1、文科微積分學習指導(5)第五章定積分及其應用第一部分重點、難點及分析定積分的定義及幾何意義:bn由定積分的定義，Jf(x)dx=lim SaAT 0 j =1一個函數的定積分是由“和式的極限”來定義的，許多現實問題都 歸結于求和式極限問題，如：不規則圖形的面積、變力作功、變速 直線運動的距離，都可通過定積分的計算找到答案。因此定積分的 概念來源于實際，是解決實際問題的有力“工具”。定積分的概念由四個步驟構成：分割、近似代替、求和、取極限。n求和lim 5： f(工Xj得到了近似的結果，而取極限改善了近似程度, 幾 T0i=1極限值給出了所要測度的量的精確定義。應該說，定積分是一種特 殊的“和式

2、極限”，它比數列、函數的極限要復雜一些，在它的定義 中，沒有簡單地寫成以n或入為自變量。b定積分f f (x)dx的值與積分區間a, b和被積函數f有關，而與積分ab變量用哪一個字母來表示無關，也就是說在Jf(x)dx中將x改寫成ua或t得到的結果是一樣的。b在f (x) > 0時，Jf(x)dx的數值，在幾何上等于曲線f (x), x=a,ax=b和x軸所圍成的曲邊梯形的面積。在一般情況下，表示曲線f (x),直線x=a, x=b和x軸之間各部分3定積分的存在性，可以是等距離的,上任意取一點£，可以在區間端點上取，也可以在區間內取。面積的代數和。其中，在x軸上方的面積取正號，

3、在x軸下方的面 積取負號。要滿足兩個任意性：一是對區間a, b的任意劃分,也可以是非等距離的；另一個是在小區間XjjXj在“微分學”中知道函數的連續性不能保證可導性，連續性是可導 的必要條件，但不是充分條件。而連續函數的定積分一定存在，這 是定積分存在的充分條件，而不是必要條件，也就是說對不連續的 函數，定積分也可能存在。事實上，在區間a, b上若有有限個間斷點，而有界的函數f (X)在這個區間上的定積分是存在的。定積分的性質:bbb1)運算性質:2f(X)± Pg(x)dx = a jf(x)dx± P Jg(x)dxaaa2)區間性質:b? Jf (x)dx= Jf (

4、x)dx+ If (x)dxaac?若f(x)在a,b上可積，則在a,b上的部分區間引匕匸a,b上也是可積的。不等式性質:?若 f(X), g (X)在a, b上可積，且 f(X)蘭 g (x),那么bbJf (x)dx 蘭 Jg(x)dxaa?若 f(X)在a, b上可積，m < f(X)< M，則bm(b a) < ff (x)dx < M (b a)ab4)積分中值性質：Jf(x)dx= fG)(b-a),仝a,ba定積分的計算:b重要的公式：Jf (x)dx= F(b) - F(a),F(x)是f (x)的原函數a計算定積分也可以象計算不定積分時使用換元法， 在

5、用第二換元法時，一是，要注意引進的新變量應是單值、單調、連續的函數,且要有連續的導數。二是進行變量代換要換積分限。如:1111 111 - 722dx,如果令 x = -,J dx=二2t_11 + x2卄丄1t2dt,1 1從而f dx=O,這是一個錯誤的結果，關鍵的是引進的_11 + X2在-1,1上是不連續的，也不是計算定積分時，常用的幾個結論：單調的函數。1)aJf(x)dx = O。當被積函數f(X)是奇函數時，-a2)aaJ f(x)dx=2Jf(x)dx當被積函數f (X)是奇函數時，-a03)a+TTJf (x)dx= Jf (x)dx若f f X)是以T為周期的連續函數，a0

6、文科微積分學習指導（5）-1)四、定積分的應用：在幾何中的應用：求平面圖形的面積、求旋轉體的體積；在物理方面的應用：變力作功。第二部分書后習題b用定義證明：Jkdx = k（b-a）。將a，b分成n等分，分點為xiI（b-a），1,2, n-1, n;每個小區間b a.的長度 AXj =，取©j = Xj；而f( x)k，從而 f( xi) = knnn b - a b - a禾n式 S f (xi HXj = Z k= nk= k(b - a)i=1i=i n求極限:nli Z f(xi 工Xj = k(b - a).n 0 i =112) K2eX + 1)dx = 2 Jexd

7、 Jdx = 2ex10 = 2(e-1) + 1 = 2e-11 dx=arcta nx_11 +x21 2X兀JI-14( 4)_ 2J 201 +x2dx = J"/dxJjdx - I1 + x201 + x2dx= x0 - arctanx3)aJf (x)dx =Jf(x)dx+ Jf(x)dx7a-a對第一個積分，0 0J- f(-t)dt = -Jf(t)dt= Jf(x)dx-aaa二 Jf (x)dx = 2 Jf (x)dx。-a4)-1-1_2(11 +5x)3dx5/(11 + 5X)3d(11+5xq6dU11廣2)|6510 (3610(36)722td

8、t =dt = 2ft 1 dt =01 + to1+t丄 dt - id - t)dt)=2ln(1+t)0仆 01 2+ 0(1一t)22 10 = 2l n3 + 2(1-1) = 2I n3e In x 2i tdx = Hnxd(lnx)衛J ftdt 1 xJsin 2 xcosxdx=fsi n2xdsi nx空匕-*Jt2dt1t32 arcs inxdx = /arcsin xd (arcsin x) arcsin t-?匕（36-0) =72Cl22 .討4 - x2dx-仝-竺吋0注：第一個等號是根據 書P.90例8的結果。T gcos'tdt = 2(t + s

9、intcost)0In 2In 2J xeXdx= J xdex = xex0 0= eln 2 (I n2T) +1=2I n2TIn 20In 2-J exd In 2eln -0ex2(?In 20-0) =In 2eln 2 - (eln 2 T)= 0兀冷宀）2 x 2 x 1 x fex cosxdx = cosxdeX = -(cosx + sin x) ex 26)lim鷗aT F X3=lim (-lx-2) aT2=lim aT "1一丄+1) J2a22)2-He收斂.X3e X 處J +=1 lim (-12aT 亦 ©a-He2/. J xeX d

10、x收斂.07）所圍面積:x3)S = J(Vx - x2)dx = (-X0 3所圍圖形有兩個交點:|y-2x二y = X - 4(X - 4)2 = 2x= X = 2,8 ；兩個交點:(2, 2)、(8, 4)文科微積分學習指導（5）119可以用兩種方法計算所圍面積:4y21 2s= Hy+ 4-）dy =（加2-222訂y3)642 + 4(4 + 2) = 6- 12 + 24 = 182 8 _S =-(-72X)dx + J J2x -(X - 4)dx = 242 -01 2X28+4(8 - 2)= 38)球的方程是：X230+F24 242 + (1672 2冋-(64 -

11、4) + 24 = 182+ y2 + Z2 = R2，它可以看成是由圓"一、-（R2 -x2）X2 + y2 =R2繞X軸旋轉得到的A(x)=兀f (X)=兀 rV =兀(R2-r4 _3=兀R3-X2) dx= R2R-(-R)-X33R -R=2花 r3-1(兀 r3 +;ir3)9JL兀V =兀 fsin2 xdx = 1(1 - cos2x)dx =兀-(-兀)-2汀2一兀一兀rin2X兀一兀=一21）求定積分:第三部分附加習題1 - - , 2j|2Xdx = J- 2xdx + J2xdx = -2 Jxdx+ 2Jxdx = -2 x2-1-10-10=-(OT) +

12、 (1-0) =2-12干exldx 藥xdx+e如dxJx21 X11 X21 2 1 2=-(e2 -1) + (ln e - In1) = (e2 + 1)e+ 2Jlnxd(ln x)dx =+ 2 ；(ln X)2文科微積分學習指導(5)25兀 fcos xsinxdx,令 t = cosx,dt = - sixdx 當 x= 0 時,t = 1,當 x = 2時,t= 0-°t5dt1=76Fdx,令J2x + 1 =t,0 j2x+1t =3；t2 -1x =, dx = tdt，當 x = 0時，t= 1,當 x = 4時,2原積分化為:1(13 +tdt 二1/ +

13、 3)dt =二(二+ 3t)1 t212 323332)設f (X)在0 , 1上連續并單減，試證:£Jf (x)dxk 匕0對任何E(0, 1 )有J f (x)dx.0匕1匕Jf (x)dx -匕 J f (x)dx = J f (x)dx - E J f (x)dx - r J f (x)dx =110匕1=(1 -匕)Jf (x)dx-© Jf(x)dx©因為f (X)在0 , 1上單減,所以 J f (x)dx 二 J f (E )dx = E f C)1Jf(x)dx< Jf(©)dx = (1-r)f(©)JJ匕1故 J

14、f (x)dx - © J f (x)dx > (1 - © )©f G ) - © (1 - © ) f G) = 0匕1即 Jf(x)dx>E Jf(x)dx.01町223)設 f(x) 0 , 1，且 Jf(x)dx = 3,求 J f (cos2 x) sin2xdx。2J f (cos x)sin 2xdx0瑪2-Jf (t)dt = Jf (t)dt =3 。-J f (cos2 x)d (cos2 x)04)求 1 J2x- x2dx。該積分的被積函數非負,積分上限大于下限，該積分值為積分區間上曲邊梯形的面積，由于2x

15、-X2 = 1 -(X-1)2，積分值應為圓(x-Dn 的面積的四分之一于是積分值是45)設非負二階可導函數f (X)在0 , +乂內，廣(x)>0,對常數 a>0,證明:Jf (x)dxx af(!)x設 F(X)= J f (t)dt - xf(3), X 亡0, a文科微積分學習指導（5）F（X）= f（x） 一 f（2）冷f切由拉各郎日中值旦T f 2擰f百）=xx=丿©)-(2)其中 © - (|,x)由 f ”(x)>0,知 f '(X)單增，即 f 徉)> f -),于是 F'(X)XO,即 F(x)單增, 由2二 F(

16、a) > F(0) =0aa因此 J f (x)dx 3 af ()021p+2P + nP(PO)np十1limP + 2P 十+ nPnp+1ink=lim 1 送(k)pnT 處 n k =1 n6)計算：lim 12n157）計算:X xlim J f (t)dt，f(t)是連續函數XT a X - a aXT aX - a令 F(x) = xjf (t)dt , lim(t)dt = lim F(x F(a Fa)XT a X a a而 F x) = xf(X)+ J f (t)dt , F '(a) = af (a)X x所以，lim Jf (t)dt = af (a

17、)。XT a X - a aX8)設 f (t)連續，g(x) = xH(t)dt,求g"(0)og'(x) = Jf(t)dt +xf(x),g(O) = Og“(0)= iim gw ggXT 0 X 0XIf (t)dt=liml0xt0 x-f(0) + f(0)9)設 J dtJf(t)dt+ xf(x)-g'(O)lim -XT 0x-0+ f(X) = limXJf (t)dt02f(0)XT 0十 f(0) =limXT 0f(x)+ f(0) =In 2 寸 et - 1=,求 X。6先計算:dtu =加-1,tdn(1+u2)TJ 2udu

18、1; =Vet -12u 1+u22Jdu1 + u2=2arctanJet -c所以,X dt =2arctanJr；!兀In 2 Jet T-12G=2arcta桃J-?蔦,二 arctan JeX T = -( +2 2兀6)由此可得，x=ln4。10）求極限:文科微積分學習指導(5)217XJin cosxdxlim 0xt0 +x3,這是0型不定式limXT 0十Jin cosxdx0x3limXT 0 +In cosx3x2-sin x lim 1 sin X 1 -lim 11)若當X > 0,F (x) = Jdt01 +12xt0+ cosx+ J01 +1dt2,求

19、F(x)6x6 3 0中 X cosx2宀2a(x)所以，F(x)是常數。"AX=0,請注意:Jf(t)dt、a'(x)fa(x)F(x) = F(1)01 +t2arctantXJtf (t)dt12)設 F(x) = 2x2cxH0，其中f (X)是一個在x=0處連續的已知函數，如果F(x)在x=0處連續，求c。由 lim F (x)=XT 0limXT 0XJtf (t)dt0x2lim = 1 f(0) = F(0) = cXT 0 2x1即 c=- f (0)文科微積分學習指導（5）113)已知 Jf (xt)dt = nf(x),求 f(X).01xdu令 u =

20、 tx, f f (tx)dt = f f (u)=nf (x)00XX即 J f (u)du = nxf(X),0對等式兩邊關于x求導，f（X）= nf（X）+ nxf 丫x） 另寫成方程：匸兇=上n丄f （x） n X兩邊求積分：J匚兇dx = 口匸dx= 口 4 dx1 nnf (x)n Xf (x) n Xlnf(x) = 口 lnx + lnc= f(x) = c|x n14）設 f（X）T丹dt，求f（x）+心1 +t1u=-dttTln1 1T-丄 11( u2u1 x lntX Intf(x) + H1) = jULdt+ J 竺7dt =X1tx12=JI ntd(l nt)

21、= (I nt)2121t +t21X = 2(|nx)2附dt=第四部分英語參考資料Fundamental theorem of calculus A sound approach to integrationbdefined the integral J f (x)dxaas the limit ,in a certa in sense of a sum .That this can be evaluated ,whe ns continu ous, by finding an an tiderivative of f , is the result embodied in the so-

22、called Fun dame ntal Theorem of Calculus. It establishes that in tegrati on is the reverse p rocess to differe ntiatio n:Theorem: If f is continuous on a,b and * is a function such thatb* '(X)= f (x) for all x in a,b,then / f (x)dx = © (b) - * (a)aArea under a curve Suppose that the curvez=

23、f (x) lies above thex-axis, so that f(X)二 0 for all x in a,b. The area under a curve, that is ,the area of the regi on boun ded by the curve , the(-axis and the lin esx=a andbx=b,equalsjf (x)dx.aThe defi niti on of i ntegral is made pr ecisely in order to achive this resultff(X)蘭 0 for all x in a,b,

24、the intagral above is negative. However, it is still the case that its absolute value is equal to the area of the regi on boun ded by the curve, thex-axis and the lin esx=a and x=b.Integral Let f be a fun cti on defi ned on the closed in terval a, b. takepoints x0,x1,x2Xn such that a = x0 吒為吒 X2<

25、 XnT 吒 xn=b,andin each sub in terval xi, x+1 take a point g . From the sumn 1送 f(Ci)(Xi+1 - Xi); that is ,i =0f (C0)(Xi -Xo)+ f(Ci)(X2 -X1)+f(Cn_i)(xn -Xn_i). Such a sum is called a Riema n sum forf over . Geometrically, it gives the sum of the area of n recta ngles and is an app roximati on to the

26、area un der the curv(e=f (x) betwee nx=a and x=b.The integral of f over a,b is defi ned to be the limit I (in a sense that n eeds more clarificati on tha n can be giveb here) of such a Riema n sumas n , the nu mber of poin ts, in creases and the size of the sub in tervalsbbgets smaller. The value of

27、l is denoted by J f (x)dxor J f (t)dtaaWhere it is immaterial what letter ,such asx or t ,is used in the in tegral .The inten ti on is that the value of the in tegral is equal to what is in tuitively un derstood to be the area un der the curve=f (x ) .Such a limit does not always exist ,but it can b

28、e pro ved that is does if for exa mpl ef,is a con ti nu ous fun cti on on a,b.XF(x) = Jf(t)dtIf f is con ti nu ous on a,b and F is defi ned bya,the nF'(x)= f(x)for all X in a,b ,so that F is an an tiderivative of f . Moreover, if an antiderivative © of f is known the integral can be easily

29、evaluated: theTheorem of Caculus gives its value a© (b)(a). Of the two integrals b/ f (x)dx and J f (x)dx ,the first, with limits, is called a definiteain tegral and the sec ond ,which deno tes an an tiderivative off ,is anin defi nite in tegral.Intermediate value theorem The following theorem

30、stating an important property of con ti nu ous fun cti ons: If the real fun cti on f is con ti nu ous on the closed interval a ,b and n is a real number betweerf(a) and f(b), then for somec in (a,b), f(c)= n .The theorwm is useful for locat ing roots of equati on. For exa mple , suppose thatf (x )=x

31、-cosx Then f is continu ous on 0,1,a ndf(O)<O and f(1)>0 ,so it follows from the In termediate value theorem that the equati on f(x)=0 has a root in the in terval (0,1).Density The average density of a body is the ratio of its mass to its volume.In gen eral ,the den sity of a body may not be c

32、on sta nt throughout the body.The density at a pointP, denoted by p(P) , is equal to the limit as mV T 0 of ,where 也V is the volume of a small region containingP Vand 也 m is the mass of the part of the body occupying that small region.Consider a rod of lengthl , with density p(x) at the point a distancex from one end of the rod. The n the masm of the rod is give n bylm = JP(x)dx .In the same way , the mass of a lamina or a 3-dimensional0rigid body, with den sity p (r) at the point with p ositi on