The area of the region bounded by the curves \( y=1+|x-1|\ ,x=-1\ ,x=2\ ,y=0 \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{11}{2} \) square units
The modulus changes form at x = 1:
For x < 1: y = 1 + (1 − x) = 2 − x
For x ≥ 1: y = 1 + (x − 1) = x
The curve is above the x-axis throughout, so
\( \displaystyle \text{Area}=\int_{-1}^{1}(2-x)\,dx+\int_{1}^{2}x\,dx \)
\( \displaystyle =\left[2x-\frac{x^2}{2}\right]_{-1}^{1}+\left[\frac{x^2}{2}\right]_{1}^{2} \)
\( \displaystyle =4+\frac{3}{2} \)
\( \displaystyle =\frac{11}{2}\ \text{square units} \)
Shortcut: The region is made of two trapeziums, so the area can be found without integration.
From x = −1 to x = 1, y = 2 − x goes from 3 down to 1. Area = ½ × (3 + 1) × 2 = 4.
From x = 1 to x = 2, y = x goes from 1 up to 2. Area = ½ × (1 + 2) × 1 = 3/2.
Total area = 4 + 3/2 = 11/2 square units.