The area (in sq. units) of the region bounded by the curves y = |x −…

The correct answer is Option (3) → 12

The curve y = 5 − |x| has its corner at x = 0 and y = |x − 1| has its corner at x = 1. The curves meet as follows.

For \(x \ge 1\): \(x-1 = 5-x\), so x = 3 and y = 2.

For \(x \lt 0\): \(1-x = 5+x\), so x = −2 and y = 3.

For \(0 \le x \lt 1\): \(1-x = 5-x\) has no solution, and 5 − x is above 1 − x.

So the region runs from x = −2 to x = 3 with y = 5 − |x| on top, and the integral is split at both corners, x = 0 and x = 1.

For \(-2 \le x \le 0\) the height is \((5+x)-(1-x) = 4+2x\):

    \( \displaystyle \int_{-2}^{0}(4+2x)\,dx = \left[4x+x^{2}\right]_{-2}^{0} = 0-(-8+4) = 4 \)

For \(0 \le x \le 1\) the height is \((5-x)-(1-x) = 4\):

    \( \displaystyle \int_{0}^{1}4\,dx = 4 \)

For \(1 \le x \le 3\) the height is \((5-x)-(x-1) = 6-2x\):

    \( \displaystyle \int_{1}^{3}(6-2x)\,dx = \left[6x-x^{2}\right]_{1}^{3} = 9-5 = 4 \)

    \(A = 4+4+4 = 12\)

Why the other options are incorrect

• Option (1) 8 — Incorrect. This leaves out one of the three pieces, each of area 4; the middle strip from x = 0 to x = 1 is easy to miss.

• Option (2) 16 — Incorrect. The region has three pieces of area 4 each, not four.

• Option (4) 10 — Incorrect. The three pieces give 4 + 4 + 4 = 12.