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

The correct answer is Option (3) → 12

The line is \(y = \displaystyle\frac{x+6}{2}\). It meets the two arms of y = |x| as follows.

For \(x \ge 0\), y = x:

    \( \displaystyle x = \frac{x+6}{2},\ \ x = 6 \)

For \(x \lt 0\), y = −x:

    \( \displaystyle -x = \frac{x+6}{2},\ \ x = -2 \)

Between x = −2 and x = 6 the line lies above y = |x|. Since |x| changes its form at x = 0, the integral is split at x = 0.

Left part (\(-2 \le x \le 0\)):

    \( \displaystyle \int_{-2}^{0}\left(\frac{x+6}{2}+x\right)dx = \int_{-2}^{0}\left(\frac{3x}{2}+3\right)dx \)

    \( \displaystyle = \left[\frac{3x^{2}}{4}+3x\right]_{-2}^{0} = 0-(3-6) = 3 \)

Right part (\(0 \le x \le 6\)):

    \( \displaystyle \int_{0}^{6}\left(\frac{x+6}{2}-x\right)dx = \int_{0}^{6}\left(3-\frac{x}{2}\right)dx \)

    \( \displaystyle = \left[3x-\frac{x^{2}}{4}\right]_{0}^{6} = 18-9 = 9 \)

    \(A = 3+9 = 12\)

Why the other options are incorrect

• Option (1) 9 — Incorrect. This is only the part to the right of the y-axis; the region also has a part of area 3 on the left.

• Option (2) 15 — Incorrect. The two parts add up to 3 + 9 = 12, not 15.

• Option (4) 3 — Incorrect. This is only the part to the left of the y-axis.