定积分是常数,所以设∫[0 1] f(x)dx=A
则
f(x)=e^x+2∫[0 1] f(x)dx
=e^x+2A
两边在区间[0,1]进行定积分得
∫[0 1] f(x)dx=∫[0 1] (e^x+2A)dx
A=∫[0 1] (e^x+2A)dx
=(e^x+2Ax)[0 1]
=e+2A-1
A=1-e
f(x)=e^x+2(1-e)