Let \([x]\) denote the greatest integer less than or equal to x. Which of the following statements are correct?
(A) \(\int_{0}^{2}\displaystyle[x]\,dx = 1\)
(B) \(\int_{-1}^{1}\displaystyle[x]\,dx = 0\)
(C) \(\int_{0}^{3}\displaystyle\left(x-[x]\right)dx = \frac{3}{2}\)
(D) \(\int_{0}^{2}\displaystyle x[x]\,dx = \frac{3}{2}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
\([x]\) is constant between two consecutive integers, so each integral is split at the integers:
\( [x] = -1 \) on \([-1, 0)\), \( [x] = 0 \) on \([0, 1)\), \( [x] = 1 \) on \([1, 2)\), \( [x] = 2 \) on \([2, 3)\)
\( \displaystyle \int_{0}^{2}[x]\,dx = \int_{0}^{1}0\,dx + \int_{1}^{2}1\,dx = 1 \)
\( \displaystyle \int_{-1}^{1}[x]\,dx = \int_{-1}^{0}(-1)\,dx + \int_{0}^{1}0\,dx = -1 \)
\( \displaystyle \int_{0}^{3}[x]\,dx = 0 + 1 + 2 = 3 \), so \( \displaystyle \int_{0}^{3}\left(x-[x]\right)dx = \frac{9}{2} - 3 = \frac{3}{2} \)
\( \displaystyle \int_{0}^{2}x[x]\,dx = \int_{0}^{1}0\,dx + \int_{1}^{2}x\,dx = \frac{4-1}{2} = \frac{3}{2} \)
So the values are (A) 1, (B) −1, (C) \(\frac{3}{2}\) and (D) \(\frac{3}{2}\).
Checking the statements
(A) \(\int_{0}^{2}[x]\,dx = 1\) — Correct. Only the part from 1 to 2 counts, with \([x] = 1\).
(B) \(\int_{-1}^{1}[x]\,dx = 0\) — Incorrect. \([x] = -1\) on \([-1, 0)\), so the value is \(-1\).
(C) \(\int_{0}^{3}\left(x-[x]\right)dx = \frac{3}{2}\) — Correct. \(\frac{9}{2} - 3 = \frac{3}{2}\).
(D) \(\int_{0}^{2}x[x]\,dx = \frac{3}{2}\) — Correct. \(\int_{1}^{2}x\,dx = \frac{3}{2}\).