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The value of $\int\limits_{1}^{e} \log x \, dx$ is: |
$0$ $1$ $e$ $e \log e$ |
$1$ |
The correct answer is Option (2) → 1 $I = \int\limits_{1}^{e} \log x \, dx$ $= [x \log x - x]_1^e$ $= [e \log e - e] - [1 \log 1 - 1]$ $=e\log e-e-\log 1+1$ {Since $\log e = 1, \log 1 = 0$} $= e - e - 0 + 1$ $I = 1$ |
