设f(x)在[a，b]上有连续的导数，证明
【正确答案】正确答案：可设
｜f(x)｜=｜f(x)｜，即证 (b一a)｜f(x
0
)｜≤｜∫
a
b
f(x)｜+(b一a)∫
a
b
｜f"(x)｜dx， 即｜∫
a
b
f(x
0
)dx｜—｜∫
a
b
f(x)dx｜≤(b—a)∫
a
b
｜f"(x)｜dx． 事实上， ｜∫
a
b
f(x
0
)dx｜—｜∫
a
b
f(x)dx｜≤｜∫
a
b
[f(x
0
)—f(x)]dx｜ =｜∫
a
b
[
｜f(x)｜=｜f(x)｜，即证 (b一a)｜f(x
0
)｜≤｜∫
a
b
f(x)｜+(b一a)∫
a
b
｜f"(x)｜dx， 即｜∫
a
b
f(x
0
)dx｜—｜∫
a
b
f(x)dx｜≤(b—a)∫
a
b
｜f"(x)｜dx． 事实上， ｜∫
a
b
f(x
0
)dx｜—｜∫
a
b
f(x)dx｜≤｜∫
a
b
[f(x
0
)—f(x)]dx｜ =｜∫
a
b
[
【答案解析】