Consider the region bounded by the line y = x − 1, the x-axis and the lines x = −1 and x = 3. Which of the following statements are correct?
(A) The line meets the x-axis at x = 1
(B) The area of the region is \(\int_{-1}^{3}\displaystyle (x-1)\,dx\)
(C) The area of the region is 4 sq. units
(D) The part of the region below the x-axis has a larger area than the part above the x-axis
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
The line y = x − 1 is zero at x = 1. It is below the x-axis for x < 1 and above it for x > 1, so the two parts are found separately and the part below is taken with its absolute value.
\( \displaystyle \int_{-1}^{1}(x-1)\,dx = \left[\frac{x^{2}}{2}-x\right]_{-1}^{1} = -\frac{1}{2}-\frac{3}{2} = -2 \)
\( \displaystyle \int_{1}^{3}(x-1)\,dx = \left[\frac{x^{2}}{2}-x\right]_{1}^{3} = \frac{3}{2}+\frac{1}{2} = 2 \)
So the single integral from −1 to 3 is
\(-2+2 = 0\)
while the area is
\(|-2|+2 = 4\)
So the line crosses the x-axis at x = 1, the integral from −1 to 3 is 0, the area is 4, and the parts below and above the x-axis have area 2 each.
Checking the statements
(A) The line meets the x-axis at x = 1 — Correct. y = 0 gives x = 1.
(B) The area of the region is \(\int_{-1}^{3}(x-1)\,dx\) — Incorrect. This integral is 0, but the area is 4.
(C) The area of the region is 4 sq. units — Correct. The two parts give 2 + 2 = 4.
(D) The part below the x-axis has a larger area than the part above it — Incorrect. Both parts have area 2.