设函数f(x)与g(x)在区间[a，b]上连续，证明：
[∫
a
b
f(x)g(x)dx]
2
≤∫
a
b
f
2
(x)dx∫
a
b
g
2
(x)dx． (*)
【正确答案】正确答案：把证明定积分不等式 (∫
a
b
f(x)g(x)dx)
2
≤∫
a
b
f
2
(x)dx∫
a
b
g
2
(x)dx (*) 转化为证明重积分不等式． 引入区域D={(x，y)｜a≤x≤b，a≤y≤b} (*)式左端=∫
a
b
f(x)g(x)dx.∫
a
b
f(y)g(y)dy =
[f(x)g(y).f(y)g(x)]dxdy≤
[f
2
(x)g
2
(y)+f
2
(y)g
2
(x)]dxdy =
f
2
(x)g
2
(y)dxdy+
f
2
(y)g
2
(x)dxdy =
∫
a
b
f
2
(x)dx∫
a
b
g
2
(y)dy+
[f(x)g(y).f(y)g(x)]dxdy≤
[f
2
(x)g
2
(y)+f
2
(y)g
2
(x)]dxdy =
f
2
(x)g
2
(y)dxdy+
f
2
(y)g
2
(x)dxdy =
∫
a
b
f
2
(x)dx∫
a
b
g
2
(y)dy+
【答案解析】