VII. The Integral. 50. Area under a Graph. y = f(x)

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1 VII. The Integrl In this chpter we efine the integrl of function on some intervl [, b]. The most common interprettion of the integrl is in terms of the re uner the grph of the given function, so tht is where we begin. 5. Are uner Grph y = f() b Let f be function which is efine on some intervl b n ssume it is positive, i.e. ssume tht its grph lies bove the is. How lrge is the re of the region cught between the is, the grph of y = f() n the verticl lines y = n y = b? One cn try to compute this re by pproimting the region with mny thin rectngles. To o this you choose prtition of the intervl [, b], i.e. you pick numbers < < n with = < < < < n < n = b. These numbers split the intervl [, b] into n sub-intervls whose lengths re [, ], [, ],..., [ n, n] =, =,..., n = n n. In ech intervl we choose point c k, i.e. in the first intervl we choose c, in the secon intervl we choose c,..., n in the lst intervl we choose some number n c n n. See figure 9. We then efine n rectngles: the bse of the k th rectngle is the intervl [ k, k ] on the -is, while its height is f(c k ) (here k cn be ny integer from to n.) The re of the k th rectngle is of course the prouct of its height n with, i.e. its re is f(c k ) k. Aing these we see tht the totl re of the rectngles is (44) R = f(c ) + f(c ) + + f(c n) n. This kin of sum is clle Riemnn sum. If the prtition is sufficiently fine then one woul epect this sum, i.e. the totl re of ll rectngles to be goo pproimtion of the re of the region uner the grph. Replcing the prtition by finer prtition, with more ivision points, shoul improve the pproimtion. So you woul epect the re to be the limit of Riemnn-sums like R s the prtition becomes finer n finer. A precise formultion of the efinition goes like this: 8

2 9 = c c c 3 3 c 4 4 c 5 5 c 6 6 c 7 b = 7 Figure 9. Forming Riemnn sum b Figure 3. Refining the prtition. Definition. 5.. If f is function efine on n intervl [, b], then we sy tht b f() = I, i.e. the integrl of f() from = to b equls I, if for every ε > one cn fin δ > such tht f(c) + f(c ) + + f(c n) n I < ε hols for every prtition ll of whose intervls hve length k < δ.

3 5. When f chnges its sign If the function f is not necessrily positive everywhere in the intervl b, then we still efine the integrl in ectly the sme wy: s limit of Riemnn sums whose mesh size becomes smller n smller. However the interprettion of the integrl s the re of the region between the grph n the -is hs twist to it. b Figure 3. Illustrting Riemnn sum for function whose sign chnges Let f be some function on n intervl b, n form the Riemnn sum R = f(c ) + f(c ) + + f(c n) n tht goes with some prtition, n some choice of c k. When f cn be positive or negtive, then the terms in the Riemnn sum cn lso be positive or negtive. If f(c k ) > then the quntity f(c k ) k is the re of the corresponing rectngle, but if f(c k ) < then f(c k ) k is negtive number, nmely minus the re of the corresponing rectngle. The Riemnn sum is therefore the re of the rectngles bove the -is minus the re below the is n bove the grph. Tking the limit over finer n finer prtitions, we conclue tht b f() = re bove the -is, below the grph minus the re below the -is, bove the grph. 5. The Funmentl Theorem of Clculus Definition. 5.. A function F is clle n ntierivtive of f on the intervl [, b] if one hs F () = f() for ll with < < b. For instnce, F () = is n ntierivtive of f() =, but so is G() = +8.

4 Theorem 5.. If f is function whose integrl R b f() eists, n if F is n ntierivtive of f on the intervl [, b], then one hs (45) b f() = F (b) F (). ( proof ws given in lecture.) Becuse of this theorem the epression on the right ppers so often tht vrious bbrevitions hve been invente. We will bbrevite F (b) F () ef = ˆF () b = = ˆF () b. 5.. Terminology In the integrl b f() the numbers n b re clle the bouns of the integrl, the function f() which is being integrte is clle the integrn, n the vrible is integrtion vrible. The integrtion vrible is ummy vrible. If you systemticlly replce it with nother vrible, the resulting integrl will still be the sme. For instnce, n if you replce by ϕ you still get = ˆ 3 3 = = 3, ϕ ϕ = ˆ 3 ϕ3 ϕ= = 3. Another wy to pprecite tht the integrtion vrible is ummy vrible is to look t the Funmentl Theorem gin: b f() = F (b) F (). The right hn sie tells you tht the vlue of the integrl epens on n b, n oes not nything to o with the vrible. Eercises 5. Fin n ntierivtive F () for ech of the following functions f(). Fining ntierivtives involves fir mount of guess work, but with eperience it gets esier to

5 guess ntierivtives. () f() = + (b) f() = 3 (c) f() = + () f() = 4 (e) f() = (f) f() = (g) f() = e (h) f() = (i) f() = e (j) f() = + (k) f() = e e (l) f() = + (m) f() = e + e (n) f() = (o) f() = sin (p) f() = (q) f() = cos (r) f() = cos (s) f() = sin( π/3) (t) f() = sin + sin (u) f() = sin() + (v) f() = ( + ) 5 In ech of the following eercises you shoul compute the re of the inicte region, n lso of the smllest enclosing rectngle with horizontl n verticl sies. Before computing nything rw the region. 5. The region between the verticl lines = n =, n between the -is n the grph of y = The region between the verticl lines = n =, n between the -is n the grph of y = n (here n >, rw for n =,,, 3, 4). 5.4 The region bove the grph of y =, below the line y =, n between the verticl lines =, = The region bove the -is n below the grph of f() = The region bove the -is n below the grph of f() = The region bove the -is n below the grph of f() = The region bove the -is, below the grph of f() = sin, n between = n = π. 5.9 The region bove the -is, below the grph of f() = /( + ) ( curve known s Mri Agnesi s witch), n between = n =. 5. The region between the grph of y = / n the -is, n between = n = b (here < < b re constnts, e.g. choose = n b = if you hve something ginst either letter or b.) 5. The region bove the -is n below the grph of f() = + +.

6 3 5. Compute p without fin n ntierivtive for (you cn fin such n ntierivtive, but it s not esy. This integrl is the re of some region: which region is it, n wht is tht re?) 5.3 Compute () / p (b) (c) without fining ntierivtives. 53. The inefinite integrl The funmentl theorem tells us tht in orer to compute the integrl of some function f over n intervl [, b] you shoul first fin n ntierivtive F of f. In prctice, much of the effort require to fin n integrl goes into fining the ntierivtive. In orer to simplify the computtion of the integrl (46) b f() = F (b) F () the following nottion is commonly use for the ntierivtive: (47) F () = f(). For instnce, = 3 3, sin 5 = cos 5, etc... 5 The integrl which ppers here oes not hve the integrtion bouns n b. It is clle n inefinite integrl, s oppose to the integrl in (46) which is clle efinite integrl. You use the inefinite integrl if you epect the computtion of the ntierivtive to be lengthy ffir, n you o not wnt to write the integrtion bouns n b ll the time. It is importnt to istinguish between the two kins of integrls. ifferences: Here is list of Inefinite integrl Definite integrl R f() is function of. By efinition R f() is ny function of whose erivtive is f(). is not ummy vrible, for emple, R = +C n R tt = t +C re functions of iffferent vribles, so they re not equl. R b R b f() is number. f() ws efine in terms of Riemnn sums n cn be interprete s re uner the grph of y = f(), t lest when f() >. R is ummy vrible, for emple, =, n R tt =, so R = R tt.

7 You cn lwys check the nswer Suppose you wnt to fin n ntierivtive of given function f() n fter long n messy computtion which you on t relly trust you get n nswer, F (). You cn then throw wy the ubious computtion n ifferentite the F () you h foun. If F () turns out to be equl to f(), then your F () is inee n ntierivtive n your computtion isn t importnt nymore. For emple, suppose tht we wnt to fin R ln. My cousin Louie sys it might be F () = ln. Let s see if he s right: ` ln = + ln = ln. Who knows how Louie thought of this 7, but it oesn t mtter: he s right! We now know tht R ln = ln + C About +C Let f() be function efine on some intervl b. If F () is n ntierivtive of f() on this intervl, then for ny constnt C the function F () = F ()+C will lso be n ntierivtive of f(). So one given function f() hs mny ifferent ntierivtives, obtine by ing ifferent constnts to one given ntierivtive. Theorem 53.. If F () n F () re ntierivtives of the sme function f() on some intervl b, then there is constnt C such tht F () = F () + C. Proof. Consier the ifference G() = F () F (). Then G () = F () F () = f() f() =, so tht G() must be constnt. Hence F () F () = C for some constnt. It follows tht there is some mbiguity in the nottion R f(). Two functions F () n F () cn both equl R f() without equling ech other. When this hppens, they (F n F ) iffer by constnt. This cn sometimes le to confusing situtions, e.g. you cn check tht sin cos = sin sin cos = cos re both correct. (Just ifferentite the two functions sin n cos!) These two nswers look ifferent until you relize tht becuse of the trig ientity sin + cos = they relly only iffer by constnt: sin = cos +. To voi this kin of confusion we will from now on never forget to inclue the rbitrry constnt +C in our nswer when we compute n ntierivtive. 7 He took mth n lerne to integrte by prts.

8 Stnr Integrls Here is list of the stnr integrls everyone shoul know. f() = F () + C n = n+ n + + C for ll n = ln + C (Note the bsolute vlues) e = e + C = ln + C (on t memorize: use = e ln ) sin = cos + C cos = sin + C tn = ln cos + C (Note the bsolute vlues) = rctn + C + = rcsin + C The following integrl is lso useful, but not s importnt s the ones bove: cos = + sin ln sin + C for π < < π. All of these integrls shoul be fmilir from the ifferentition rules we hve lerne so fr, ecept for for the integrls of tn n of. You cn check those by ifferentition cos (using ln = ln ln b simplifies things bit). b 54. Properties of the Integrl Just s we h list of properties for the limits n erivtives of sums n proucts of functions, the integrl hs similr properties. Suppose we hve two functions f() n g() with ntierivtives F () n G(), respectively. Then we know tht F () + G() = F () + G () = f() + g(), in wors, F + G is n ntierivtive of f + g, which we cn write s (48) f() + g() = f() + g(). Similrly, (49) `cf () = cf () = cf() implies tht cf() = c f() if c is constnt.

9 6 These properties imply nlogous properties for the efinite integrl. For ny pir of functions on n intervl [, b] one hs (5) b b ˆf() + g() = f() + n for ny function f n constnt c one hs (5) b cf() = c b f(). b g(), Definite integrls hve one other property for which there is no nlog in inefinite integrls: if you split the intervl of integrtion into two prts, then the integrl over the whole is the sum of the integrls over the prts. The following theorem sys it more precisely. Theorem 54.. Given < c < b, n function on the intervl [, b] then (5) b f() = c f() + b c f(). Proof. Let F be n ntierivtive of f. Then so tht c f() = F (c) F () n b c b f() = F (b) F () f() = F (b) F (), = F (b) F (c) + F (c) F () = c f() + b c f(). So fr we hve lwys ssume thet < b in ll inefinitie integrls R b.... funmentl theorem suggests tht when b <, we shoul efine the integrl s (53) For instnce, b f() = F (b) F () = `F () F (b) = b = =. f(). The 55. The efinite integrl s function of its integrtion bouns Consier the epression I = t t. Wht oes I epen on? To see this, you clculte the integrl n you fin I = ˆ 3 t3 = = 3 3. So the integrl epens on. It oes not epen on t, since t is ummy vrible (see 5. where we lrey iscusse this point.) In this wy you cn use integrls to efine new functions. For instnce, we coul efine I() = t t,

10 7 which woul be rounbout wy of efining the function I() = 3 /3. Agin, since t is ummy vrible we cn replce it by ny other vrible we like. Thus I() = efines the sme function (nmely, I() = 3 3 ). α α The previous emple oes not efine new function (I() = 3 /3). An emple of new function efine by n integrl is the error-function from sttistics. It is given by (54) erf() ef = π e t t, so erf() is the re of the she region in figure 3. The integrl in (54) cnnot be com- y = π e Are = erf() Figure 3. Definition of the Error function. pute in terms of the stnr functions (squre n higher roots, sine, cosine, eponentil n logrithms). Since the integrl in (54) occurs very often in sttistics (in reltion with the so-clle norml istribution) it hs been given nme, nmely, erf(). How o you ifferentite function tht is efine by n integrl? The nswer is simple, for if f() = F () then the funmentl theorem sys tht n therefore i.e. A similr clcultion gives you f(t) t = F () F (), f(t) t = F () F () = F () = f(), b f(t) t = f(). f(t) t = f(). So wht is the erivtive of the error function? We hve erf () = n o π e t t = e t t π = e. π

11 8 56. Metho of substitution 56.. Emple The chin rule sys tht so tht F (G()) = F (G()) G (), F (G()) G () = F (G()) + C. Consier the function f() = sin( + 3). It oes not pper in the list of stnr ntierivtives we know by hert. But we o notice 8 tht = ( + 3). So let s cll G() = + 3, n F (u) = cos u, then n so tht (55) 56.. Leibniz nottion for substitution F (G()) F (G()) = cos( + 3) = sin( + 3) {z } F (G()) {z} = f(), G () sin( + 3) = cos( + 3) + C. The most trnsprent wy of computing n integrl by substitution is by following Leibniz n introuce new vribles. Thus to o the integrl f(g())g () where f(u) = F (u), we introuce the substitution u = G(), n gree to write Then we get u = G() = G (). f(g())g () = f(u) u = F (u) + C. At the en of the integrtion we must remember tht u relly stns for G(), so tht f(g())g () = F (u) + C = F (G()) + C. As n emple, let s o the integrl (55) using Leibniz nottion. We wnt to fin sin( + 3) n ecie to substitute z = + 3 (the substitution vrible oesn t lwys hve to be clle u). Then we compute so tht z = ` + 3 = n sin( + 3) = sin z, sin( + 3) = sin z z = cos z + C. Finlly we get ri of the substitution vrible z, n we fin sin( + 3) = cos` C. 8 You will strt noticing things like this fter oing severl emples.

12 9 When we o integrls in this clculus clss, we lwys get ri of the substitution vrible becuse it is vrible we invente, n which oes not pper in the originl problem. But if you re oing n integrl which ppers in some longer iscussion of rel-life (or rel-lb) sitution, then it my be tht the substitution vrible ctully hs mening (e.g. the effective stoichiometric molity of CQF self-inhibition ) in which cse you my wnt to skip the lst step n leve the integrl in terms of the (meningful) substitution vrible Substitution for efinite integrls For efinite integrls the chin rule implies `F (G()) = F (G())G () = f(g())g () b which you cn lso write s f(g())g () = F (G(b)) F (G()). (56) b f(g())g () = G(b) = u=g() f(u) u Emple of substitution in efinite integrl Let s compute +, using the substitution u = G() = +. Since u =, the ssocite inefinite integrl is + {z {z} = u u. } u u To fin the efinite integrl you must compute the new integrtion bouns G() n G() (see eqution (56).) If runs between = n =, then u = G() = + runs between u = + = n u = + =, so the efinite integrl we must compute is (57) + = which is in our list of memorble integrls. So we fin + = Sometimes the integrls in (57) re written s = + = u u, u u = ˆln u = ln. u= u u, to emphsize (n remin yourself) to which vrible the bouns in the integrl refer.

13 Eercises 56. Compute these erivtives () (c) (e) s θ s sin θ ` + t 4 t (b) () + (f) t t ln z z s s e 56. Compute the secon erivtive of the error function. How mny inflection points oes the grph of the error function hve? Compute (some of) the following integrls / 5 + 4e (/ + / ) e ` (3 5) (hm... ) t t (!) t (!!!) ( 3 ) ( ) (5y 4 6y + 4) y (y 9 y 5 + 3y) y 3/ t «t t 4 t 6 t t t 4 + ( 3 ) u( u + 3 u) u ( + /) p 5 + ( )(3 + ) ( t / t) t 5 e r + 3 r «r ( + ) «

14 π/3 π/4 π/ π/ π π/3 π/ π/ ln 6 ln e 3 e 3 sin t t (cos θ + sin θ) θ (cos θ + sin θ) θ tn cos cot sin 6 + (/) 8e t t π π 56.5 Compute ( ) ( ) f() where f() = ( 4 if <, 5, if. f() where (, if π, f() = sin, if < π. I = ` + 3 in two ifferent wys: () Epn (+ ) 3, multiply with, n integrte ech term. (b) Use the substitution u = Compute I n = ` + n If f () = / n f() = / fin f() Sketch the grph of the curve y = + n etermine the re of the region enclose by the curve, the -is n the lines =, = Fin the re uner the curve y = n bove the -is between = n =. Drw sketch of the curve Grph the curve y =, [, ], n fin the re enclose between the curve n the -is. (Don t evlute the integrl, but compre with the re uner the grph of y =.) Determine the re uner the curve y = n between the lines = n = Grph the curve y = 9 n etermine the re enclose between the curve n the -is Grph the re between the curve y = 4 n the line = 3. Fin the re of this region Fin the re boune by the curve y = 4 n the lines y = n y = Fin the re enclose between the curve y = sin, π/4 n the es Fin the re enclose between the curve y = cos, π/4 n the es.

15 56.6 Grph y + =, n fin the re enclose by the curve n the line = Fin the re of the region boune by the prbol y = 4 n the line y = Fin the re boune by the curve y = ( ) n the line = y Fin the re boune by the curve = 4y n the line = 4y Clculte the re of the region boune by the prbols y = n = y Fin the re of the region inclue between the prbol y = n the line + y = Fin the re of the region boune by the curves y = n y = Here re the grphs of function f(), its erivtive f () n n ntierivtive F () of f(). Unfortuntely the grphs re not lbele. Ientify which grph is which. 3 Eplin your nswer Below is the grph of function y = f() y y=f() b c Which mong the following sttements true? (b) F () = F (c)? True/Flse Reson: (c) F (b) =? True/Flse Reson: () F (b) > F (c)? True/Flse Reson: (e) The grph of y = F () hs two inflection points? True/Flse Reson:

16 3 Use substitution to evlute the following integrls π u u + u + + s s 3 s + + cos(θ + π 3 )θ sin π + 5 sin + cos π/3 π/4 sin θ cos θ θ ξ= 3 r ln r, r sin + cos sin + sin z p z z ln ξ( + ξ ) ξ sin ρ`cos ρ) 4 ρ αe α α e t t t