6.3 Definite Integrals and Antiderivatives

Transcription

1 Section 6. Definite Integrls nd Antiderivtives 8 6. Definite Integrls nd Antiderivtives Wht ou will lern out... Properties of Definite Integrls Averge Vlue of Function Men Vlue Theorem for Definite Integrls Connecting Differentil nd Integrl Clculus nd wh... Working with the properties of definite integrls helps us to understnd etter the definite integrl. Connecting derivtives nd definite integrls sets the stge for the Fundmentl Theorem of Clculus. Properties of Definite Integrls In defining f() s limit of sums gc k k, we moved from left to right cross the intervl [, ]. Wht would hppen if we integrted in the opposite direction? The integrl would ecome f() d gin limit of sums of the form gf(c k ) k ut this time ech of the k s would e negtive s the -vlues decresed from to. This would chnge the signs of ll the terms in ech Riemnn sum, nd ultimtel the sign of the definite integrl. This suggests the rule f() d = - f() d. Since the originl definition did not ppl to integrting ckwrds over n intervl, we cn tret this rule s logicl etension of the definition. Although [, ] is technicll not n intervl, nother logicl etension of the definition is tht f() d =. These re the first two rules in Tle 6.. The others re inherited from rules tht hold for Riemnn sums. However, the limit step required to prove tht these rules hold in the limit (s the norms of the prtitions tend to zero) plces their mthemticl verifiction eond the scope of this course. The should mke good sense nonetheless. TABE 6. Rules for Definite Integrls. Order of Integrtion: f() d f() d A definition. Zero: f() d Also definition. Constnt Multiple: kf() d k f() d An numer k. Sum nd Difference: (f() g()) d f() d g() d. Additivit: f() d f() d f() d 6. M-Min Inequlit: If m f nd min f re the mimum nd minimum vlues of f on [, ], then # f() d f() d k = - min f ( ) f() d m f ( ).. Domintion: f g() on [, ] f() d g d f on [, ] f d g = c # c

2 Chpter 6 The Definite Integrl EXAMPE Using the Rules for Definite Integrls Suppose f() d =, - f() d = -, nd h() d =. - Find ech of the following integrls, if possile. () f() d () f() d (c) [f() + h()] d - - (d) f() d (e) h() d (f) [f() + h()] d - - SOUTION () () (c) f() d = - f() d = -(-) = f() d = f() d + f() d = + (-) = - - [f() + h()] d = f() d + h() d = () + () = (d) Not enough informtion given. (We cnnot ssume, for emple, tht integrting over hlf the intervl would give hlf the integrl!) (e) Not enough informtion given. (We hve no informtion out the function h outside the intervl [-, ].) (f) Not enough informtion given (sme reson s in prt (e)). Now Tr Eercise. EXAMPE Finding Bounds for n Integrl Show tht the vlue of + cos d is less thn >. SOUTION The M-Min Inequlit for definite integrls (Rule 6) ss tht min f # ( - ) is lower ound for the vlue of nd tht m f # f() d ( - ) is n upper ound. The mimum vlue of + cos on [, ] is, so Since + cos d is ounded ove (which is. Á ), it is less thn >. Now Tr Eercise. + cos d # ( - ) =. Averge Vlue of Function The verge of n numers is the sum of the numers divided n. How would we define the verge vlue of n ritrr function f over closed intervl [, ]? As there re infinitel mn vlues to consider, dding them nd then dividing infinit is not n option. Consider, then, wht hppens if we tke lrge smple of n numers from regulr

3 Section 6. Definite Integrls nd Antiderivtives suintervls of the intervl [, ]. One w would e to tke some numer of the n suintervls of length The verge of the n smpled vlues is f(c ) + f(c ) + Á + f(c n ) n = -. n = # n n f(c k ) k= = - n from ech Does this lst sum look fmilir? It is >( - ) times Riemnn sum for f on [, ]. Tht mens tht when we consider this verging process s n : q, we find it hs limit, nmel >( - ) times the integrl of f over [, ]. We re led this remrkle fct to the following definition. = k= n # - f(c k ). k= f(c k ) c k n = - DEFINITION Averge (Men) Vlue If f is integrle on [, ], its verge (men) vlue on [, ] is v(f) = f() d. - Figure 6. The rectngle with se [, ] nd with height equl to (the verge vlue of the function f() = ( - ) hs re equl to the net re etween f nd the -is from to. (Emple ) = f() c f(c) Figure 6. The vlue f(c) in the Men Vlue Theorem is, in sense, the verge (or men) height of f on [, ]. When f Ú, the re of the shded rectngle f(c)( - ) = f() d, is the re under the grph of f from to. EXAMPE Appling the Definition Find the verge vlue of f() = - on [, ]. Does f ctull tke on this vlue t some point in the given intervl? SOUTION v(f) = f() d - = = - # - = Using NINT The verge vlue of f() = - over the intervl [, ] is. The function ssumes this vlue when - = or =. Since = lies in the intervl [, ], the function does ssume its verge vlue in the given intervl (Figure 6.). Now Tr Eercise. Men Vlue Theorem for Definite Integrls It ws no mere coincidence tht the function in Emple took on its verge vlue t some point in the intervl. ook t the grph in Figure 6. nd imgine rectngles with se ( - ) nd heights rnging from the minimum of f ( rectngle too smll to give ( - ) d

4 Chpter 6 The Definite Integrl the integrl) to the mimum of f ( rectngle too lrge). Somewhere in etween there is just right rectngle, nd its topside will intersect the grph of f if f is continuous. The sttement tht continuous function on closed intervl lws ssumes its verge vlue t lest once in the intervl is known s the Men Vlue Theorem for Definite Integrls. THEOREM The Men Vlue Theorem for Definite Integrls If f is continuous on [, ], then t some point c in [, ], f(c) = f() d. - EXPORATION How ong is the Averge Chord of Circle? r r Figure 6.6 Chords perpendiculr to the dimeter [-r, r] in circle of rdius r centered t the origin. (Eplortion ) [-r, r] Suppose we hve circle of rdius r centered t the origin. We wnt to know the verge length of the chords perpendiculr to the dimeter on the -is.. Show tht the length of the chord t is r - (Figure 6.6).. Set up n integrl epression for the verge vlue of r - over the intervl [-r, r].. Evlute the integrl identifing its vlue s n re.. So, wht is the verge length of chord of circle of rdius r?. Eplin how we cn use the Men Vlue Theorem for Definite Integrls (Theorem ) to show tht the function ssumes the vlue in step. Connecting Differentil nd Integrl Clculus Before we move on to the net section, let us puse for moment of historicl perspective tht cn help ou to pprecite the power of the theorem tht ou re out to encounter. In Emple we used NINT to find the integrl, nd in Section 6., Emple, we were fortunte tht we could use our knowledge of the re of circle. The re of circle hs een round for long time, ut NINT hs not; so how did people evlute definite integrls when the could not ppl some known re formul? For emple, in Eplortion of the previous section we used the fct tht p sin d =. Would Newton nd einiz hve known this fct? How? The did know tht quotients of infinitel smll quntities, s the put it, could e used to get velocit functions from position functions, nd tht sums of infinitel thin rectngle res could e used to get position functions from velocit functions. In some w, then, there hd to e connection etween these two seemingl different processes. Newton nd einiz were le to picture tht connection, nd it led them to the Fundmentl Theorem of Clculus. Cn ou picture it? Tr Eplortion.

5 Section 6. Definite Integrls nd Antiderivtives EXPORATION Finding the Derivtive of n Integrl f () Figure 6. The grph of the function in Eplortion. Group Activit Suppose we re given the grph of continuous function f, s in Figure 6... Cop the grph of f onto our own pper. Choose n greter thn in the intervl [, ] nd mrk it on the -is.. Using onl verticl line segments, shde in the region etween the grph of f nd the -is from to. (Some shding might e elow the -is.). Your shded region represents definite integrl. Eplin wh this integrl cn e written s. (Wh don t we write it s f(t) dt f() d?). Compre our picture with others produced our group. Notice how our integrl ( rel numer) depends on which ou chose in the intervl [, ]. The integrl is therefore function of on [, ]. Cll it F.. Recll tht F ( ) is the limit of F> s gets smller nd smller. Represent F in our picture drwing one more verticl shding segment to the right of the lst one ou drew in step. F is the (signed) re of our verticl segment. 6. Represent in our picture moving to eneth our newl drwn segment. Tht smll chnge in is the thickness of our verticl segment.. Wht is now the height of our verticl segment? 8. Cn ou see wh Newton nd einiz concluded tht F () = f()? If ll went well in Eplortion, ou concluded tht the derivtive with respect to of the integrl of f from to is simpl f. Specificll, d f(t) dt = f(). d This mens tht the integrl is n ntiderivtive of f, fct we cn eploit in the following w. If F is n ntiderivtive of f, then for some constnt C. Setting in this eqution equl to gives Putting it ll together, f(t) dt = F() + C f(t) dt = F() + C = F() + C C = -F(). f(t) dt = F() - F().

6 Chpter 6 The Definite Integrl The implictions of the previous lst eqution were enormous for the discoverers of clculus. It ment tht the could evlute the definite integrl of f from to n numer simpl computing F() - F(), where F is n ntiderivtive of f. EXAMPE Finding n Integrl Using Antiderivtives Find p using the formul sin d f(t) dt = F() - F(). SOUTION Since sin is the rte of chnge of the quntit F() = -cos, tht is F () = sin p sin d = -cos (p) - (-cos()) = -(-) - (-) =. This eplins how we otined the vlue for Eplortion of the previous section. Now Tr Eercise. Quick Review 6. (For help, go to Sections.,., nd..) Eercise numers with gr ckground indicte prolems tht the uthors hve designed to e solved without clcultor. In Eercises, find d> d.. = -cos. = sin. = ln (sec ). = ln (sin ). = ln (sec + tn ) 6. = ln - n+. = (n Z -) 8. = n + +. = e. = tn - Section 6. Eercises The eercises in this section re designed to reinforce our understnding of the definite integrl from the lgeric nd geometric points of view. For this reson, ou should not use the numericl integrtion cpilit of our clcultor (NINT) ecept perhps to support n nswer.. Suppose tht f nd g re continuous functions nd tht f() d = -. f() d = 6, g() d = 8. Use the rules in Tle 6. to find ech integrl. () (c) g() d f() d (e) [f() - g()] d (f) [f() - g()] d. Suppose tht f nd h re continuous functions nd tht f() d = -, f() d =, h() d =. () (d) g() d f() d Use the rules in Tle 6. to find ech integrl. () -f() d () [f() + h()] d (c) [f() - h()] d (d) f() d (e) f() d (f) [h() - f()] d. Suppose tht f() d =. Find ech integrl. () f(u) du () f(z) dz (c) f(t) dt (d) [-f()] d. Suppose tht g(t) dt =. Find ech integrl. - - () g(t) dt () g(u) du - g(r) (c) [-g()]d (d) - dr -

7 Section 6. Definite Integrls nd Antiderivtives. Suppose tht f is continuous nd tht f(z) dz = nd f(z) dz =. Find ech integrl. () f(z) dz () f(t) dt 6. Suppose tht h is continuous nd tht h(r) dr = nd h(r) dr = Find ech integrl. () h(r) dr () - h(u) du. Show tht the vlue of sin ( ) d cnnot possil e. 8. Show tht the vlue of + 8 d lies etween.8 nd.. Integrls of Nonnegtive Functions Use the M-Min Inequlit to show tht if f is integrle then f() Ú on [, ] Q f() d Ú.. Integrls of Nonpositive Functions Show tht if f is integrle then f() on [, ] Q f() d. In Eercises, use NINT to find the verge vlue of the function on the intervl. At wht point(s) in the intervl does the function ssume its verge vlue?. = -, [, ]. = -, [, ]. = - -, [, ]. = ( - ), [, ] In Eercises 8, find the verge vlue of the function on the intervl without integrting, ppeling to the geometr of the region etween the grph nd the -is. +, - -,. f() = e on [-, ] - +, - 6, 6. f(t) = - - t, [-, ] s s t t. f(t) = sin t, [, p] p 8. f(u) = tn u, c -, p d In Eercises, interpret the integrnd s the rte of chnge of quntit nd evlute the integrl using the ntiderivtive of the quntit, s in Emple. p. sin d. p. e d.. d. d - 6. d 6. 8 d - > d - + d e d d In Eercises 6, find the verge vlue of the function on the intervl, using ntiderivtives to compute the integrl.. = sin, [, p].. = sec, c, p. = +, [, ] d. = +, [-, ] 6. = sec tn,. Group Activit Use the M-Min Inequlit to find upper nd lower ounds for the vlue of + d. 8. Group Activit (Continution of Eercise ) Use the M- Min Inequlit to find upper nd lower ounds for the vlues of. Add these to rrive t n improved estimte for. Writing to ern If v(f) rell is tpicl vlue of the integrle function f() on [, ], then the numer v(f) should hve the sme integrl over [, ] tht f does. Does it? Tht is, does Give resons for our nswer. + d nd p> p> + d. =, [e, e] v(f) d = f() d?. cos d sec d + d. c, p d

8 6 Chpter 6 The Definite Integrl. Writing to ern A driver verged mph on -mile trip nd then returned over the sme miles t the rte of mph. He figured tht his verge speed ws mph for the entire trip. () Wht ws his totl distnce trveled? () Wht ws his totl time spent for the trip? (c) Wht ws his verge speed for the trip? (d) Eplin the error in the driver s resoning.. Writing to ern A dm relesed m of wter t m / min nd then relesed nother m t m /min. Wht ws the verge rte t which the wter ws relesed? Give resons for our nswer.. Use the inequlit sin, which holds for Ú, to find n upper ound for the vlue of sin d.. The inequlit sec Ú + ( >) holds on (-p>, p>). Use it to find lower ound for the vlue of sec d.. Show tht the verge vlue of liner function () on [, ] is () + (). [Cution: This simple formul for verge vlue does not work for functions in generl!] Stndrdized Test Questions You m use grphing clcultor to solve the following prolems.. True or Flse The verge vlue of function f on [, ] lws lies etween f() nd f(). Justif our nswer. 6. True or Flse If f() d =, then f() = f(). Justif our nswer.. Multiple Choice If nd f() d = g() d =, then ll of the following must e true ecept (A) (B) (C) f()g() d = [f() + g()] d = 8 f() d = (D) [f() - g()] d = (E) [g() - f()] d = 8. Multiple Choice If f() d = nd f() d =, then ll of the following must e true ecept (A) (B) (C) f() d = 6 8 f() d - f() d = f() d = - -8 (D) - f() d = - (E) 6 8 f() d + f() d = 6 6. Multiple Choice Wht is the verge vlue of the cosine function on the intervl [, ]? (A) -. (B) -. (C) -.8 (D). (E).8. Multiple Choice If the verge vlue of the function f on the intervl [, ] is, then f() d = (A) f() + f() (B) - (C) - (D) - f() + f() (E) Eplortion 8. Compring Are Formuls Consider the region in the first qudrnt under the curve = (h>) from = to = (see figure). () Use geometr formul to clculte the re of the region. () Find ll ntiderivtives of. (c) Use n ntiderivtive of to evlute () d. h h 8

9 Section 6. Definite Integrls nd Antiderivtives Etending the Ides. Grphing Clcultor Chllenge If k, nd if the verge vlue of k on [, k] is k, wht is k? Check our result with CAS if ou hve one ville.. Show tht if F () = G () on [, ], then F() - F() = G() - G(). Quick Quiz for AP* Preprtion: Sections Multiple Choice If, then f() d = + (f() + ) d = (A) + + (B) - (C) - (D) - (E) -. Multiple Choice The epression A + A + A + Á + A is Riemnn sum pproimtion for (A) A d (C) A d (E) d (B) d (D) d. Multiple Choice Wht re ll vlues of k for which k d =? (A) - (B) (C) (D) - nd (E) -,, nd. Free Response et f e function such tht f () = 6 +. () Find f() if the grph of f is tngent to the line - = t the point (, -). () Find the verge vlue of f() on the closed intervl [-, ].