注意到
∫[0,1] f(x)dx是一个定值,设∫[0,1] f(x)dx=B
∫[0,2] f(x)dx是一个定值,设∫[0,2] f(x)dx=A
f(x)=x^2-Ax+2B
两边求定积分得
B=∫[0,1] f(x)dx
=∫[0,1] (x^2-Ax+2B)dx
=(x^3/3-Ax^2/2+2Bx)[0,1]
=1/3-A/2+2B
=B
即
1/3-A/2+B=0 (1)
A=∫[0,2] f(x)dx
=∫[0,2] (x^2-Ax+2B)dx
=(x^3/3-Ax^2/2+2Bx)[0,2]
=8/3-2A+4B
8/3-3A+4B=0 (2)
(1)(2)解得
A=4/3,B=1/3
f(x)=x^2-4/3x+2/3
再问: 谢谢大神相助!!!请问第二和第三题如何