第一次实验 姓名 学号 班级 四个求极限的题
13 )lim((1?3)x?-1x?1x?1
syms x;
f=1/(x+1)-3/(x^3+1); limit(f,x,-1) ans = -1 1x(2)lim(1?)x x?? syms x
limit((1+1/x)^x,x,inf) ans = exp(1)
|x|(3)lim? x?0x
syms x
limit(abs(x)/x,x,0,’right’)
ans = 1
222sin x?1/(y?x)((1? 4)lime2x?1/yx y??
syms x y a;
f=exp(-1/(y^2+x^2))*sin(x)^2/ ... x^2*(1+1/y^2)^(x+a^2*y^2); L=limit(limit(f,x,1/sqrt(y)),y,inf) L =
exp(a^2)
1x?a2y2)2y
四个函数求导
(1)y?ex(sinx?cosx)
syms x
y=exp(x)*(sin(x)+cos(x)); diff(y) ans =
exp(x)*(sin(x)+cos(x))+exp(x)*(cos(x)-sin(x))
x3?1(2)f(x)?ln2x?1syms x
f=log((x^3+1)/(x^2+1)); diff(f) ans =
(3*x^2/(x^2+1)-2*(x^3+1)/(x^2+1)^2*x)/(x^3+1)*(x^2+1)
(3)设x2?y2?R2,求y?
syms x y;
f=solve('x^2+y^2-R^2=0',y); diff(f,x)
ans =-1/(-x^2+R^2)^(1/2)*x1/(-x^2+R^2)^(1/2)*x
1(4)已知函数f(x)?,求f?(1),f?(2)x
syms x;
f=1/x; f1=diff(f,x); ff=inline(f1) ff =
Inline function: ff(x) = -1./x.^2 x=1; ff(1) ans =-1 x=-2; ff(-2) ans =-0.2500
四个定积分
(1)
?101?x2dxsyms x;
f=sqrt(1-x^2); int(f,x,0,1) ans = 1/4*pi
(2)?|x?1|dx02syms x;
int(abs(x-1),0,2) ans= 1 9 (3)x(1?x)dx4
syms x; f=sqrt(x)+x; I=int(f,x,4,9) I =
271/6
e?2t ?2x2?1dx (4)I(t)?cost(2x2?3x?1)2
syms x t;
f=(-2*x^2+1)/(2*x^2-3*x+1)^2 I=simple(int(f,x,cos(t),exp(-2*t))) I =
-(2*exp(-2*t)*cos(t)-1)*(exp(-2*t)-cos(t))/(exp(-2*t)-1)/(2*exp(-2*t)-1)/(cos(t)-1)/(2*cos(t)-1)
??四个不定积分
2(1)(x?3x?2)dx?
syms x; y=x^2-3*x+2; int(y) ans =
1/3*x^3-3/2*x^2+2*x
(2)?cos(3x?4)dxsyms x;
int(cos(3*x+4)) ans =
sin(3*x + 4)/3
(3)
syms x;
f=exp(x)*(5^x+(exp(-x)/tan(x))); int(f,x) ans =
log(tan(x)) -log(tan(x)^2 + 1)/2 + (5^x*exp(x))/(log(5) + 1)
(4)
syms x; f=1/(x^3); int(f,x) ans = -1/(2*x^2)
dx
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