Mastering Integration by Parts: Tips and Techniques Explained
School
Florida State College at Jacksonville**We aren't endorsed by this school
Course
MAC 2313
Subject
Mathematics
Date
Dec 11, 2024
Pages
2
Uploaded by CountInternetGoldfinch16
Homework 5 Section 7.1: 1 / . 2 Letu=Inz,dv=\xdr = du= < dr,v = %.rS' 2. Then by Equation 2, / Vzlhaeder = 22*? nx - / 2257 % de = 32*?Inx - / 22 de = 22% ma — $2°% 4 C. Note: A mnemonic device which is helpful for selecting « when using integration by parts is the LIATE principle of precedence for u: Logarithmic Inverse trigonometric Algebraic Trigonometric Exponential If the integrand has several factors, then we try to choose among them a w which appears as high as possible on the list. For example, in [ e’ dx the integrand is z:¢>*, which is the product of an algebraic function () and an exponential function (€2). Since Algebraic appears before Exponential, we choose u = x. Sometimes the integration tums out to be similar regardless of the selection of u and dv, but it is advisable to refer to LIATE when in doubt. 3. Letu =x, dv =cosbrdr = du=dr.v= %sin 5. Then by Equation 2, frcusf).rd.r = %rsin-fil‘ - f%sin&‘rdr = %.l'sinfi.r-{- %cosSI+(‘. — M. Letu = Int, dv = t*dt = du= %dt. v= gts. Then by Equation 2, 4 1s 151 1s 1 1s ls, ~ /t llltdt=3t Int — t '?dt=gt Int — gt dt=3f hlt—%t +C. 5 2. Fistletu =2’ + 1, dv=e"dr = du=2zdr,v=—e". By (6). a2+ 1)e Fda = [—(a® + 1)e %]} + [} 2ze " da = -2 + 1+ 2 [} ze* da. fo 0 0 0 NextletU = 2,dV =e “dr = dU =dx,V = —e~*. By (6) again, lre*dr = [-ze™*] + [efdr = —e 1+ [—e*] ) = el —e" 1 +1 =21 +1. So 0 0 0 0 fl@®+1)e " de=—2e"+142(-2e7+1) =21 +1-de ' +2=—6e7" +3. 1 1 1 2]. Letu = In R, dv = fidR = du= T?dR' V=g By (6). *InR 1 ’ * 1 11,5 1)* 11,5 1 4 _ 115 [?dR_[_T?IHR]l_,/I —EdR——gln-J—O—[T?]l——gll]-)—(g—l)—g—g]no, 37. Lett = \/z. so that t* = z and 2t dt = dx. Thus, fe‘/; dx = [ €'(2t)dt. Now use parts with u = t, dv = e* dt, du = dt, andv =e‘toget2 [te'dt =2te' —2 [e'dt =2te' — 2" +C =2,/TeV" — 2V +C. 4. Lety = 1 + . so that dy = dx. Thus, fl‘lll(l +z)dr = f(y ~1)Inydy. Now use parts with u = Iny. dv = (y — 1) dy. du=§dy.v=3y° —ytoget [y=1mydy= (3>~ v)ny — [ (3y 1) dy= 3yl 2y~ 3? +y+ € =11+a)@-)h(l+2)-31+2)°+1+2+C, which can be writtenas 3(2> — 1)In(1+2) — 42 + o+ 3 + C. 57. The curves y = 22 Inz and y = 4In intersect when 2° Inz = 4lnz < 2’Inz—4lnr=0 = (2’-4)hr=0 = z=1or2 [sincex > 0]. For1 < & < 2.4Inz > 2° Inz. Thus, area = ff(-l Inz — 2’ Inz)de = f12[(4 —2?)Inz]dr. Letu = Inz. dv=(4-2%)dr = du=1dr v=4z— }2° Then 0|‘—7' area = [(Inz)(4z — %1‘3)]? - /2 [(41‘ - %13)%} dr = (In2)(%) -0- /2 (4-32%) da 1 3 1 — 16, 1,312 _ 161, ¢ 64 _ 35\ _ 16|, ¢ 29 —71112—[41‘—3.1 ]1—?1112—(?—?)—71112—?
m=((_’%+%w%= W (e ek -9+ 50T (Gpngvam ) +[gJomos ] 0P (%&xrf x,glwf(mn—n?—,;y + "f'};j) ¥- Yoy e R gif e o0 3T = SAS oo T = [ [ ey o iy ey Rpes - i - T =y PR N = = s 4 2V S gg° 7 b~ go 2k B L | ogIs-Son ‘(o' s Ty o v o vh v om0 by o5 13 9k hwemd Ay 2 Tz g s GORA o S it Jo pen o ok 25 = [1+cue - (G = [z +cuv — (cu)g)s = {:[z]zfluw - Z(Z“I)Z}-'L = {(JP J- :[ru{ r])a - L(zfll)z}u = r=a wpi=np veare [ e L p=op wu=n pruz (e pr = r=a zpf-zug=np o e [ = ap = ap (zu)=n Py [)-z =A ISIXE-2 31 0 SYSIP 3s7) (q) (¢ - zme)ee = {{[:2%) - (0 -z pog = {wpat }f - [ww 28] fug = . . . s =a wpionp B [‘t“:«p -z.q:,,] aprureg /—_\ @ stxe-fl a1 0qe S[IRYS 3s7) (2) ‘69