Calculus and linear algebra for biomedical engineering Week 11: The Riemann integral and its properties

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1 Clculus nd liner lgebr for biomedicl engineering Week 11: The Riemnn integrl nd its properties Hrtmut Führ Lehrstuhl A für Mthemtik, RWTH Achen Jnury 9, 2009

2 Overview 1 Motivtion: Computing the re under the grph 2 Approximtion by piecewise constnt functions 3 Denition of the Riemnn integrl 4 Properties of the Riemnn integrl

3 Motivtion: Computing ow from ow rtes We observe the ow of wter through drin, which vries with time. The result is ow rte, in litres/second, continuously recorded over time intervl [, b]. From these dt, we wnt to determine the totl mount A of wter tht hs pssed through the vlve during the intervl. This vlue corresponds to the re under the grph of f.

4 Answer for constnt rte If the ow rte is constnt, sy equl to c, the nswer is esily obtined: A = (length of the intervl) (ow rte) = (b ) c This corresponds to the formul for rectngulr res. re = width height. The ide to clculte the re under rbitrry grphs is to pproximte the grph by piecewise constnt functions.

5 Are under the grph: Piecewise constnt functions A piecewise constnt function or step function is function f : [, b] R tht consists of nitely mny constnt pieces Here, the region under the grph is mde up out of rectngles nd its re is computed by summing the res of the rectngles.

6 Prtition Denition. Let I = [, b] R be some intervl. A prtition of I is given by nite subset P = {x 0,..., x n } stisfying {, b} P. Without loss of generlity, = x 0 < x 1 < x 2 <... < x n = b. Exmple: The set P = {0, 0.3, 0.5, 0.8, 1.0} denes prtition of the intervl [0, 1].

7 Approximtion by step functions Denition. Let f : [, b] R be function, nd P = {x 0, x 1,..., x n } prtition. We dene M k (f ) = sup{f (x) : x k < x < x k+1} M k (f ) = inf{f (x) : x k < x < x k+1} Interprettion: M k nd M k provide optiml pproximtion of the grph of f by step functions with jumps in P, one from bove, one from below.

8 Exmple: Approximtion from bove A function dened on [0, 3], prtition P = {0, 1, 2, 3}. Blue: Function grph, Blck: Step function ssocited to M k

9 Exmple: Approximtion from below A function dened on [0, 3], prtition P = {0, 1, 2, 3}. Blue: Function grph, Blck: Step function ssocited to M k

10 Upper nd lower sum Denition. Let f : [, b] R, nd let P = {x 0, x 1,..., x n } be prtition of [, b], with = x 0 < x 1 <... < x n = b. We write S(P) = S(P) = n M k 1(x k x k 1) k=1 n M k 1(x k x k 1) k=1 Interprettion: The re below the step function with vlues M k 1 contins the re below the grph of f. Hence S(P) is greter or equl to the re below the grph of f. Likewise: S(P) is smller or equl to the re below the grph of f.

11 Grphicl interprettion of upper nd lower sum The dierence S(P) S(P) is the re between upper nd lower step function pproximtion

12 Renement of prtition Denition. Let P 1, P 2 be two prtitions of [, b]. Then P 1 is clled renement of P 2 if P 1 P 2. Interprettion: If P 1 P 2, then S(P 2 ) S(P 1 ) S(P 1 ) S(P 2 ) Hence the re between upper nd lower pproximtion decreses. The two should pproximte the sme vlue, s the prtition gets ner nd ner.

13 Illustrtion for renement A function f : [0, 3] R, prtition {0, 1, 2, 3}, lower nd upper pproximtion

14 Illustrtion for renement The sme function, lower nd upper pproximtion for the renement {0, 0.5, 0.7, 0.8, 0.9, 1, 1.3, 1.5, 1.6, 1.7, 1.8, 1.9, 2, 2.2, 2.4, 2.6, 2.8, 3}.

15 Riemnn integrble function Denition. The function f : [, b] R is clled (Riemnn) integrble if for every ɛ > 0 there is prtition P of [, b] such tht S(P) S(P) < ɛ Note: This implies S(P ) S(P ) < ɛ for every renement P of P.

16 Convergence of upper nd lower sums Theorem 1. Let f be Riemnn integrble function. Let P n be sequence of prtitions stisfying δ n 0, where δ n is the mximl distnce of two neighboring elements of P n. Then I (f ) = lim n S(P n) exists, with I (f ) = lim n S(P n). Moreover, I (f ) is the sme for ll sequences of prtitions with δ n 0.

17 Denition of the Riemnn integrl Denition. If f is integrble, I (f ) s in Theorem 1. I (f ) is clled the (Riemnn) integrl of f over [, b], nd denoted s b f (x)dx. is clled lower bound of the integrl, b is clled upper bound of the integrl, nd f is clled the integrnd. Furthermore, we dene, for < b, b b f (x)dx = f (x)dx s well s f (x)dx = 0

18 Criteri for Riemnn integrbility Sucient conditions: If f is continuous on [, b], then f is integrble. If f is monotonic nd bounded on [, b], then f is integrble. Exmple: A bounded function tht is not Riemnn integrble: { 1 x Q f : [0, 1] R, f (x) = 1 x Q For every prtition P, one nds S(P) = 1 1 = S(P).

19 Properties of the Riemnn integrl Theorem 2. Let f, g be integrble over the intervl with bounds, b, let s R sf is integrble, with b sf (x)dx = s b f (x)dx. f + g is integrble, with b f (x) + g(x)dx = b f (x)dx + b g(x)dx. Let c in R be such tht f is integrble over [b, c]. Then f is integrble over [, c], with c f (x)dx = b c f (x)dx + b f (x)dx. If f is integrble, then f is integrble s well, with b b f (x)dx f (x) dx

20 Monotonicity of integrls Theorem 3. Let b, let f : [, b] R be integrble nd bounded, with Then m f (x) M, for ll x [, b] m(b ) b f (x)dx M(b ). This pplies in prticulr, when f is continuous on [, b], nd m = min f (x), M = mx f (x). x [,b] x [,b] More generlly, if f, g : [, b] R re integrble, with f (x) g(x) for ll x [, b], then b b f (x)dx g(x)dx.

21 Illustrtion for the estimte

22 Fundmentl Theorem of Clculus Theorem 4. Let f : [, b] R be continuous. We dene F : [, b] R, F (y) = y f (x)dx Then F is continuous on [, b], dierentible on (, b), with F (x) = f (x), x (, b). Conversely, suppose tht G : [, b] R is continuous, dierentible on (, b) with G = f. Then the integrl is computed s b f (x)dx = G b := G(b) G()

23 Integrtion nd ntiderivtives Remrks: Let f : [, b] R be continuous function. A dierentible function F with F = f is clled ntiderivtive or primitive of f. Hence f hs primitive given by F (y) = y f (x)dx. Two primitives F, G of f only dier by constnt: F (x) = G(x) c, with c R xed. By letting F (y) = y f (x)dx one obtins the unique primitive of f stisfying F () = 0. It is customry to denote primitives s F = f (x)dx (without bounds), nd refer to them s indenite integrls of f.

24 Appliction: The length of curve Denition. Let f : [, b] R n be given, i.e., f (x) = (f 1 (x), f 2 (x),..., f n (x)) T. The set C = {f (x) : x [, b]} is clled curve in R n, nd f is clled prmeteriztion of C. We ssume tht ll f i re continuously dierentible on (, b) nd continuous on [, b]. We dene the length of C s l(c) = b f 1 (x)2 + f 2 (x) f n(x) 2 dx

25 Exmple: Circumference of the circle We consider the mp f : [0, 2π] R 2, with f (x) = (sin(x), cos(x)). The resulting curve is the unit circle. We compute nd thus, using sin 2 + cos 2 = 1, 2π 0 f 1(x) = cos(x), f 2(x) = sin(x) 2π f 1 (x)2 + f 2 (x)2 dx = 1dx = 2π. 0

26 Exmple: Length of grph We wnt to determine the length of the grph G f t [0, 1]. G f is prmeterized by of f (t) = t 2, for g : [0, 1] R 2, g(t) = (t, t 2 ) T. Using g 1 (t) = 1, g 2 (t) = 2t, we obtin 1 l(g f ) = 1 + 4t 2 dt. One cn check tht F (t) = ( 2t 1 + 4t 2 + ln(2t + ) 1 + 4t 2 )) is primitive of g(t) = 1 + 4t 2. Hence, l(g f ) = F 1 0 = 1 ( ln(2 + ) 5) 0 4

27 Summry Denition nd interprettion of integrls; re under the grph Integrbility criteri: Continuity, monotonicity Properties of the integrl: Linerity, monotonicity Evlution of integrls vi ntiderivtives ( New problem: How to obtin ntiderivtives) Appliction of integrls: Curve length

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