MA Exam 2 Study Guide, Fall u n du (or the integral of linear combinations

Transcription

1 LESSON 0 Chpter 7.2 Trigonometric Integrls. Bsic trig integrls you should know. sin = cos + C cos = sin + C sec 2 = tn + C sec tn = sec + C csc 2 = cot + C csc cot = csc + C MA 6200 Em 2 Study Guide, Fll 204 tn = ln sec + C = ln cos + C sec = ln sec + tn + C cot = ln sin + C csc = ln csc cot + C 2. Products nd powers of trig functions. Usully the ide is to chnge the integrl to the form of u n ). u n du (or the integrl of liner comintions Integrnds contining powers of sine nd cosine functions. sin m cos n. Odd powers of sin fctor odd fctor of sin nd use the identity sin 2 = cos 2 to convert remining even powers of sin to cos. Then let u = cos. Odd powers of cos fctor odd fctor of cos nd use the identity cos 2 = sin 2 to convert remining even powers of cos to sin. Then let u = sin. No odd powers of sin or cos use cos 2 = cos(2) or sin2 = 2 2 cos(2). After integrting the trig identity sin 2 = sin cos my e useful. Integrnds contining tngent nd secnt functions. tn m sec n. fctor sec 2 nd chnge remining fctors of sec 2 to tngent using sec 2 = tn 2 +. fctor tn sec nd chnge remining fctors of tn 2 to secnt using tn 2 = sec 2.

2 LESSONS nd 2 Chpter 7.3 Trig Susitution If integrnd contins Use this sustitution And use this identity 2 (u()) 2 u() = sin θ sin 2 θ = cos 2 θ 2 + (u()) 2 u() = tn θ + sin 2 θ = sec 2 θ (u())2 2 u() = secθ sec 2 θ = tn 2 θ * Sometimes the integrnd will contin powers of these roots. Completing the squre is sometimes necessry to get the forms 2 (u()) 2, 2 +(u()) 2, or (u()) 2 2. LESSONS 3 nd 4 Chpter 7.3 Prtil Frctions Let P() e rtionl function (P() nd Q() re oth polynomils) where P nd Q hve no Q() common fctors nd the degree of P is less thn the degree of Q. If P nd Q hve common fctors, cncel them. If degree of P not less thn degree of Q, then use long division to divide P y Q. Fctor the denomintor Q() into liner fctors (+) nd irreducile qudrtic fctors ( 2 ++c, where 2 4c < 0). m liner fctors ( + ) m give rise to m prtil frctions of the form A + + A 2 ( + ) 2 + A 3 ( + ) A m ( + ) m + A m ( + ) m A,A 2,A 3,...,A n re constnts. When you set up your prtil frctions use A,B,C,D,... in the numertors insted. With the emples we use, you won t run out of letters! n irreducile qudrtic fctors ( c) n give rise to n prtil frctions of the form A + B c + A 2 + B 2 ( c) 2 + A 3 + B 3 ( c) A n + B n ( c) n + A n + B n ( c) n A,A 2,A 3,...,A n nd B,B 2,B 3,...,B n re constnts. When you set up your prtil frctions use A + B, C + D, E + F,... in the numertors insted. With the emples we use, you won t run out of letters! Set the rtionl function P() equl to its sum of prtil frctions. Net multiply oth sides of this Q() eqution y the common denomintor Q(). Then solve for the constnts y either choosing vlues of tht will eliminte ll ut one constnt, nd/or equting the coefficients of like powers of to get system of liner equtions where the unknowns re the constnts. Solve tht system using your fvorite method.

3 When integrting the prtil frctions, you will usully e deling with the following kinds of integrls: = ln + C ( )n+ = ( ) n n = tn + C + C, n LESSON 5 Chpter 7.6 Integtion using tles When using tle of integrls to evlute f(), rememer tht is not lwys equl to u. Sometimes you first must use sustitution u = u(). If you don t, your nswer will e incorrect y constnt fctor. Chpter 7.7 Approimte Integrtion ( ) To pproimte f(), sudivide the intervl [,] into n equl suintervls. Let i = +i. n Note tht 0 = nd n = n n Midpoint Rule: ( ) ( ( ) ( ) ( ) ( )) n + n f() M n = f + f + f + + f n Trpeziodl Rule: ( ) f() T n = (f( 0 ) + 2f( ) + 2f( 2 ) + + 2f( n ) + f( n )). 2n Simpson s Rule: Note: n must e even f() ( ) S n = (f( 0 ) + 4f( ) + 2f( 2 ) + 4f( 3 ) + 2f( 4 ) + + 2f( n 2 ) + 4f( n ) + f( n )). 3n

4 LESSON 6 Chpter 7.8 Improper Integrls The integrl = = f() is improper if t lest one of the following is true: f hs n infinite discontinuity t some point c in the intervl [,] (the grph of f hs verticl symptote = c). We determine whether n improper integrl hs vlue y evluting the limit of proper integrls s follows: ( rememer to evlute the proper integrl on the right hnd side efore evluting the limit) Improper integrl of Type I t f() = lim f(). t f() = lim f(). t t Improper integrl of Type II If f hs discontinuity t, then If f hs discontinuity t, then f() = lim f(). t + t t f() = lim f(). t If the limit eists, then the limit vlue is the vlue of the improper integrl. If the limit does not eist, then the improper integrl is divergent. If f hs n infinite discontinuity t c where < c <, then f() = c f() + c f() provided BOTH improper integrls on the right hnd side eist. If either diverges, then diverges. An importnt integrl: is convergent if p > nd divergent if p. p f() Comprison Test for Improper Integrls When you cn t determine n nti-derivtive for n improper integrl, comprison with known convergent or divergent integrl is useful. Suppose tht f nd g re continuous functions with 0 g() f() for.

5 If If LESSON 7 Chpter 8. Arc length f() is convergent, then g() is divergent, then If y = f() nd, then L = If = g(y) nd c y d, then L = d Chpter 8.2 Are of surfce of revolution c g() is convergent. f() is divergent. + (f ()) 2 + (g (y)) 2 dy Rotte out the is y y = f() Rotte out the y is R y R y = f() R = rdius = f() R = rdius = Surfce Are = 2π(f()) + ( ) 2 dy Surfce Are = 2π() + ( ) 2 dy