y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求(1,2)区间的定积分?要数值解
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 04:37:38
![y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求(1,2)区间的定积分?要数值解](/uploads/image/z/6919348-4-8.jpg?t=y%3D1%2F%28x%5E4+%2B+x%5E3+%2B+x%5E2+%2B+x+%2B+1%29%E7%94%A8matlab%E6%80%8E%E4%B9%88%E6%B1%82%EF%BC%881%2C2%EF%BC%89%E5%8C%BA%E9%97%B4%E7%9A%84%E5%AE%9A%E7%A7%AF%E5%88%86%3F%E8%A6%81%E6%95%B0%E5%80%BC%E8%A7%A3)
y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求(1,2)区间的定积分?要数值解
y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求(1,2)区间的定积分?要数值解
y=1/(x^4 + x^3 + x^2 + x + 1)用matlab怎么求(1,2)区间的定积分?要数值解
用数值积分函数,可以直接得到结果:
>> f=@(x)1./(x.^4+x.^3+x.^2+x+1);
>> y=quad(f,1,2)
y =
0.0888
>> syms x
>> int(1/(x^4 + x^3 + x^2 + x + 1),1,2)
ans =
-1/10*(5^(1/2)*log(6-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^...
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>> syms x
>> int(1/(x^4 + x^3 + x^2 + x + 1),1,2)
ans =
-1/10*(5^(1/2)*log(6-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*(10-2*5^(1/2))^(1/2)+2*atan(9/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10-2*5^(1/2))^(1/2)-5^(1/2)*log(6+5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)-10*atan(9/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*(10+2*5^(1/2))^(1/2)-2*atan(9/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10+2*5^(1/2))^(1/2)-5^(1/2)*log(5-5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)+10*atan(5/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*(10-2*5^(1/2))^(1/2)-2*atan(5/(10+2*5^(1/2))^(1/2)-1/(10+2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10-2*5^(1/2))^(1/2)+5^(1/2)*log(5+5^(1/2))*(10+2*5^(1/2))^(1/2)*(10-2*5^(1/2))^(1/2)+10*atan(5/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*(10+2*5^(1/2))^(1/2)+2*atan(5/(10-2*5^(1/2))^(1/2)+1/(10-2*5^(1/2))^(1/2)*5^(1/2))*5^(1/2)*(10+2*5^(1/2))^(1/2))/(10+2*5^(1/2))^(1/2)/(10-2*5^(1/2))^(1/2)
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