题目内容
设函数y=f(x)在区间[0,2]上是连续函数,那么∫02f(x)dx( )
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试题答案
C
| A.∫01xdx+∫12f(x)dx | B.∫01f(t)dt+∫02f(x)dx |
| C.∫01f(t)dt+∫12f(x)dx | D.∫01f(x)dx+∫0.52f(x)dx |
| A.∫01xdx+∫12f(x)dx | B.∫01f(t)dt+∫02f(x)dx |
| C.∫01f(t)dt+∫12f(x)dx | D.∫01f(x)dx+∫0.52f(x)dx |
| x4 |
| 12 |
| mx3 |
| 6 |
| 3x2 |
| 2 |
| x4 |
| 12 |
| mx3 |
| 6 |
| 3x2 |
| 2 |
