°定理1(积分上限函数的导数) 如果函数/(x)在区间ab上连续,则函数(x)=Cf(x)dx 在[a,b]上可导,并且 d(x)= dx f(t)dt=f(x)(a≤x≤b) 简要证明取Ax使x+△x∈(a,b) x+△x △=x+△x)-(x)=f()t-f(tr f Px+△x (t+.(t-f(0t x+△r r f(dt=f(sAx 其中生在x与x+△x之间.因为Ax→>0时,xx,所以 d(x)=lim Ad=lim f(5)=lim f(s)=f(x) △x→>0△X△x->0 2→>X 上页 下页
上页 返回 下页 简要证明 其中x在x与x+Dx之间. 因为Dx→0时, x→x, 所以 取Dx使x+Dx(a, b). •定理1(积分上限函数的导数) 如果函数 f(x)在区间[a, b]上连续, 则函数 x f x dx x a ( ) ( ) = 在[a, b]上可导, 并且 ( ) f (t)dt f (x)(a x b) dx d x x a = = . D=(x+Dx)−(x) f t dt f t dt x a x x a ( ) ( ) = − +D f t dt f t dt f t dt x a x x x x a ( ) ( ) ( ) = + − +D f t dt f x x x x = = D +D ( ) (x) , D=(x+Dx)−(x) f t dt f t dt x a x x a ( ) ( ) = − +D f t dt f x x x x = = D +D ( ) (x) , 返回 (x) lim lim ( ) lim ( ) ( ) 0 0 f f f x x x x x = = = D D = D → D → → x x x (x) lim lim ( ) lim ( ) ( ) . 0 0 f f f x x x x x = = = D D = D → D → → x x x (x) lim lim ( ) lim ( ) ( ) . 0 0 f f f x x x x x = = = D D = D → D → → x x x (x) lim lim ( ) lim ( ) ( ) . 0 0 f f f x x x x x = = = D D = D → D → → x x x
