl to y nd 0 + tke the ln of oth sides of the eqution. Emple 3: lim( os ) 0 Emple 4: lim + ln NOTE: When possile, try heking these limits using grphing lultor. J

19 8.8 Improper Integrls You will rell tht when we evlute definite integrl, the intervl [,] is losed intervl. In other words, the limits of integrtion re usully two onstnts. In ddition, the Fundmentl Theorem of Clulus requires tht funtion e everywhere ontinuous in order to integrte the funtion. There re two types of integrls tht defy these typil requirements for integrtion. We ll them Improper Integrls. We will first onsider integrls tht hve one or oth limits of integrtion equl to or -. Then we will onsider integrls tht hve disontinuity t or etween the limits of integrtion. Infinite Limits of Integrtion Cse I: If f is ontinuous on the intervl [, ), then f () = lim f () Cse II: If f is ontinuous on the intervl (-,], then f () = lim f () Cse III: If f is ontinuous on the intervl (-, ), then f () = f () + f () where is ny rel numer. Rewrite s the sum of two limits = lim f () + lim f () In the first two ses, the improper integrl onverges if the limit eists - otherwise, the improper integrl diverges nd we n not evlute it. In the third se, the improper integrl on the left onverges only when oth integrls on the right hve limit nd re therefore onvergent. Emple : Rewrite the integrl s limit: lim Now, evlute the integrl nd then determine the limit. Emple : Rewrite the integrl s limit: lim Compre the grphs of f ( ) = nd f ( ) =. The grphs re very similr nd yet one integrl onverges while the other diverges. Cn you eplin why?

20 Emple 3: e vlue of, let's sy 0 in this se. lim e Rewrite the integrl s the sum of two limits, hoosing onvenient 0 + lim e 0 Integrls with Disontinuity t or within the Limits of Integrtion Cse I: If f is ontinuous on the intervl [, ) nd hs n infinite disontinuity t, then f () = lim f (). Cse II: If f is ontinuous on the intervl (, ] nd hs n infinite disontinuity t, then f () = lim + f (). Cse III: If f is ontinuous on the intervl [,] eept for some in (,) t whih f hs n infinite disontinuity, then Rewrite s the sum of limits f () = f () + f (). = lim d d f () + lim e + e f () In the first two ses, the improper integrl onverges if the limit eists - otherwise, the improper integrl diverges. In the third se, the improper integrl on the left onverges only if oth integrls on the right onverge. Emple 4: 4 Rewrite s limit: lim 4 4 4

21 Emple 5: Rewrite s limit: lim Emple 6: 8 Rewrite s the sum of two limits: 3 lim lim

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