^e_1 dx= 1. 1 2. e - 1 3. e + 1 4. 0 - Vidyadip

Question

$\int\limits^\text{e}_1\log\text{x}\text{ dx}=$

1
e - 1
e + 1
0

✓

Answer

$1$

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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