Let \(f(x) = \begin{cases} x^{2}, & 0 \le x \le 1 \\ 2x-1, & 1 \lt x \le 2 \end{cases}\)
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Definite integral | Value | ||
| (A) | \(\int_{0}^{2}\displaystyle|x-1|\,dx\) | (I) | 2 |
| (B) | \(\int_{-2}^{1}\displaystyle|x|\,dx\) | (II) | \(\displaystyle\frac{5}{2}\) |
| (C) | \(\int_{0}^{2}\displaystyle f(x)\,dx\) | (III) | 1 |
| (D) | \(\int_{0}^{\pi}\displaystyle|\cos x|\,dx\) | (IV) | \(\displaystyle\frac{7}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
Each integral is found by splitting the interval where the expression inside the modulus changes sign, or where the formula for f changes.
(A) \(\int_{0}^{2}|x-1|\,dx\)
This matches (III) 1:
\( \displaystyle \int_{0}^{1}(1-x)\,dx + \int_{1}^{2}(x-1)\,dx = \frac{1}{2} + \frac{1}{2} = 1 \)
(B) \(\int_{-2}^{1}|x|\,dx\)
This matches (II) \(\frac{5}{2}\):
\( \displaystyle \int_{-2}^{0}(-x)\,dx + \int_{0}^{1}x\,dx = 2 + \frac{1}{2} = \frac{5}{2} \)
(C) \(\int_{0}^{2}f(x)\,dx\)
This matches (IV) \(\frac{7}{3}\):
\( \displaystyle \int_{0}^{1}x^{2}\,dx + \int_{1}^{2}(2x-1)\,dx = \frac{1}{3} + \left[x^{2}-x\right]_{1}^{2} \)
\( \displaystyle = \frac{1}{3} + 2 = \frac{7}{3} \)
(D) \(\int_{0}^{\pi}|\cos x|\,dx\)
This matches (I) 2. Since \(\cos x \ge 0\) on \([0, \frac{\pi}{2}]\) and \(\cos x \le 0\) on \([\frac{\pi}{2}, \pi]\):
\( \displaystyle \int_{0}^{\pi/2}\cos x\,dx - \int_{\pi/2}^{\pi}\cos x\,dx = 1 - (-1) = 2 \)