Lecture 7: Integration Techniques

Transcription

1 Lecture 7: Integrtion Techniques Antiderivtives nd Indefinite Integrls. In differentil clculus, we were interested in the derivtive of given rel-vlued function, whether it ws lgeric, eponentil or logrithmic. Here we re concerned with the inverse of the opertion of differentition. Tht is, the opertion of serching for functions where derivtives re given function.. Consider ny ritrry rel-vlued function f: X R defined on suset of the rel line, i.e. X R. By the ntiderivtive (primitive) of f; we men ny differentile function F: X R whose derivtive is the given function f. Hence, df d = F () = f().. Let c e ny rel numer c R, (which is constnt). Now since the derivtive of c with respect to or dc/d =, we hve: Theorem. If function F: X R is n ntiderivtive of f: X R, so is the function F + c: X R defined y (F + c) () = F() + c, for every X R. Remrk : Hence, the derived function f() is trcele ck to n infinite numer of possile primry, ntiderivtive or primitive functions of the form F() + c. Remrk : However, here there is sutle point. Every ntiderivtive fn of f is of the form F() + c if X R is connected suset of the rel line R.

2 Def : A set X R is connected if nd only if, for ny two rel numers, X with <, X contins the closed intervl [, ]. Emple: Consider the function f() =, where f: X R, X = [, ). It hs n ntiderivtive function defined y F() = +, since F () = df/d =. However, it lso hs n ntiderivtive function defined y G() = / +, since dg ( ) =. In fct, ny function of the form d F() = + c, for ny c R is n ntiderivtive function of f() =. Theorem. If F: X R is n ntiderivtive function of rel-vlued function f: X R, where X R is connected, then {F + c: c R} is the set of ll ntiderivtives of f. Def.. The set of ll ntiderivtive functions of rel vlued function f: X R, X R, is clled the indefinite integrl of f nd is denoted y f( ) d. Remrk : The symol is the integrl sign. If f hs no ntiderivtive (nonintegrle), then its indefinite integrl is. The f() prt is the integrnd. The f() d my e tken s the differentil df of primry or ntiderivtive function. Remrk 4: The process of determining the indefinite integrl of given rel-vlued function sy f: X R is termed the indefinite integrtion of f. We hve tht d d F() = f() implies f ( ) d = F ( ) + c

3 for ny rel c, where X connected. The constnt will e referred to s the ritrry constnt of integrtion. If X is not connected, then f( ) d = F + c mens only tht F + c is n ntiderivtive of f for some. Counteremple. Let denote the suset of R which consists of ll non-zero rel numers: X = { R: } () Consider the fn f() = /, f: X R. Then the fn. F() is n ntiderivtive of f. F() = - -, F: X R () Consider the function G(), G: X R, + if > G ( ) if < Note dg/d = G () is given y: G () = / = f(). G() is lso n ntiderivtive of f. However, G - F = ; if > Hence G - F = + ; if < G - F = + if > if <

4 Thus, G - F c, where c represents constnt function, for ll. Hence, G is not of form F() + c, for ll. Bsic Rules of Integrtion. Theorem. If F is n ntiderivtive of f nd G is n ntiderivtive of g, then F + G is n ntiderivtive of f + g. Hence, [f() + g()] d = f() d + g() d. (holds for finite sum). Theorem. If F is n ntiderivtive of f, then cf is n ntiderivtive of cf for every c R. Thus, cf() d = c f() d.. Theorem. For every N, N+ /N + is the ntiderivtive of N. Thus, N d = N + XN+ + c. 4. Emples: # Let f() = 4, then check: 4 d = c. 5 ( / 5) d d = 4. # Let f() = + 5 4, then [ ] d = d d = 4 /4 + 5( 5 /5) + c = 4 / c check. 4

5 # Let f() = - d d (4 /4 + 5 ) = [ - ] d = d + - d = / - d + c = / - ( /) + c = / - + c. 5. Theorem 4. The ntiderivtive of e is given y e d = e + c. Theorem 5. The ntiderivtive of -, is given y d = ln + c. Remrk : (Why ln insted of ln.) Note tht the function f() = - is such tht f: R - {} R. Tht is, f() = - is defined on the set of non-zero rel numers. Now, the nturl logrithmic function G() = ln, is such tht G: P R, where P {: R, > }. Tht is, ln is defined only on the positive rel numers. Thus, G() = ln, G: P R, cn not e n ntiderivtive function of f() = -, where f: R - {} R. This is true since the ntiderivtive of f must mp from the sme domin. In this cse, clerly, R- {} P. To e le to define n ntiderivtive function of -, we must etend the function ln to the set R - {} somehow. Thus, we define the function F() = ln( ), where, of course F: R-{} R. Tht is, ln ( ) is defined on ll non-zero rel numers. Emple: Let f() = e - -. f() d = e d - - d = e - ln + c. 5

6 The Methods of Integrtion y Sustitution nd y Prts.. Method of Sustitution: To determine the indefinite integrl of function f(), we choose y inspection suitle differentile fn u() such tht the given function f() cn e epressed s the product g(u()) u (), where g(u()) is function of u nd constnts. Then we hve f() d = g(u) u () d = g(u) du. Remrk : We hve tht u = u(); hence, du = u () d.. Emples # Let f() = ( + ) /, find f() d. Choose u = u() = +, then we hve tht u () =, such tht du = u () d = d Now f() = g(u) u () f() = ( + ) / = / = / u ( ) u u Here, g(u) = (/)(u -/ ). Thus f() d = ( + ) ut du = u () d, hence f() d = g(u) du = Now integrte the lst result. f() d = u -/ du / d = / u ( ) d u u / du. 6

7 = u / + c + = u/ + c = u / + c Sustitute for u = +, then ( + ) To check, find / d = ( +) / + c. d d ( +) / = ½ ( +) -/ = ( + ) / # Let f() = ( + ) 4 () Choose u = ( ) /. Find f d. u () = 4 -, such tht du = (4 - ) d. () Construct the product g(u) u () such tht f() = g(u) u () f() = ( + ) 4 ( ) u ( ) = u = u ( ) u / = / u Here g(u) = u 7

8 () Thus we hve tht f() d = g(u) u () d = u u () d = u du Tke lst integrl: f() d = u-/ du = u + c = 4 u + c To check tke d/d 4 = ( 4 ) + + c = (4 - + ) -/ (4 - ) = 4 ( + ) 4 / = ( + ) 4 /. Method of Integrtion y Prts. To determine the indefinite integrl of the fn f(), we choose y inspection two differentile functions u(), v() such tht f() d = u dv = uv - v du. Remrk. If we cn find u nd v such tht f() = u v (), then dv = v () d nd f() d = u v () d = u dv. To see tht u dv = uv - v du, let z vu, then dz = u dv + v du 8

9 Integrte oth sides dz = u dv + v du z = u dv + v du u dv = z - u du u du = uv - v du. Emples. # Let f() = e () Choose dv so tht it is the most complicted epression, ut is esy to integrte Let dv = e d Let u = such tht du = d () u dv = e - v du Now since dv = e d dv = v = e d v = e Also since u = du = d u dv = e - v du = e - e d 9

10 = e - e d = e - e + c e d = e - e + c. check d d e e e e = + e = e # Let f() = 6 e + () Let v = e +. Integrte y prts. then dv = e + d Hence, let u = du = d = () Then f() d = u dv Hence, = e + - e + du = To check: 6 e + d = e + + c d ( e d + ) = 6 e +. Definite Integrls. Let f() e continuous on n intervl X R, where f: X R. Let F() e n ntiderivtive of f, then f() d = F() + c.

11 . Now choose, X such tht <. Form the difference [F() + c] - [F() + c] = F() - F(). (note indep. of c) This difference F() - F() is clled the definite integrl of f from to. The point is termed the lower limit of integrtion nd the point, the upper limit of integrtion.. Nottion: We would write fd ( ) F ( )] F ( ) = F ( ) F ( ) 4. All tht we hve descried thus fr is known s the fundmentl theorem of clculus. We stte these results formlly s follows: Theorem : For ny ntiderivtive function F(), F: X R, X R, of continuous function f(), f: X R defined on closed intervl I = [, ], we hve fd ( ) = F() - F(). Emples # Let f() =, find 4 d () ( ) 4 4 = = = 4. 5, # f() = e, find fd ( ) 5 5 e = e = e 5 - e 9 5. The definite integrl represents the re under f() etween the points nd. 6. Properties of Definite Integrls. P. If f() is such tht f ( ) d over [, ], then

12 fd ( ) = fd ( ). P. If f() is defined nd continuous t the point, then fd ( ) =. P. If f() is defined nd continuous on ech of the closed intervls [, ],, [ N, N+ ], where N, the numer of suintervls, is finite nd i [ i, i+ ] = [, N+ ] U, then N+ N+ fd ( ) = fd ( ) + fd ( ) + K+ fd ( ) P. 4 If f() nd g() re such tht f ( ) d nd g ( ) Improper Integrls (i) kf( ) d = k f( ) d, for ny k R [ ] ( ) ( ) (ii) f( ) + g( ) d = f d + g d.. Consider fn. f() nd the definite integrl fd ( ) undefined for some [, ], the ove epression is termed improper. N d, then. If or or oth re infinite or if f() is Def. If f() is defined for [, ), then the epression f ( ). d is defined s lim f ( ) d. If f() is defined for (-, ], then f ( ) d is defined s lim f ( ) Def. If f() is defined for [, ] then the epression f ( ) + () lim fd. + d. d is defined s

13 Def. If f() is defined for [, ), R, ut not defined for =, then fd ( ) c defined s lim fd ( ) c. If f() is defined for (, ], R, ut not defined for =, then the epression fd ( ) is defined s lim fd () c c +. Def 4. If the limit of the improper integrls clled for in Def., or eists, the improper integrl is sid to e convergent, otherwise divergent.. Emples: 7 # Evlute d. Hence, y Definition, 7 d 6 = lim lim = = 6 6 lim 6 = lim 7 d is = 6 6 our integrl is convergent nd = ( ) 6 6 # Evlute d = lim+ [ ] d lim d= lim = lim + lim = Hence our integrl is divergent nd hs no vlue. + # Do # for d

14 + d = d + d ut we found ove tht lim d= lim = it diverges. #4 Let f() e continuous proility density function with support (domin) (-,). We know tht the following improper integrl is convergent. f ()d =. Both the epecttion nd the vrince of the rndom vrile re convergent improper integrls. E() = f ()d, Vr() = ( E()) f ()d. Likewise, the distriution function of is convergent improper integrl. This function is defined y ' F(') = f ()d = Pr o( '). Differentition of n Integrl. The following rule pplies q(y) q f (,y)d = f y (, y)d + f (q, y)q'(y) f (p,y)p'(y) y. p(y) p 4

15 . Emple: In Economics, we study consumer's demnd function in inverse form p = p(q), where Q is quntitiy demnded nd p denotes the mimum uniform price tht the consumer is willing to py for given quntity level Q. We ssume tht p' is negtive. The definite integrl Q p (z)dz = TV(Q) is clled totl vlue t Q. It gives us the mimum revenue tht could e etrcted from the consumer for Q units of the product. The dollr mount TV(Q) p(q)q = CS(Q) is clled consumer's surplus. Let C(Q) e the cost of supplying Q units. If firm could etrct mimum revenue from the consumer, its profit function would e Q p (z)dz C(Q). The output level mimizing the firm's profit sets the derivtive of the previous definite integrl equl to zero. This implies p(q) = C'(Q). Some Notes on Multiple Integrls. In this section, we will consider the integrtion of functions of more thn one independent vrile. The technique is nlogous to tht of prtil differentition. When performing integrtion with respect to one vrile, other vriles re treted s constnts. Consider the following emple: d c f (, y)ddy. 5

16 We red the integrl opertors from the inside out. The ounds, refer to those on, while the ounds c,d refer to y. Likewise, d ppers first nd dy ppers second. The integrl is computed in two steps: #. Compute f (, y)d = g(y). #. Compute (y)dy = d c d g c f (, y)ddy. If there were n vriles, you would follow the sme recursive steps n times. Ech successive integrtion elimintes single independent vrile.. Some Emples. Emple : Suppose tht z = f(,y). We wish to compute integrls of the form d c f (,y)ddy. Consider the emple f = y, where c = = nd d =, =. We hve yddy. Begin y integrting with respect to, treting y s constnt y d = y = y. Net, we integrte the ltter epression with respect to y. ydy = y Emple : Compute (y + = 4 6 z)ddydz, =. where the limits of integrtion re nd, in ech cse. Begin with. 6

17 y y ( + z) = + z. Net, y y ( 4 + zy) = + z. 4 Finlly, z z ( + z 4 ) = 7. 4 Emple. Let f(,y) e joint proility density function defined on the continuous rndom vriles (,y) with support R. The unconditionl epecttion of is defined s the doule integrl E() = f (, y) dyd. This epecttion cn e rewritten in terms of the mrginl proility density function for Thus, f () = E() = f (,y) dy f () d.. The conditionl epecttion of for given vlue of y is quite different. The conditionl density of, given y, is g ( y) = f(,y)/f (y), nd the conditionl epecttion is E( y) = g ( y)d. 7