Which of the following statements are correct?
(A) \(\int_{0}^{2}\displaystyle\max\{x, 1\}\,dx = \frac{5}{2}\)
(B) \(\int_{0}^{2}\displaystyle\min\{x, 1\}\,dx = \frac{3}{2}\)
(C) \(\int_{0}^{\frac{\pi}{2}}\displaystyle\max\{\sin x, \cos x\}\,dx = \sqrt{2}\)
(D) \(\int_{-1}^{1}\displaystyle\min\{x, 0\}\,dx = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Each integral is split at the point where the two functions are equal, and on each part the larger (or smaller) one is integrated.
For (A) and (B), the crossing point is \(x = 1\); \(x \lt 1\) on [0, 1) and \(x \gt 1\) on (1, 2]:
\( \displaystyle \int_{0}^{2}\max\{x,1\}\,dx = \int_{0}^{1}1\,dx + \int_{1}^{2}x\,dx = 1 + \frac{3}{2} = \frac{5}{2} \)
\( \displaystyle \int_{0}^{2}\min\{x,1\}\,dx = \int_{0}^{1}x\,dx + \int_{1}^{2}1\,dx = \frac{1}{2} + 1 = \frac{3}{2} \)
For (C), \(\cos x \ge \sin x\) on \([0, \frac{\pi}{4}]\) and \(\sin x \ge \cos x\) on \([\frac{\pi}{4}, \frac{\pi}{2}]\):
\( \displaystyle \int_{0}^{\pi/4}\cos x\,dx + \int_{\pi/4}^{\pi/2}\sin x\,dx = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \)
For (D), \(\min\{x, 0\} = x\) on [−1, 0] and 0 on [0, 1]:
\( \displaystyle \int_{-1}^{0}x\,dx + 0 = -\frac{1}{2} \)
So the values are (A) \(\frac{5}{2}\), (B) \(\frac{3}{2}\), (C) \(\sqrt{2}\) and (D) \(-\frac{1}{2}\).
Checking the statements
(A) \(\int_{0}^{2}\max\{x, 1\}\,dx = \frac{5}{2}\) — Correct. \(1 + \frac{3}{2} = \frac{5}{2}\).
(B) \(\int_{0}^{2}\min\{x, 1\}\,dx = \frac{3}{2}\) — Correct. \(\frac{1}{2} + 1 = \frac{3}{2}\).
(C) \(\int_{0}^{\pi/2}\max\{\sin x, \cos x\}\,dx = \sqrt{2}\) — Correct. \(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\).
(D) \(\int_{-1}^{1}\min\{x, 0\}\,dx = 0\) — Incorrect. Only \(\int_{-1}^{0}x\,dx = -\frac{1}{2}\) remains.