Question
$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$
- 1
- e - 1
- e + 1
- 0
Solution:
$\int\limits^\text{e}_1\log\text{x}\text{ dx}$
$=\int\limits^\text{e}_1\log\text{x}\text{ x}^0\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\int\limits^\text{e}_1\frac{1}{\text{x}}\text{dx}$
$=\big[\text{x}\log\text{x}\big]^\text{e}_1-\big[\text{x}\big]^\text{e}_1$
$=(\text{e}-0)-(\text{e}-1)$
$= \text{e}-\text{e}+1$
$=1$
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| $X=x$ | 0 | 1 | 2 | 3 | 4 |
| $P(X=x)$ | $k$ | $2k$ | $4k$ | $2k$ | $k$ |
$($ where $det(B)$ denotes determinant of Matrix $B) -$