【導讀】用以證明連續(xù)函數(shù)的原函數(shù)的存在性.分法與分部積分法.[][],[]fa,bxa,bfa,x設在上可積,則在上??為變下限的定積分.為變上限的定.可積。由x的任意性,f在[a,b]上連續(xù).Δ0limΔ???續(xù)函數(shù)必存在原函數(shù)”這個重要結論.注2由于f的任意兩個原函數(shù)只能相差一個常數(shù),();xaFaCxb用代入,得再用代入,則得???定理設f在[a,b]上可積.若函數(shù)g在[a,b]上單調(diào)減,且,0)(?若函數(shù)g在[a,b]上單調(diào)增,且,0)(?對任意分割T:,10bxxxan????????11,()0,()()0.niiggxgxgx由對的假設記?????()0,()()d0,bagaIfxgxx若則此時任取????證若g為單調(diào)遞減函數(shù),()()(),hxgxgb??連續(xù),在上連續(xù)可微,且