The value of \(\int_{1}^{e}\displaystyle\left(\log_e x\right)^{2}dx\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(e-2\)
Using integration by parts with \(\left(\log_e x\right)^{2}\) as the first function and 1 as the second:
\( \displaystyle \int\left(\log_e x\right)^{2}dx = x\left(\log_e x\right)^{2} - \int x\cdot\frac{2\log_e x}{x}\,dx \)
\( \displaystyle = x\left(\log_e x\right)^{2} - 2\int\log_e x\,dx \)
\( \displaystyle = x\left(\log_e x\right)^{2} - 2\left(x\log_e x - x\right) \)
At \(x = e\): \(e(1) - 2(e - e) = e\).
At \(x = 1\): \(0 - 2(0 - 1) = 2\).
\( \displaystyle \int_{1}^{e}\left(\log_e x\right)^{2}dx = e - 2 \)
Why the other options are incorrect
Option (1) e — Incorrect. It is only the value at the upper limit; the value 2 at \(x = 1\) is not subtracted.
Option (2) \(e-1\) — Incorrect. The value at \(x = 1\) is 2, not 1.
Option (3) \(2-e\) — Incorrect. This is negative, but \(\left(\log_e x\right)^{2} \ge 0\), so the integral cannot be negative.