Question

∫-2~2[(x^3cos(x/2)+1/2)根号下(4-x^2)]dx

∫-2~2[(x^3cos(x/2)+1/2)根号下(4-x^2)]dx求问一道定积分题，如题，感激大佬们！

Answer 1

f(x)=cosx(sinxcosπ/3+cosxsinπ/3)-√3cos2x+√3/4 =(1/2)sinxcosx+(√3/2)cos2x-√3cos2x+√3/4 =(1/2)sinxcosx-(√3/2)cos2x+√3/4 =(1/4)sin2x-(√3/4)cos2x =(1/2)sin(2x-π/3) 所以T=π x∈[-π/4,π/4]时，则2x-π/3∈[-5π/6,π/6] 则sin(2x-π/3)∈[-1, 1/2] 所以y∈[-1/2, 1/4] 所以函数最大值为1/4，最小值为-1/2.

Answer 2

>> syms x
>> Fx=((x^3)*cos((1/2)*x)+1/2)*sqrt(4-x^2)
Fx =
(4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2)
>> FUN=int(Fx,x,-2,2)
Warning: Explicit integral could not be found.
FUN =
int((4 - x^2)^(1/2)*(x^3*cos(x/2) + 1/2), x = -2..2)
%八位精度
>> vpa(FUN,8)
ans =
3.1415927
%二十位精度
>> vpa(FUN,20)
ans =
3.1415926535897932385
-----------------------------------------------------------
积出来应该是π

Answer 3

答案是π圆周率3.1415926..

Answer 4

图中博主的答题方式可以，但是∫（-2，2）是不可以变为∫（0，2）这样是会改变积分的值的，总结来说，我主答题有一点点小错误望采纳，如有错误可以纠正