Matlab習(xí)題與答案5

1、11計(jì)算下列不定積分X +1解：syms xin t(xA2/(x+1) ans =1/2*xA2-x+log(x+1)Rsin2xdx71 + sin2解in t(si n(2*x)/sqrt(1+si n(x)A2)ans =2*(2-cos(x)A2)A(1/2)J+5解：in t(1/sqrt(xA2+5)ans =asi nh(1/5*5A(1/2)*x)心 + X +1解：int(x+1)/(xA2+x+1)ans =1/2*log(xA2+x+1)+1/3*3A(1/2)*ata n(1/3*(2*x+1)*3A(1/2)2計(jì)算下列定積分xln 沁Ji解in t(x*log(x)

2、,1,exp(1)ans =9366741398929500034245406117369/2535301200456458802993406410752*log(3060513257434037)-234168534973237500856135152934225/1267650600228229401496703205376*log (2)-809909079870120270632748702911993/507060240091291760598681282150403sin2x解：int(x/sin(x)A2,pi/4,pi/3)ans =-1/9*pi*3A(1/2)-1/2*lo

3、g(2)+1/2*log (3)+1/4*pi-1解:int(sin(log(x),1,exp(1)ans =-3060513257434037/2251799813685248*cos(log(3060513257434037)*cos(50*log(2)-3060513257434037/2251799813685248*s in(log(3060513257434037)*si n(50*log (2) )+3060513257434037/225 1799813685248*sin( log(3060513257434037)*cos(50*log(2)-306051325743403

4、7/2251799813685248*cos(log(3060513257434037)*s in(50*log(2)+1/2解：int(xA3*si門儀)人2)心人4+2*乂人2+1),-1,1)ans =t1 亠 l n x3.求i時(shí)血并用師對結(jié)果求導(dǎo).解:syms t; f=int(1+log(x)/(x*log(x)A2),x,1,t)diff(f,x)f =infans =*舶山T I4+2x2+ 144計(jì)算二重積分JJ(x + y)必妙解: syms a syms r in t(i nt(rA2*cos(a)+rA2*si n( a),r,0,1),a,0,2*pi)ans =fj

5、 (x2+滬陽和解： int(int(A3,r,0,cos(a),a,-pi/2,pi/2)ans =3/32*pi5.計(jì)算皿L為圓周2+y2=訛 o) 解：in t(a*sqrt(a*x)/(a*x-xA2),x,0,a)ans =2*aA27.求微分方程的通解2x2yy=y21解:syms ydsolve(2*xA2*y*Dy=yA2+1,x)ans =1/exp(1/x)*(-exp(1/x)*(exp(1/x)-C1)A(l/2)-1/exp(1/x)*(-exp(1/x)*(exp(1/x)-C1)A(l/2)業(yè)dx yx解：dsolve(Dy=(y+x)/(y-x),x)ans =

6、5x+(2*xA2+exp(C1)A2)A(1/2)x-(2*xA2+exp(C1)A2)A(1/2)，y 丄 xy cos x x解：dsolve(Dy=cos(y/x)+y/x,x)ans =ata n(-(-xA2+exp(C1)A2)/(xA2+exp(C1)A2),2*x*exp(C1)/(xA 2+exp(C1)A2)*x(xcosysin 2y)y = 1解：dsolve(x*cos(y)+si n(2*y)*Dy=1,x)ans =asi n(-lambertw(-1/2*C1*exp(-1/2*x-1)-1/2*x-1)8.給出函數(shù)f (x)二exsinx - 2xsinx

7、2xcosx在點(diǎn) x = 0 的7階taylor開式以及在x=1處的5階taylor展開式.解： f=exp(x)*si n(x)+2Ax*s in (x)+2Ax*cos(x); taylor(f,0,8)ans =1+(2+log(2)*x+(1/2+1/2*log (2) A2+log(2)*xA2+(-1/2*log(2 )+1/6*log (2)A3+1/6+1/2*log(2)A2)*xA3+(-1/4*log(2)A2+1/24+1/ 24*log(2)A4-1/6*log(2)+1/6*log(2)A3)*xA4+(-1/40+1/24*log(2)A4-1/12*log(2)

8、A2+1/120*log(2)A5+1/24*log(2)-1/12*log(2)A3)* xA5+(-1/80+1/48*log (2)A2+1/720*log (2) A6-1/48*log(2)A4+1/120 *log(2)+1/120*log(2)A5-1/36*log(2)A3)*xA6+(-1/560-1/240*log (2) A5+1/5040*log(2)A7+1/144*log (2)A3-1/720*log(2)-1/144*log A4+1/720*log (2) A6+1/240*log(2)A2)*xA7 taylor(f,1,6)ans =exp(6)*si n

9、(6)+64*si n( 6)+64*cos(6)9.判別下列級(jí)數(shù)的斂散性，若收斂求其和.6 syms n1.1.17 limit(1/(2* n-1)/(1/n), n,i nf)ans =1/2所以，級(jí)數(shù)發(fā)散。 tgn =4 limit(ta n(pi/(2* n*sqrt( n+1)/(1/nA(3/2), n,i nf)ans =1/2*pi所以，級(jí)數(shù)收斂。 symsum(ta n(pi/(2* n*sqrt( n+1), n,1,i nf)ans =sum(ta n(1/2*pi/n/( n+1)A(1/2), n = 1 . i nf)解：Iimit(1/( n*log( n),

10、n,i nf)ans =0所以，級(jí)數(shù)收斂。 symsum(-1)A n/(n *log( n), n,2,i nf)ans =sum(-1)A n/n/log( n),n = 2 . i nf)五、實(shí)驗(yàn)總結(jié)請根據(jù)本次實(shí)驗(yàn)的內(nèi)容寫出實(shí)驗(yàn)總結(jié)，總結(jié)一下Matlab的求解微積分問題的特點(diǎn)解: (-1)n -21n In n8及其基本操作。(要求不少于300字)實(shí)驗(yàn)總結(jié)1應(yīng)用diff(s)沒有指定微分變量和微分階數(shù)，則系統(tǒng)按findsym函數(shù)指示的默認(rèn)變量對符號(hào)表達(dá)式s求一階微分。應(yīng)用diff(s， )或diff(s，sym( V格式, 表示以v為自變量，對符號(hào)表達(dá)式s求一階微分。應(yīng)用diff(s，n

11、)格式，表示對 符號(hào)表達(dá)式s求n階微分，n為正整數(shù)。 應(yīng)用diff（s，v，n）diff（s，n，v） 格 式，表示以v為自變量，對符號(hào)表達(dá)式s求n階微分2應(yīng)用int（s）格式，表示沒有指定積分變量和積分階數(shù)時(shí)，系統(tǒng)按findsym函數(shù)指示的默認(rèn)變量對被積函數(shù)或符號(hào)表達(dá)式s求一階積分。應(yīng)用int（s，v）格式，表示以v為自變量，對被積函數(shù)或符號(hào)表達(dá)式s求一階不定積分。應(yīng)用積分函數(shù)時(shí)，如果給定a、b兩項(xiàng)， 表示是進(jìn)行定積分運(yùn)算。a、b分別表示定積分的下限和上限。不指定積分的下限和上限表示求不定積分。3將符號(hào)多項(xiàng)式s用嵌套形式表示，即用多層括號(hào)的形式表示。Horner函數(shù)可以實(shí)現(xiàn)此功能。該函數(shù)的格式horne4為了了解函數(shù)引用過程中使用的符號(hào)變量個(gè)數(shù)及變量名，可以用findsym函數(shù)查詢默認(rèn)的變量。該函數(shù)的引用格式為：findsym（f,n）。說明：f為用戶定義的符號(hào)函數(shù)，n為正整數(shù)，表示查詢變量的個(gè)數(shù)。n=i,表示