1 The Riemann Integral


 Mervin Davidson
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1 The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re in generl? bse height), polygon Exmple Find the re A under the grph of the function f (x) = x between x = nd x = (see Figure ). f(x) = x y x Figure : Approximting the re A under the grph of f(x) = x on [, ] by L 4 (red) nd R 4 (blue). To pproximte this re, you could could proceed like this. Divide the intervl [, ] into four pieces (smller subintervls), for exmple considering the (eqully spced) division points <.5 <.5 <.75 <. Next, note tht the sum L 4 of the res of the rectngles below the grph of the function is smller thn the re A you re looking for, nd the sum R 4 of the res of the rectngles bove the grph of the function is lrger thn the re A, tht is L 4 < A < R 4. nd A simple computtion shows nd you obtin the estimte L 4 = (.5) +.5 (.5) +.5 (.75) =.4375 R 4 = (.5) +.5 (.75) +.5 () =.9375,.875 < A < for the re under the grph of f (x) = x on [, ]. Clerly, the bove pproximtion of the re A is not gret, but if you consider more division points (very close to ech other), the method bove will produce quite good results. This is the min idee in defining the Riemnn integrl (re under the grph of function).
2 . Definition of the Riemnn integrl Terminology: division/prtition of the intervl [, b] is : = x < x <... < x n < x n = b, nd we denote by x i = x i x i the length of the i th subintervl [x i, x i ]. The norm/length of the prtition is the length of the longest subintervl of the prtitition = mx i n x i. A set of intermedite points for is sequence x,..., x n, where ech x i is chosen in the subintervl [x i, x i ]. Given function f : [, b] R we define:  the left Riemnn sum of f corresponding to the division by L n = L n (f, ) =  the right Riemnn sum of f corresponding to the division by R n = R n (f, ) = f (x i ) x i () f (x i ) x i ()  the Riemnn sum of f corresponding to the division nd to the intermedite points x i [x i, x i ] by R (f,, x i ) = f (x i ) x i. (3) Exmple Evlute the left/right Riemnn sums for the function f (x) = x 3 6x defined on the intervl [, 3] considering n = 6 equl subintervls. We hve x = 3 6 =.5, so x =, x =.5, x =, x 3 =.5, x 4 =, x 5 =.5, nd x 6 = 3. The right Riemnn sum is R 6 = 6 f (x i ) x = (f (.5) + f () + f (.5) + f () + f (.5) + f (3)) , nd the left Riemnn sum is L 6 = 6 f (x i ) x = (f () + f (.5) + f () + f (.5) + f () + f (.5)) , With this premble, we cn now introduce the definition of Riemnn integrl of function, s follows. Definition We sy tht the function f : [, b] R is integrble on the intervl [, b] if the following limit exists lim R (f,, x i ) = L R. The number L is clled the Riemnn integrl of f on [, b] nd is denoted by f (x) dx = L. (4)
3 Remrk The bove definition sys tht the function f is integrble if the Riemnn sums get rbitrrily close to L. Explicitly, this mens tht for ny ε > there exists δ (ε) > such tht for ny prtition with < δ (ε) nd ny intermedite points x i, we hve R (f,, x i ) L < ε. (5) Remrk 3 If the function f is integrble, then f (x) dx gives the signed/net re under the grph of the function f (x) dx = A + A, where A + is the re bove the xxis, A is the re below the xxis. Exmple 3 Evlute the given integrls by interpreting them s res. ) 7dx b) 3 x dx c) x dx Theorem 4 If f : [, b] R is continous, then f is integrble on [, b] (i.e. f (x) dx exists). Proof. A totl mystery! Red the notes if you re curios why is this so. Remrk 5 If the function f is integrble, then you cn use ny Riemnn sum (for exmple L n or R n ) nd ny prtition (esiest is to work with is subintervls of equl length) in order to find the vlue of the integrl, for exmple (using right Riemnn sums) f (x) dx = lim f (x i ) x, where x = b n nd x i = + i x. n Exmple 4 Show tht the function f (x) = x is integrble on [, ] nd compute x dx. Since f is continuous on [, ], b y the bove theorem f is lso integrble on [, ]. Considering the equidistnt prtition with x = n = n we hve x i = + i n = i n, nd we obtin the right Riemnn sum R n = f (x i ) x = n ( ) i = ( n n n ) = n (n + ) (n + ) (n + ) (n + ) n 3 = 6 6n. Since f is integrble, we cn use ny Riemnn sum to compute its integrl. We hve ( x dx = lim R n = lim + ) ( + ) = n n 6 n n 6 = 3..3 Properties of the Riemnn integrl Theorem 6 (Properties of the Riemnn integrl) Let f, g : [, b] R. ) If c [, b] nd f is integrble on [, c] nd on [c, b], then f is integrble on [, b] nd In prticulr, f (x) dx = f (x) dx = nd c f (x) dx + b c f (x) dx = b) If f is integrble on [, b] nd k R is constnt, then kf is integrble on [, b] nd kf (x) dx = k f (x) dx 3
4 c) If f nd g re integrble on [, b] then f ± g is integrble on [, b] nd f (x) ± g (x) dx = f (x) dx ± d) If f nd g re integrble on [, b] nd f (x) g (x) for every x [, b], then f (x) dx In prticulr, f f f, so f f f, or f (x) dx g (x) dx. f (x) dx. g (x) dx. Proof. The bove properties come from the sme properties of limits nd sums (recll tht the integrl is the limit of (Riemnn) sums). Exercise Use the properties of integrls to compute ( ) 4 + 3x dx. Exercise If f (x) dx = 7 nd 8 f (x) dx =, find 8 Exercise 3 Show tht x for x [, 4], nd use it in order to estimte 4 xdx..4 Evluting definite integrl Theorem 7 (LeibnizNewton formul) If f is continuous on [, b], then where F is n ntiderivtive of f on [, b] (i.e. F = f). f (x) dx = F (x) b = F (b) F (), Proof. F (b) F () = n F (x i) F (x i ) Lgrnge = n F (x i ) (x i x i ) = n f (x i ) x i = R (f,, x i ). Since f is continuous, it is integrble, nd tking limits we hve F (b) F () = lim R (f,, x i ) = Exmple 5 An ntiderivtive of f (x) = x + 3 cos x is F (x) = x + 3 sin x (check!), so the integrl of f over the intervl [π, π] is π π x + 3 cos xdx = x + 3 sin x ) ((π) π π = + 3 sin (π) ( π + 3 sin π ) = 3π. Exercise 4 Evlute the given integrls: ) x dx b) x3 dx c) π/ cos xdx d) π/ sin xdx For given function f, there re mny ntiderivtives F (e.g. x, x +, x 3,... re ll ntiderivtives of x), but on ny intervl they ll differ by just constnt (why?). The ntiverivtives of f re denoted by f (x) dx, clled indefinite integrl of f. For exmple, xdx = x + C, where C stnds for n rbitrry constnt. Exercise 5 Find the indicted indefinite integrls (ntiderivtives): ) x dx b) cos x dx c) x dx d) x dx e) ( x 4 sec x ) dx f) Exercise 6 Evlute the given definite integrls: ) 3 ( x 3 6x ) dx b) (x sin x) dx c) 9 t +t t t dt 4
5 .5 The fundmentl theorem of Clculus It is clled so becuse estblishes connection between two importnt concepts in Anlysis (Clculus): derivtive nd integrl. The connection is tht they re inverse opertions to ech other, s shown by the following exmples. Exmple 6 Consider g (x) = x f (t) dt. Verify tht g (x) = f (x). Theorem 8 (Fundmentl Theorem of Clculus) If f : [, b] R is continuous, then ) d x dx f (t) dt = f (x) for x [, b] b) f (t) dt = F (b) F (), where F = f (or F (t) dt = F (b) F ()). Proof. ) If F is n ntiderivtive of f, then d dx x f (t) dt = d dx (F (x) F ()) = F (x) = f (x). Exercise 7 Repet the bove proof to help you show the following formul d dx (x) (x) f (t) dt = f (b (x)) b (x) f ( (x)) (x). Exercise 8 Find the derivtives of the given functions: ) g (x) = x + t dt b) g (x) = x x + t dt c) g (x) = x sec tdt d) g (x) = x sec tdt x Definition 9 The verge vlue of the function f over the intervl [, b] is b Exercise 9 Find the verge vlue of the function f (x) = + x on the intervl [, ]. Theorem (Men vlue theorem for Riemnn integrl) If f : [, b] R is continuous on [, b], then there exists c [, b] such tht b f (x) dx = f (c). (6) b Proof. Since f is continuous on [, b], by Weierstrss boundedness theorem, there exists x m, x M [, b] such tht so f (x m ) f (x) f (x M ), x [, b], f (x m ) b f (x) dx f (x M ), nd by the intermedite vlue theorem it follows tht there exists c between x m nd x M such tht b f (x) dx = f (c). Exercise Apply the MVT to the function f (x) = + x on the intervl [, ]. Wht is/re the possible vlues of c? 5
6 .6 Techniques of integrtion There re minly three methods for computing n integrl f (x) dx: using formul, using substitution (in fct this method just simplifies the integrl, so tht it becomes esier ), or integrtion by prts. Using formul To be ble to compute integrls, you should strt by lerning the ntiderivtives of some common functions (see the tble of ntiderivtives t the end of the notes, for exmple). Exmple 7 Using the fct tht the ntiverivtive of sin x is cos x nd the LeibnizNewton formul, we hve Using substitution π/ sin x = cos x π/ = cos π ( cos ) = + =. If you hve to compute n integrl, mybe substitution will help you to simplify (nd hence compute) the integrl. There re minly two possibilities. The first is to try to substitute x = g (u) for convenient function g (u) (u is the new vrible). Since x = g (u), dx du = g (u), which cn be written in differentil form dx = g (u) du. The substitution formul is in this cse f (x) dx = f (g (u)) g (u) du. If the resulting integrl in u is esier (cn be computed), it mens you hve the right substitution. If not... try gin (or use integrtion by prts). Exmple 8 Compute x x + dx. Consider the substitution x = u, so dx = du. The integrl becomes x x + dx = (u ) u + du = u u udu = u 3/ u / du = u5/ 5/ u3/ 3/ + C = 5 (x + )5/ 3 (x + )3/ + C. A second possibility is to consider substitution of the form g (x) = u. Since u = g (x), du dx = g (x) which cn be written in differentil form du = g (x) dx. The substitution formul is in this cse f (g (x)) g (x) dx = f (u) du. If the resulting integrl in u is esier (cn be computed), it mens you hve the right substitution. If not... try gin (or use integrtion by prts). Exmple 9 Compute x e x3 dx. Note tht if you consider x 3 = u, then 3x dx = du, so x dx cn be replced by 3du. The given integrl cn be computed s follows x e x3 dx = e u du = 3 3 eu + C = 3 ex3 + C. Exmple Compute x ( x + 3 ) 4 dx. Note tht if you consider x + 3 = u then xdx = du, or xdx = du. The given integrl cn be computed s follows x ( x + 3 ) 4 dx = u 4 du = ( u 5 x 5 + C = + 3 ) 5 + C. 6
7 Using integrtion by prts The integrtion by prts formul is the following u v = uv uv. Choosing u in the following order works in most cses:. Exponentil (e x,, 3 x, etc). Sine or cosine (sin (x), cos x, sin x, etc) 3. Power (x, x 3/, x 5, etc) 4. (the constnt function ) Exmple Compute xe x dx. Since there is n exponentil inside the integrl, this should be chosen s u. So u = e x (hence u = e x dx = ex ) nd v = x (so v = x = ). Using the integrtion by prts formul we obtin ( ) xe x dx = ex xdx = ex x ex dx = xex = xex = xex e x dx ex + C ex 4 + C. Exmple Compute ln xdx. The integrl does not contin exponentils, sine or cosine, powers. So choose u = (so u = dx = x) nd v = ln x (so v = (ln x) = x ). Using the integrtion by prts formul we obtin ln xdx = (x) ln xdx = x ln x x dx = x ln x dx = x ln x x + C. x 7
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