设连续函数f﹙x﹚满足lim﹙x→0﹚f﹙x﹚/x=2 ,令F﹙x﹚=∫﹙0,1﹚f﹙xt﹚dt ,则F′﹙0﹚= __
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设连续函数f﹙x﹚满足lim﹙x→0﹚f﹙x﹚/x=2 ,令F﹙x﹚=∫﹙0,1﹚f﹙xt﹚dt ,则F′﹙0﹚= _______
lim(x→0) ƒ(x)/x = 2
F(x) = ∫(0→1) ƒ(xt) dt
令u = xt,du = x dt
t = 0,u = 0
t = 1,u = x
F(x) = ∫(0→x) ƒ(u) * (1/x du)
F(x) = (1/x)∫(0→x) ƒ(u) du
F'(x) = (1/x)ƒ(x) - (1/x²)∫(0→x) ƒ(u) du
F'(0) = lim(x→0) ƒ(x)/x - lim(x→0) [∫(0→x) ƒ(u) du]/x²
= 2 - lim(x→0) ƒ(x)/(2x)
= 2 - (1/2)(2)
= 1
F(x) = ∫(0→1) ƒ(xt) dt
令u = xt,du = x dt
t = 0,u = 0
t = 1,u = x
F(x) = ∫(0→x) ƒ(u) * (1/x du)
F(x) = (1/x)∫(0→x) ƒ(u) du
F'(x) = (1/x)ƒ(x) - (1/x²)∫(0→x) ƒ(u) du
F'(0) = lim(x→0) ƒ(x)/x - lim(x→0) [∫(0→x) ƒ(u) du]/x²
= 2 - lim(x→0) ƒ(x)/(2x)
= 2 - (1/2)(2)
= 1
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