Question

高等数学定积分一题证明：设函数f(x)在区间[a,b]上连续,g(x)在[a,b...

高等数学定积分一题证明：设函数f(x)在区间[a,b]上连续,g(x)在[a,b]上连续且不变号,则在[a,b]存在一点E 使得∫(a,b)f(x)g(x)dx=f(e)∫(a,b)g(x)dx

Answer 1

函数f(x)在区间[a,b]上连续,所以有最大值与最小值,分别设为M,N.不妨设g(x)≥0N≤f(x)≤M
Ng(x)≤f(x)g(x)≤Mg(x)∫[a,b]
Ng(x)dx≤
∫[a,b]f(x)g(x)dx≤
∫[a,b]Mg(x)dxN∫[a,b]
g(x)dx≤
∫[a,b]f(x)g(x)dx≤
M∫[a,b]g(x)dxN≤
{∫[a,b]f(x)g(x)dx}/∫[a,b]g(x)dx≤
M,所以存在ξ∈[a,b],使得,f(ξ)={∫[a,b]f(x)g(x)dx}/∫[a,b]g(x)dx,即f(ξ)∫[a,b]g(x)dx,=∫[a,b]f(x)g(x)dx