《高数双语》课件section 5-1.pptx

1、Concepts and Properties of Definite Integrals定积分 12Example of Definite Integral ProblemsOab()yf x?A xyComputing the area surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb 0.y ,and the horizontal line the vertical lines()()0),yf xf x3Example of Definite Integral ProblemsOab(

2、)yf x xyComputing the area surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb 0.y ,and the horizontal line the vertical lines()()0),yf xf xAS 4Example of Definite Integral ProblemsOab()yf x xyOab()yf x xyComputing the area surrounded by the curve Example(Area of a trapezoid

3、with a curved top)xa xb 0.y ,and the horizontal line the vertical lines()()0),yf xf x5Example of Definite Integral ProblemsOab()yf x xy1x2x1kx kx121naxxxb LArbitrarily insert n-1 points of division,Between a and b,such thatSolution(1)“partition”121,nx xx L,1,1,2,kkkxxxkn L0 xa nxb If we denote and,a

4、nd Computing the area surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb y0.,and the horizontal line the vertical lines yf xf x()()0),6Example of Definite Integral ProblemsOab()yf x xy1x2x1kx kxk 1,kkkxx ()kf Solution:becomes very small.and as its height.(2)“homogenization”1

5、,kkxx is very small,While1,kkxx the variance of function onapproximate the area of trapezoid by choosing any pointSo,we can use the rectangle to Computing the area surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb 0.y ,and the horizontal line the vertical lines()()0),yf xf

6、x()kkkAfx 7Example of Definite Integral ProblemsOab()yf x xy1x2x1kx kxDoing the same thing for each subinterval,and then combiningall the approximate values,we have11()nnkkkkkAAfx Solution(3)“summation”Computing the area surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb 0.y

7、 ,and the horizontal line the vertical lines()()0),yf xf x8Example of Definite Integral ProblemsOab()yf x xy1x2x1kx kx1()nkkkfx 1max.kk ndx If we refine the partitions of a,b,the sum,is a closer approximation to the total area.Therefore,where 01lim()nkkdkAfx ,Solution:(4)“precision”Computing the are

8、a surrounded by the curve Example(Area of a trapezoid with a curved top)xa xb 0.y ,and the horizontal line the vertical lines()()0),yf xf xThe definition of definite integral901lim()()nbkkdakAfxf x dx Oab()yf x?A xy10The definition of definite integralDefinition(Definite Integral定积分定积分)Let()f x be a

9、 function on an interval ,a b.Insert arbitrarily 1n points of division in the interval(,)a b:011nnaxxxxb L so that the interval ,a b is partitioned into n subintervals,the length of the kth subinterval is 1,1,2,kkkxxxkn L On each subinterval 1,kkxx,select arbitrarily a point k,1kkkxx ,and form the p

10、roduct:(),1,2,.kkfxkn L Add the n products()kkfx to form the sum 1()nkkkfx Riemann sum11The definition of definite integralDefinition(definite integral)(continued)If any partition of ,a b and any selection of k on 1,kkxx,the limit of the Riemann sum exists as 0d,where 1maxkk ndx ,then we say that f

11、is integrable over ,a b,and the limit value is called the definite integral of f over ,a b,denoted by 01lim()()nbkkdakfxf x dx which is read“integral of f from a to b”.Hence()f x is called the integrand function or integrand,()f x dx is called the integrand representation,a b is called the interval

12、of integration,b and a are called the upper limit 积分上限 and lower limit 积分下限 of integration,respectively.The symbol is called an integration sign.12The definition of definite integralNote It is obvious that n as 0d,but the converse is not necessarily true.Therefore,in general,we can not use n to repl

13、ace 0d.Note The Riemann sum may be different if the partition of ,a b or the selection of k is changed.The existence of the limit of the Riemann sum means that the limit of the Riemann sums is the same value as 0d,no matter how ,a b is partitioned and no matter how k is selected.Note In the definiti

14、on,the restriction of()0f x is not necessary,that is,the sign of()f x may change on ,a b.13The definition of definite integralNote For a given function f and interval ,a b,the definite integral()baf x dx is a definite value which depends only on the integrand f and the interval of integration.Also,i

15、t is independent of the variable of integration x.Hence the following integrals are the same ()()()bbbaaaf x dxf t dtf u du.()()baabf x dxf x dx Note We also define()0aaf x dx 14Conditions for existence for the definite integralabxyOcTheorem(necessary condition for integrability)()f x,a b()f xIf is

16、integrable over the interval,then must be.,a bbounded on ()f x,a bIn other words,if is unbounded on then it is not integrable.15Conditions for existence for the definite integralTheorem(sufficient condition for integrability)a finite number of first type discontinuous points,then()f x,a b must be in

17、tegrable over the interval.()f x,a b is continuous on the interval If,or has onlyxyOabc16Conditions for existence for the definite integral01211210,1nnnxxxxxnnn L1,1,2,.kxknnLSolution:0,1We partition into n equal subintervals,then the points of the and0,1,is continuous on the interval 2xSince the fu

18、nctionit is integrable.the Riemann sum.partition areThen we can select a special partition to construct Find 120 x dx by the definition of the definite integral.Conditions for existence for the definite integral17,1,kkxkn L2231111nnkkkknnn212011limnnkkx dxnn Solution(continued)then the Riemann sum i

19、sTherefore,Let 22311nnL.31(1)(21)1lim63nn nnn31(1)(21)6n nnn.Find 120 x dx by the definition of the definite integral.18Conditions for existence for the definite integralExample:Find 211dxx by the definition of the definite integral.1,kkq 11111111()(1)(1)(1)nnnnkkkkkkkkkkfxxqqqn qq 231,nq q qq L(1,2

20、)We choose between It easy to see that If we choose then2nq If we let,we have 11()(21)nnkkkfxn .Solution and these points 1,2ndivide the interval into subintervals.11(1)kkkkxqqqq.19Conditions for existence for the definite integralExample:Find 211dxx by the definition of the definite integral.ln2 11

21、21lim(21)lim1xxxxxx 1lim(21)ln2nnn1210111limlim(21)nnkdnkkdxxnx Solution(continued)so,we have and this means thatFinish.Sinceln2.20Geometric interpretation of the definite integralIf()0f x ,then()baf x dx is the area A of the trapezoid with the curved top()yf x.That is()baf x dxA .()0f x 01()lim().n

22、bkkdakf x dxfxA If,then)(kkkAfx ()yf x Oabxy21Properties of Definite IntegralsProperty 1(Linearity property)If,f gR a b,k is a constant,then kf,fgR a b and ()()bbaakf x dxkf x dx ()()()()bbbaaaf xg xdxf x dxg x dx.,a bBy the definition of definite integral,we call this kind of definite integral as R

23、iemann integral and we denote the set of all functions which are integrable over the interval by ,R a b.This property can be easily proved by the definition of the definite integral.22Properties of Definite Integrals,()()()kkkkkka ba cc bfxfxfx0d cWe select a partition such that is always kept as a

24、point of the partition.Let,we obtain()()()bcbaacf x dxf x dxf x dx.Proofacb(1)Assume that.ThusProperty 2(Additivity for intervals)If f is integrable on an interval containing the points a,b and c,then ()()()bcbaacf x dxf x dxf x dx no matter what the order of a,b and c.23Properties of Definite Integ

25、ralsabc()()cbbcf x dxf x dxsay,then we haveSo thatSince,thus we have the conclusion.Proof(continued),a bc(2)Suppose that is out side the interval,()()()cbcaabf x dxf x dxf x dx.()()()bccaabf x dxf x dxf x dx.Property 2(Additivity for intervals)If f is integrable on an interval containing the points

26、a,b and c,then ()()()bcbaacf x dxf x dxf x dx no matter what the order of a,b and c.24Geometric interpretation of the definite integral1234()baf x dxAAAA 1A2A3A4A()yf x abProperties of Definite Integrals25 Find 2101xdx by its geometric meaning.yOx12101.4xdx 26Properties of Definite IntegralsProperty

27、 3(Inequalities for integrals)Suppose that ab,f gR a b.(1)If()()f xg x,xa b,then ()()bbaaf x dxg x dx;(2)()()bbaaf x dxf x dx;(3)If()mf xM,xa b,then ()()()bam baf x dxM ba.27Properties of Definite Integrals31(),3sinf xx 0,x 30sin1,x3111,433sin x 3000111,433sindxdxdxx 301.433sindxx Solution：Find the

28、bound for 301.3sindxx Properties of Definite Integrals28Property 4(Mean value theorem for definite integrals)If ,fC a b,then there exists at least one point ,a b such that ()()()baf x dxfba .,a b()()()bam baf x dxM ba Proof:of the function on the closed interval,and we have Since we have known thatw

29、e have()baf x dxmMba .,fC a b mM,there must have the maximum and minimum Since()mf xM.Property 4(Mean value theorem for definite integrals)If ,fC a b,then there exists at least one point ,a b such that ()()()baf x dxfba .Properties of Definite Integrals29Mean value()baf x dxba a b,()()baf x dxfba Pr

30、oof(continued)()f xminimum and maximum of.By the intermediate value theorem,we such thatand this implies the conclusion.And this means that()baf x dxmMba is a value between the have that there exists at least one 30Properties of Definite Integralsxyoab)(f()baf x dxba Mean value of function f(x)31Properties of Definite Integrals23sin()xxtf t dtt 3sin()(2),fxx 23limsin()xxxtf t dtt 32 limsin()f 2 lim 3()f 6 ,2x x()f x,a blim()1.xf x 23limsin()xxxtf t dtt Suppose that is derivable on an interval and Find By the mean value theorem of definite integrals,there must,such that Solutionhave a