wftree
幼苗
共回答了23个问题采纳率:95.7% 举报
∫a g(x)f’(x)dx+∫1 f(x)g’(x)dx
0 0
=∫a g(x)f’(x)dx+∫a f(x)g’(x)dx+∫1 f(x)g’(x)dx
0 0 a
=∫a(g(x)f’(x)+ f(x)g’(x))dx+∫1 f(x)g’(x)dx
0 a
=f(a)g(a)-f(0)g(0)+∫1 f(x)g’(x)dx
a
由在[0,1]上f'(x)≥0得,f(x)在[0,1]上单调递增,而f(0)=0
∴ 对于 0≤a≤x≤1,有 f(0)=0≤f(a)≤f(x)≤f(1),加上 g'(x)≥0
∫1 f(x)g’(x)dx ≥ ∫1 f(a)g’(x)dx =f(a)(g'(1)-g'(a))
a a
因此 ∫a g(x)f’(x)dx+∫1 f(x)g’(x)dx
0 0
≥f(a)g(a)-f(0)g(0)+f(a)(g(1)-g(a))
=f(a)g(1)
1年前
5