$\int_{1}^{e} \log x d x =$

(a) 1
(b) e − 1
(c) e + 1
(d) 0

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Solution

(a) 1

$\int_{1}^{e} \log x d x = \int_{1}^{e} \log x x^{0} d x = {[x \log x]}_{1}^{e} - \int_{1}^{e} \frac{1}{x} x d x = {[x \log x]}_{1}^{e} - {[x]}_{1}^{e} = (e - 0) - (e - 1) = e - e + 1 = 1$

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