Area bounded by the curve y=log x , the x-axis, x = 1 and x = 2 in sq…

The correct answer is Option (2) → \(\log\left(\displaystyle\frac{4}{e}\right)\)

For 1 ≤ x ≤ 2, \(\log x \ge 0\), so the curve lies on or above the x-axis and

    \( \displaystyle \text{Area} = \int_{1}^{2}\log x\,dx \)

Integrating by parts, with \(\log x\) as the first function and 1 as the second:

    \( \displaystyle \int\log x\,dx = x\log x - \int x \cdot \frac{1}{x}\,dx = x\log x - x \)

So

    \( \displaystyle \text{Area} = \left[x\log x - x\right]_{1}^{2} \)

    \(= (2\log 2 - 2) - (0 - 1)\)
    \(= 2\log 2 - 1\)

Since \(2\log 2 = \log 4\) and \(1 = \log e\):

    \( \displaystyle \log 4 - \log e = \log\left(\frac{4}{e}\right) \)