TOPIC 5: INTEGRATION

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1 TOPIC 5: INTEGRATION. Th indfinit intgrl In mny rspcts, th oprtion of intgrtion tht w r studying hr is th invrs oprtion of drivtion. Dfinition.. Th function F is n ntidrivtiv (or primitiv) of th function f on th intrvl I if F () = f() for ll I. Thus, both F () = nd F () = 3 r ntidrivtivs of f() = 3 in ny intrvl. Thorm.. If F nd F r two rbitrry ntidrivtivs of f on I, thn F () F () = const. on I. Proof. By dfinition of ntidrivtiv F = F = f on I, thus (F F ) () = for vry I. Sinc function with null drivtiv on n intrvl is constnt, w hv F () F () = const. Corollry.3. If F is on of th ntidrivtivs of f on I, ny othr ntidrivtiv G of th function f on I hs th form G() = F () + C, whr C is constnt. Dfinition.4. Th st of ll ntidrivtivs of th function f on th intrvl I is clld th indfinit intgrl of f on I, nd it is dnotd f() d. Notic tht by Corollry??, f() d = F () + C, whr F is on of th ntidrivtivs of f on I, nd C is n rbitrry constnt. Somtims th symbol f() d dnots not th whol st of ntidrivtivs but ny on of thm... Proprtis of th Indfinit Intgrl. () F () d = F () + C; () For ny functions f, g nd constnts, b, b g() d... Bsic Indfinit Intgrls. () d = C; () d = + C; (3) d = C ( ); d (4) = ln + C ( ); (5) d = + C ( < ), ln (f() + bg()) d = f() d + d = + C;

2 TOPIC 5: INTEGRATION (6) (7) (8) (9) () sin d = cos + C; cos d = sin + C; cos d = tn + C ( d d = rctn + C. + π + kπ, k intgr); = rcsin + C ( < < );.3. Intgrtion by Chng of Vribl. Somtims th tsk of finding th intgrl f() d is simplifid by mns of chng of vribl = ϕ(t). Th formul of chng of vribl in n indfinit intgrl is f() d = =ϕ(t) Empl.5. Find tn d. f(ϕ(t))ϕ (t) dt. Solution: Lt t = cos. Thn dt = sin d. Hnc, by th formul of chng of vribl sin dt tn d = cos d = = ln t + C = ln cos + C. t Empl.6. Find d. Solution: Lt t =. Thn dt = d. Hnc, t d = dt = t / dt = t 3/ 3/ + C = 3 t3/ + C = 3 ( )3/ + C. Empl.7. Find d. Solution: Lt t =. Thn dt = d. Morovr, = ( + t)/. Applying th chng of vribl formul w gt d = ( + t)t / dt = ( ) t / + t 3/ dt = t 3/ / + t5/ + C 5/ = 3 ( )3/ + 5 ( )3/ + C. ln Empl.8. Find d. Solution: Lt t = ln. Thn dt = d/ nd ln d = t dt = t + C = (ln ) + C.

3 Empl.9. Find d. TOPIC 5: INTEGRATION 3 Solution: Lt t =. Thn dt = d nd d = t dt = t + C = + C..4. Intgrtion by prts. For diffrntibl functions u nd v w hv (uv) = uv + vu. Tking intgrls nd givn tht (uv) () d = u()v(), w hv u()v () d = u()v() v()u () d. This rltion is known s th formul of intgrtion by prts. u () d = du nd v () d = dv w cn writ this formul s u dv = uv v du. Empl.. Find d. Using th idntifictions Solution: Lt u = nd dv = d. Thn du = d nd v =. Hnc d = d = ( ) + C. Empl.. Find ln d. Solution: Lt u = ln nd dv = d. Notic tht du = d/ nd v = 3 /3. Thn, using th formul of intgrtion by prts w gt ( ) 3 ( ) 3 3 ln d = ln d = ln ( ) 3 d = ln C. Empl.. Find rctn d. Solution: Lt u = rctn nd dv = d. Thn du = d/( + ) nd v =. Hnc rctn d = rctn + d. Now, obsrv tht using th chng of vribl t = w hv dt = d thrby + d = + t dt = ln + t + C = ln ( + ) + C. Plugging this vlu into th bov prssion w finlly gt rctn d = rctn ln ( + ) + C.

4 4 TOPIC 5: INTEGRATION Empl.3. Find sin d. Solution: Lt u = nd dv = sin d. Thn du = d nd v = cos. Thus sin d = cos + cos d. Applying gin intgrtion by prts to th scond intgrl, u = nd dv = cos d w hv du = d nd v = sin, hnc cos d = sin sin d = sin + cos + C. Plugging this vlu into th prvious prssion w finlly gt sin d = cos + sin + cos + C..5. Intgrtion of Rtionl Functions. A rtionl function is of th form P n() Q m (), whr P n nd Q m r polynomils of dgrs n nd m, rspctivly. If n m th frction is impropr nd cn b rprsntd P n () Q m () = P n m() + R k() Q m (), whr th dgr of th polynomil R k is k < m. Thus, th intgrtion of n impropr frction cn b rducd to th intgrtion of propr frction (k < m) Pn () Q m () d = Rk () P n m () d + Q m () d. Empl d = + bcus dividing th polynomils w find Thn Thorm.5. Suppos tht P n() Q m () ( + ) d = d, d = + ( + ) rctn + C. is n impropr frction (n < m) nd tht Q m () = ( ) α ( b) β ( + p + q) ( + r + s), whr,..., b r rl roots of multiplicity α,..., β nd + p + q,..., + r + s r qudrtic trinomils irrducibl into rl fctors. Thn thr r constnts A i, B i, M, N,

5 K nd L such tht P n () Q m () = TOPIC 5: INTEGRATION 5 A α ( ) α + A α ( ) α + + A + + B β ( b) β + B β ( b) β + + B b + M + N + p + q + + K + L + r + s. Th frctions which ppr on th right hnd sid r prtil frctions nd th rltion is th dcomposition of propr rtionl frction in to sum of prtil frctions. Ech of th prtil frctions cn b intgrtd in trms of lmntry functions: () A d = A ln + C, () A ( ) d = A α α + C, (α > ), ( ) α (3) M+N +p+q d = M ln ( + p + q) + N Mp Empl.6. Find d. p p 4 rctn + p p p 4 Solution: Notic tht = ( 3)( ). Thn ( 3)( ) = A 3 + B + C. A( ) + B( 3) =. ( 3)( ) Stting = w gt = B, whnc B = nd stting = 3 w gt. Hnc d = ln 3 ln + C Th Dfinit Intgrl Dfinition.. Th dfinit intgrl of non ngtiv continuous function f on th intrvl I = [, b] is th r, A, of th rgion boundd by th grph of f, th horizontl is, nd th two lin sgmnts =, = b. It is dnotd Empl.. If f() =, thn b f() d = A. f limitd by =, = is squr tringl with r /. f() d = /, sinc th figur undr th grph of Dfinition.3. Th dfinit intgrl of non positiv continuous function f on th intrvl I = [, b] is th r of th rgion boundd by th grph of f, th horizontl is, nd th two lin sgmnts =, = b. Hnc b f() d = A.

6 6 TOPIC 5: INTEGRATION It is strightforwrd to dfin th dfinit intgrl of function tht chng sign in th intrvl [, b]. By wy of mpl, suppos tht f is continuous on [, b] nd stisfis f in [, c], f in [c, b]. Thn th dfinit intgrl of f on [, b] is th diffrnc of th rs b f() d = A [,c] A [c,b] = c f() d b Mor compl situtions cn b similrly hndld. c ( f)() d = Empl.4 (Empl??, continud). If f() =, thn w know f() d = / nd c f() d + b c f() d (s Proprty (3) blow). f() d =, sinc ( f)() d = /. Th lttr is bcus th rgion boundd by f btwn = nd = is gin squr tringl with r /... Proprtis of th dfinit intgrl. In wht follows f nd g r continuous functions on [, b] nd α, β R. () () (3) b b f() d = ; f() d = b f() d; αf() + βg() d = α (4) For ny c [, b], b b f() d = (5) If f() g() in [, b], thn f() d + β b c b f() d + f() d b g() d. b 3. Brrow s Rul c f() d. g() d. In this sction w show th connction btwn rs nd ntidrivtivs. Dfinition 3.. Lt th function f b continuous on th intrvl [, b]. Th function F () = is clld n intgrl with vribl uppr limit. f(t) dt ( b) Thorm 3. (Fundmntl Thorm of Intgrl Clculus). If th function f is continuous on th intrvl [, b], thn th function F () = Writtn in othr trms, th thorm stblishs ( f(t) dt) = f(). f(t) dt is n ntidrivtiv of f in [, b]. Thorm 3.3 (Brrow s Rul). If th function f is continuous on th intrvl [, b], thn b f() d = G(b) G(), holds tru, whr G is n ntidrivtiv of f in [, b].

7 TOPIC 5: INTEGRATION 7 Proof. Lt G b n rbitrry ntidrivtiv of f in [, b]. Thn G F is constnt in [, b] by Thorm??, whr F () = b f() d, bcus F is lso n ntidrivtiv of f. Hnc G() F () = G(b) F (b), or G(b) G() = F (b) F () = b f() d f() d = b f() d. Most oftn w will writ G(b) G() s G() b. Thorm 3.4 (Chng of vribl). Lt f b continuous in [, b], nd lt = g(t) b continuous, togthr with th drivtiv in [α, β], whr g(α) =, g(β) = b nd g(t) b. Thn b f() d = β α f(g(t))g (t) dt. Thorm 3.5 (Intgrtion by prts). If f nd g hv continuous drivtivs in [, b], thn b f()g () d = f()g() b b f ()g() d. 3.. Th r of pln figur. Givn continuous function f, th r of th figur boundd by th curv y = f(), th is OX nd th lin sgmnts =, = b is b f() d. Empl 3.6 (Empl??, continud). Th r of th figur limitd by y = in th intrvl [, ] is d = ( ) d + ( ) d = + =. Empl 3.7. Th r of th figur limitd by th grph of y = ln nd th horizontl is nd th lin sgmnts = /, = is ln d. Th logrithm is ngtiv in [/, ] nd positiv in [, ]. thus ln d + ln d Th intgrl cn b solvd using prts u = ln, dv = d obtining ln d = ln = +, Thus, ( + /) + = ( /). ln d = ln =.

8 8 TOPIC 5: INTEGRATION Suppos tht pln figur is boundd by th continuous curvs y = f(), y = g(), b, whr g() f(), nd two lin sgmnts =, = b (th lin sgmnts my dgnrt into point). Thn th r of th figur is b (f() g()) d. Empl 3.8. Find th r of th figur boundd by th curvs y = 3, y = in th intrvl [, ]. Solution: Th curvs mt t singl point. Solving th qution 3 =, w find th bsciss of th point, =. Hnc on of th curvs rmins bov th othr in th whol intrvl. To know which, w simply substitut into 3 + n rbitrry vlu in th intrvl; for = / w gt 3 + =/ =.375 >, thus 3 is bov in [, ]. Th r is 3 ( ) d = = ( ) = y= 3.5 y= Empl 3.9. Find th r of th figur boundd by th curvs y =, y =. Solution: Th curvs mt t two points. Solving th qution = w find tht th points r =, =. Hnc on of th curvs rmins bov th othr in th intrvl [, ]. To know which, w simply substitut into = n rbitrry vlu on th intrvl [, ]; for =, = = >, thus y = is bov Y = in [, ]. Th r is d = 3 3 = ( 3 ) ( ) = 9.

9 TOPIC 5: INTEGRATION 9 3 y= A y= 3 3