定積分是常數(shù),所以設(shè)∫[0 1] f(x)dx=A
則
f(x)=e^x+2∫[0 1] f(x)dx
=e^x+2A
兩邊在區(qū)間[0,1]進行定積分得
∫[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)