Question

求解一道高数题？

Answer 1

f(x) = x(1-x)^5 +(1/2)∫(0->1) f(x) dx
let
C =(1/2)∫(0->1) f(x) dx
f(x)
= x(1-x)^5 +(1/2)∫(0->1) f(x) dx
= x(1-x)^5 +(1/2)∫(0->1) [ x(1-x)^5 +C] dx
= x(1-x)^5 +(1/2)[Cx]|(0->1) +(1/2)∫(0->1) x(1-x)^5 dx
= x(1-x)^5 +(1/2)C -(1/12)∫(0->1) xd(1-x)^6
= x(1-x)^5 +(1/2)C -(1/12)[x.(1-x)^6]| (0->1) + (1/12)∫(0->1) (1-x)^6 dx
= x(1-x)^5 +(1/2)C -0 - (1/84)[(1-x)^7]| (0->1)
= x(1-x)^5 +(1/2)C + 1/84
x(1-x)^5 +C = x(1-x)^5 +(1/2)C + 1/84
(1/2)C =1/84
C =1/42
ie
f(x) = x(1-x)^5 +1/42

追问

谢谢

Answer 2