MCQ
The value of $\int_1^2 {\log x\,dx} $ is
  • A
    $\log 2/e$
  • B
    $\log 4$
  • $\log 4/e$
  • D
    $\log 2$

Answer

Correct option: C.
$\log 4/e$
c
(c) $\int_1^2 {\log xdx = [x\log x - x]_1^2 = 2\log 2 - 2 + 1} $

$ = \log 4 - 1 = \log 4 - \log e = \log \frac{4}{e}$.

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