Section - 2 MORE PROPERTIES
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1 LOCUS Section - MORE PROPERTES n section -, we delt with some sic properties tht definite integrls stisf. This section continues with the development of some more properties tht re not so trivil, nd, when properl pplied, cn e etremel powerful. However, even the justifictions for these properties stem from ovious phsicl ( grphicl) interprettions of functions. (9) f ( ) f ( ) This propert ss tht when integrting from to, we will get the sme result whether we use the function f ( ) or f ( ). The justifiction for this propert will ecome cler from the figures elow: f() f() - f(-) As progresses from to, the vrile - progresses from to. Thus, whether we use or -, the entire intervl [, ] is still covered. Fig - 5 The function f( - ) cn e otined from the function f() first flipping f() long the -is nd then shifting it right units. Notice tht in the intervl [, ], f() nd f( - ) descrie precisel the sme re. There re two ws to look t the justifiction of this propert, s descried in the figures on the left nd right respectivel. Let us see how to ppl this propert usefull: Emple 3 Evlute / sin. sin cos Oserve tht evluting the indefinite integrl of the function ove would e ver tedious. Using Propert-9, this integrl cn e immeditel simplified; we use the sustitution Therefore, ecomes: / cos sin cos.
2 LOCUS Now, we dd the originl epression for nd this newl otined modified epression: / sin sin cos cos / (Quite simple now!) Emple Evlute ln( tn ) We gin use propert -9 nd tr to simplif this integrl; use the sustitution function to e integrted: in the ln tn tn ln tn ln tn ln ln( tn ) ln ln ln ln 8 Agin, we see tht using propert - 9 sved us from the tedious tsk of first evluting the nti-derivtive.
3 LOCUS Emple 5 Evlute / sin cos sin cos We first use propert -9 to simplif this integrl; use the sustitution integrted: in the function to e / cos sin cos sin Adding the originl nd the modified epressions of, we otin : / sin cos sin cos / tn sec tn Now we use the sustitution tn tn sec d nd when cn now e written in terms of s : d 8 8 tn 6
4 LOCUS 3 () We cn now etend propert -9 into more generl propert : f ( ) f ( ) Grphicll, the justifiction for this propert is nlogous to tht for propert As the vrile vries from to, the vrile + - vries from to. Thus, whether we use or + -, the entire intervl [, ] is covered in oth the cses nd the res will e the sme The grph of f(+-) cn e otined from the grph of f() first flipping the grph of f() long the -is nd then shifting it (+) units towrds the right; the res descried f() nd f(+-) in the intervl [, ] re precisel the sme Fig - 6 There is stright forwrd nlticl justifiction lso. Use t in the right side integrl for tht purpose. Propert - is one of the most widel used properties to simplif definite integrls. Emple 6 Evlute /3 / 6 tn / 3 / 6 sin cos /3 / 6 cos sin cos...() We now use propert -; we sustitute 6 3 integrted. Thus, ecomes: Adding () nd (), we otin / 3 / 6 /3 / 6 sin sin cos sin cos sin cos i.e. in the function to e...()
5 LOCUS / 3 / () f ( ) f ( ) f ( ) The justifiction for this propert is descried elow: To evlute f ( ) f ( ) f ( ). f ( ), we cn equivlentl use the vrile ( ) insted of, ut the limits of integrtion will chnge from ( to ) to ( to ). This is ecuse s vries from to, will vr from ( to ) covering the sme intervl [, ]. Thus, Hence, the stted ssertion is vlid f ( ) f ( ) Emple 7 f f is n even function, then prove tht / f (cos ) cos f (sin ) cos On the left side, the integrtion limits re to while on the right side, the re to. Thus, it would e pproprite to use Propert - / f (cos ) cos f cos cos f cos cos f (cos ) cos f ( cos )sin [since f is n even function, f ( cos ) f (cos ) ] f (cos ){cos sin } f (cos )sin...()
6 LOCUS 5 Now we use propert - 9 to otin the finl form tht we require; use the sustitution in the function to e integrted in () f cos sin f (sin )cos We could lso hve strted with propert -9 directl: / f (cos ) cos...() / f cos cos / f cos sin Adding () nd (3) we otin / f cos sin...(3) / f (cos )(sin cos ) Notice now tht the function eing integrted on the right side ove is smmetric out ; i.e., if we sustitute for, we will otin the sme function gin. Thus, (propert 7): f (cos )(sin cos ) f (cos ) sin f (cos )sin This is the sme epression tht we hd otined in (). From here, we cn proceed s descried erlier.
7 LOCUS 6 () Sometimes, it is convenient to chnge the limits of integrtion into some other limits. For emple, suppose we hve to dd two definite integrls nd ; the limits of integrtion for these integrls re different. f we could somehow chnge the limits of into those of or vice-vers, or in fct chnge the limits of oth nd into third (common) set of limits, the ddition could e ccomplished esil. Suppose tht f ( ). We need to chnge the limits ( to ) to (' to '). As vries from to, we need new vrile t (in terms of ) which vries from ' to '. t ' ' As vries from to, t vries from ' to '. Thus, t - ' - = ' - ' - Fig - 7 As descried in the figure ove, the new vrile t is given, t ( ). Thus, dt f ( ) f ( t ) dt The modified integrl hs the limits ( to ). A prticulr cse of this propert is modifing the ritrr integrtion limits ( to ) to ( to ) i.e., nd. For this cse, f ( ) ( ) f ( ( ) t) dt
8 LOCUS 7 (3) f f ( ) is periodic function with period T, then the re under f ( ) for n periods would e n times the re under f ( ) for one period, i.e. nt f ( ) n f ( ) T Now, consider the periodic function f ( ) sin s n emple. The period of sin is. A + Suppose we intend to clculte intervl [, ] Fig - 8 sin s depicted ove. Notice tht the drkl shded re in the cn precisel cover the re mrked s A. Thus, sin This will hold true for ever periodic function, i.e. This lso implies tht T T sin f ( ) f ( ) (where T is the period of f ()) nt nt T f ( ) f ( ) n f ( ) nt nd f ( ) f ( ) nt nd nt T f ( ) f ( ) n f ( )
9 LOCUS 8 Emple 8 Show tht n V sin (n ) cos V, where n is positive integer nd V [, ) f ( ) sin is periodic with period. Therefore, s descried in propert-3, n V V sin sin n sin V sin n sin sin for [, ) V cos n( cos ) cos V n() (n ) cosv Emple 9 Evlute n cos, where n is positive integer. n cos n sin n sin n sin (Propert-3) n n the unit on ntegrtion Bsics, we sw tht for function f ( ), the nti-derivtive g( ) ws defined s so tht g '( ) f ( ) g( ) f ( t) dt (where is constnt)
10 LOCUS 9 We now consider n integrl of the following form: ( ) h( ) f ( t) dt ( ) Tht is, the limits of integrtion re themselves functions of. The nti-derivtive g( ) is specil cse of h( ) with ( ) nd ( ). Now, how do we evlute h '( )? Leinitz s rule for differentition tells us how to do so. Since g( ) is the nti-derivtive of f ( ), we hve: ( ) h( ) f ( t) dt ( ) g( t) ( ) ( ) g ( ) g ( ) h'( ) g ' ( ) '( ) g ' ( ) '( ) Let us see n emple of this rule: f ( ) '( ) f ( ) '( ). Emple Evlute f '( ) if f ( ) t dt Let us first find out f ( ) using stright forwrd integrtion: f ( ) 3 t 3 t f '( ) Now we redo this using the Leinitz s differentition rule: f '( ) ( )( ) ( )() 5 Let us now do n emple of this rule where stright forwrd integrtion would e much more difficult.
11 LOCUS 3 Emple Determine the eqution of the tngent to the curve f ( ) t =, where f ( ) 3 t 5 dt Notice how tedious it would e to ctull crr out the integrtion. nsted, we use the Leinitz s differentition rule: Also, t =, f '( ) f '() f ( ) f () dt 5 t Thus, the tngent psses through (, ) nd hs slope. The required eqution is ( ) (5). Consider function in two vriles nd, i.e., z f (, ) Let us consider the integrl of z with respect to, from to, i.e., f (, ) For this integrtion, the vrile is onl nd not. is essentill constnt for the integrtion process. Therefore, fter we hve evluted the definite integrl nd put in the integrtion limits, will still remin in the epression of. This mens tht is function of. ( ) f (, )...()
12 LOCUS 3 Our 5 th propert ss tht the reltion () cn e differentited with respect to s follows: where f (, ) d '( ) f (, ) d f (, ) stnds for the prtil derivtive of f (, ) with respect to, tht is, the derivtive of f (, ) w.r.t., treting s constnt Let us see the justifiction for this propert: '( ) ( h) ( ) lim h h lim h f (, h) f (, ) h lim h f (, h) f (, ) h f (, ) This propert turns out to e ver useful in certin cses. Emple Evlute k ln Oserve tht will e function of k. nsted of crring out direct integrtion, we use propert -5: d ( k) dk k ln k ln ln k. k k
13 LOCUS 3 Thus, d( k) ntegrting oth sides, we otin dk k ( k) ln( k ) C...() To otin C, note from the originl definition of tht () =. Using this in (), we otin Thus, ln C C ( k) ln( k ) Oserve gin crefull the indirect route tht propert-5 offered us to solve this integrl. Emple 3 Evlute ln We gin tr to use propert -5 to solve this integrl. Let us tret s function of. Therefore, Notice tht ( ) ln ( ) ln Now, using propert-5 we otin: Thus, d ( ) Using ( ) ove, we otin d ( ) d ln ln ln d ( ) ln( ) C ( ) ln( ) C C ln( ) ln Thus, ( ) ln( ) ln( ) ln This is the required integrl!
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