_1^2 x ,dx = - Vidyadip

MCQ

$\int_1^2 {\log x\,dx} =$

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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