资源描述：
《积分与微分-余显祥》由会员上传分享，免费在线阅读，更多相关内容在应用文档-天天文库。
1、习题一右矩形法xitiyi.aforn=[202540];%对n赋予一个向量h=(3-1)./n;%计算步长x=1:h:3;%取节点k=2;f=0;whilek<=n%循环计算出ff=cos(x(k))*exp(sin(x(k)))+f;k=k+1;ends=f.*h%计算体积ends=-1.1720s=-1.1716s=-1.1706梯形法xitiyib.mforn=[202540]%对n赋予一个向量h=(3-1)./n;%计算步长x=1:h:3;%取节点y=cos(x).*exp(sin(x));s=h.*(y(1)+y(n+1))/2;%计算两个端点的数值fork
2、=2:n%循环计算出积分ss=h.*y(k)+s;endsends=-1.1663s=-1.1670s=-1.1677三点高斯积分法xitiyic.mt=[-0.774500.7745];%三点高斯积分点x=t+2;%进行变量替换f=cos(x).*exp(sin(x));s=0.5556*f(1)+0.8889*f(2)+0.5556*f(3)%用高斯积分点积分求数值解s=-1.1789精确解s=int(sym(‘cos(x)*exp(sin(x)’),’x’,1,3)%对表达式‘cos(x)*exp(sin(x)在区间[1,3]积分s=exp(sin(3))-exp
3、(sin(1))double(s)%s保留双精度ans=-1.168213988201318综合上述梯形法计算出的结果更接近精确解；但是在其他例子中梯形法不一定有好的数值解。习题二xitier.mP=rand(10000,3);%取10000行3列的每个元素均为[0,1]的随机矩阵x=-1+2*P(:,1);%取p矩阵的第一列进行运算y=-1+2*P(:,2);%取p矩阵的第二列进行运算z=P(:,3);%取p矩阵的第三列II=find(y>=-sqrt(1-x.^2)&y<=sqrt(1-x.^2)&z<=x.^2+y.^2);%找满足逻辑表达式的点M=length(
4、II);%计算出点的个数S=4*M/10000%计算出体积的估计值plot(x(II),y(II),'g.')S=1.573200000000000用蒙特卡洛法画出的取点图计算精确解xitierb.msymsxy%定义符号变量s1=x^2+y^2;s2=int(s1,x,-sqrt(1-y^2),sqrt(1-y^2));%对表达式s1积分s3=int(s2,y,-1,1)%对表达式s2积分s3=pi/2ans=1.570796326794897比较可知蒙特卡洛方法能够很好的求数值积分习题三(1)fun4.mfunctionz=fun4(x,y)%建立一个函数文件z=(
5、sin(x*y)-x^2*y^2)/cos(x)命令窗口输入[T,Y]=ode23('fun4',[0,2],1)%调用ode23计算微分方程组数值解plot(T,Y)figure(2)plot(T,log(Y))T=00.2000000000000000.4000000000000000.6000000000000000.7837721887295110.9455468174593361.1170092091986751.2704636397966091.4239180703945441.4739180703945441.5239180703945441.555327
6、6142509321.5631800002150291.5710323861791261.5780866936304721.5874015731374331.6070427677943891.6266839624513441.6468252287842361.6785930252637771.7128346653074961.7418282566030201.7711871905695161.8022605731223281.8356141425157971.8716580326351921.9107479178101041.9532148014665552.00000
7、0000000000Y=1.0000000000000001.0175558771587401.0606493153730391.1098527827963631.1312121266654611.1013022280336951.0010644143183490.8601790185456110.6998575549329360.6484510703051470.5996863945770980.5710999429715170.5645010532958260.5578883759211380.5459335853772000.520
