請(qǐng)問您的等式到底是什么?怎么等式里面還有區(qū)間?[0,x]∫f(t)dt的意思是積分的上下限是0到x首先移項(xiàng),{f'(x)+f(x)}*(x+1)=∫f(t)dt然后兩側(cè)求導(dǎo)｛f'‘(x)+f’(x)｝*(x+1)+f'(x)+f(x)=f(x)化簡上式f''(x)/f'(x)=-(x+2)/(x+1)解上述微分方程令f'(x)=p則有dp/p=-(x+2)/(x+1)*dx 化簡繼續(xù)dp/p=｛-1-1/(x+1)｝*dx兩側(cè)取積分Lnp=-x-Ln(x+1)-Lnc即：Lnp=Lne^(-x)-Ln(x+1)-Lnc則有p=e^(-x)/（c*(x+1))即f'(x)=1/（c*e^x*(x+1)）將x=0帶入已知等式得到f'(0)=-1得到c=-1即f'(x)=-1/（e^x*(x+1)）（2）顯然f'(x)恒小于0即f(x)為減函數(shù)f(x)max=f(0)=1即有f(x)<=1∫f'(x)dx=∫-e^(-x)*1/(x+1)dx 又∵-e^(-x)*1/(x+1)>=-e^(-x)(放縮)∴∫f'(x)dx=∫-e^(-x)*1/(x+1)dx>=∫-e^(-x)dx即f(x)>=∫-e^(-x)dx即f(x)>=e^(-x)到此證畢！望采納