Math 113 Exam 2 Practice


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1 Mth Em Prctice Februry, 8 Em will cover sections 6.5, nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number of prctice questions for you to work on. The third section give the nswers of the questions in section. 6.5: Averge Vlue of Function () Find the verge vlue of the following functions over the specified intervl (b) Find c in the intervl on which the function chieves its verge vlue. (Such c is gurnteed to eist, by the Men Vlue Theorem for Integrls.) 7.: Integrtion by Prts u dv = uv v du Integrtion by prts is most often useful when integrting function of the form n e, n sin, n cos, n ln. If possible, you wnt to choose u to be function tht becomes simpler when differentited, nd dv to be function tht cn be redily integrted. This usully mens you should choose u = n. (But in the cse n ln, choose u = ln ). Integrtion by prts is lso useful for integrting inverse functions such s sin, tn, ln or functions involving these s fctors. In this cse, you should choose u = sin, u = tn, or u = ln ccordingly, even if there re no other fctors in the integrnd (i.e., you cn set dv = ). 7.: Trigonometric Integrls For sin m cos n : If n is odd, sve one cos nd convert the rest to sin using cos = sin If m is odd, sve one sin nd convert the rest to cos using sin = cos If both m nd n re even, use the identities sin = cos nd cos = +cos. For tn m sec n : If n is even, sve one sec nd convert the rest to tn using sec = tn + If m is odd, sve sec tn, nd convert the rest to sec using tn = sec. If m is even nd n is odd, convert everything to sec nd integrte by prts with dv = sec. (This lst cse will require solving for the desired integrl.) For tn n, convert one tn to sec nd split the problem into two integrls.
2 sec = ln sec + tn + C. A similr strtegy pplies for cot m csc n. 7.: Trigonometric Substitution If the integrnd involves, use = sin θ. If the integrnd involves +, use = tn θ. If the integrnd involves, use = sec θ. If the integrnd involves + b + c, complete the squre to get it into the form ( h) + k. After fctoring out the nd pplying the substitution u = h, the integrnd will then fit one of the three forms bove. Avoid using trigonometric substitution when regulr usubstitution is possible. 7.4: Integrtion of Rtionl Functions by Prtil Frctions Rtionl functions consist of frctions of polynomils. We cn split rtionl functions into simpler pieces by prtil frctions. Remember tht prtil frction decompositions re bsed on liner nd qudrtic fctors in the denomintor. For ech liner fctor, we hve term with constnt in the numertor nd the fctor in the denomintor. For ech irreducible qudrtic, we hve term with liner function in the numertor nd the qudrtic in the denomintor. For emple, + ( )( + )( + + ) = A + B ( + ) + C + D + +. () We just need to determine the vlues of A, B, C nd D. This is done by plugging in vlues for : You need to plug in s mny numbers s you hve constnts. Using some numbers, (like  nd in this cse) mkes your life esier, but ny four numbers will do. Notice tht if we multiply eqution by the denomintor on the left side, we get + = A( + )( + + ) + B( )( + + ) + (C + D)( )( + ). () Letting = in eqution gives 8 = 5B, so B = 8/5. Letting = gives = 5A, so A = /5. Letting = gives = 5A B D. Knowing A nd B helps us to find D. Finlly, if =, 4 = 9A 6B + C D, llowing us to solve for C. Remember tht repeted fctors must give repeted terms with incresing eponent in the denomintor. For emple, ( ) ( + 9) = A + B ( ) + C ( ) + D + E F + G ( + 4) Finlly, remember prtil frctions only works if the degree in the numertor is less thn the degree in the denomintor. Otherwise, you need to divide nd use prtil frctions on the reminder. 7.8 Improper Integrls Since this is the only section from chpter 7 on this em, it should be pretty esy to spot Improper integrls. However, tht will not be the cse on the finl. You should mke sure you cn recognize improper integrls nd know how to solve them. Remember, there re two types of improper integrl:
3 Infinite length: Integrls of the type f(), f(). Unbounded integrnd. Integrls of the type b f(), where f hs n infinite discontinuity t. c f() Remember, in ech type, there is problem tht definite integrl cnnot hndle. We remove the problem by turning it into limit: c f() = lim b b f() = lim b b c f() f(). Some things to remember when clculting improper integrls: Do not forget to set up n improper integrl s limit. You will likely hve points deducted if you do not. It is the only wy for the grder to tell tht you know wht you re doing. Wtch out for infinite discontinuities in the middle of the intervl. You must split the integrl t the discontinuity in tht cse. Appendi G This section is somewht of deprture from Chpter 8. This section is bsiclly included to fill some gps in your mthemticl eduction. We cn understnd the nturl logrithm better if we define it s n re integrl, thn if we just define it s the inverse of the nturl eponentil. In some wys, it is more nturl wy to develop these two functions. As you prepre for the em, you should concentrte on two spects: Use the re integrl form of the nturl logrithm to pproimte it (by pproimting the re). Be ble to prove some of the properties of the nturl logrithm using the re integrl form of the nturl logrithm.
4 Questions Try to study the review notes nd memorize ny relevnt equtions before trying to work these equtions. If you cnnot solve problem without the book or notes, you will not be ble to solve tht problem on the em. For the questions to, find the verge vlue of the function over the intervl, nd find the vlue c where f(c) is equl to the verge vlue (or show why no such vlue eists).. f() = +, [, ]. f() = sin, [, π ]. f() = /( + ), [, ] For problems 4 to, evlute the integrl. 4. cos 5. π sin 6. e 7. sin 8. tn 9. ln. e cos. π π sin( ) [Hint: First use u substitution]. π 6 sin cos. π cos4 4. π 4 tn sec 5. tn sec 6. tn 6 7. tn sec [Hint: Use rtionlizing substitution] 9. π... cos [Hint: Use the substitution t = tn( )] For questions to 7, evlute the integrl, or show tht it diverges. ln ln 4 ( + ) 4. ln = 5. π sec π/ 6. ln = For questions 8 to 9, use the Comprison Theorem to show whether the improper integrl converges or diverges. tn + e
5 4. By compring the res, show tht ( ) < ln(n + ) < 5 n+ ( n ). 4. Find the eqution of the tngent line to the curve y = /t tht is prllel to the secnt line AD, where A = (, ) nd D = (, /). 4. Wht is log + ln( e ) + log log 5 6? () 5 (b) 7 (c) (d) (e) log 5 (f) 5 log 5 4. Which one is NOT right? () ln mens the re under the curve y = /t from t = to t = ; (b) lim e = lim e = ; (c) lim + ln = lim + ln = ; (d) (e) is convergent; (ln ) d = (ln ). Answers. f ve = 5, c = ±. π f ve =, c = π f ve = ln, c = ln 4. sin + cos + C 5. tn + C 6. tn tn + tn + C (sec tn ln sec + tn ) + C 6. e 8. ln ln C 7. π tn + tn + C C 9. ln. + ln C.. e (sin + cos ) + C. ln( + ). (π ). sin ( ) 4 + C
6 5. ln ln C 6. ln + 5 ln + + C 7. ln + ln + + tn + C 8. (ln + tn ) 9. (ln 5 ln ). 9 π. Diverges Diverges Diverges. 8. Converges. Mke the comprison tn < π/. 9. Diverges. Mke the comprison + e > c) 4. d)
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