设f(x)在[a，b]有连续的导数，求证：
【正确答案】正确答案：可设
|f(x)|=|f(x
0
)|，即证 (b－a)|f(x
0
)|≤|∫
a
b
f(x)dx|+(b－a)∫
a
b
|f'(x)|dx， 即证|∫
a
b
f(x
0
)dx|－|∫
a
b
f(x)dx|≤6(b－a)∫
a
b
|f'(x)|dx． 注意|∫
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
0
)|，即证 (b－a)|f(x
0
)|≤|∫
a
b
f(x)dx|+(b－a)∫
a
b
|f'(x)|dx， 即证|∫
a
b
f(x
0
)dx|－|∫
a
b
f(x)dx|≤6(b－a)∫
a
b
|f'(x)|dx． 注意|∫
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】