The value of _1^2 x d x is - Vidyadip

MCQ

The value of $\int_1^2 \log x d x$ is

A
$\log \left(\frac{2}{e}\right)$
B
$\log 4$
✓
$\log \left(\frac{4}{e}\right)$
D
$\log 2$

✓

Answer

Correct option: C.

$\log \left(\frac{4}{e}\right)$

(C)
$\int_1^2 \log x d x=[x \log x-x]_1^2$
$\begin{array}{l}=2 \log 2-2+1 \\ =\log 4-1=\log 4-\log e=\log \left(\frac{4}{e}\right)\end{array}$

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