Let \(f(x) = \begin{cases} x+2, & -2 \le x \le 0 \\ \sqrt{4-x^{2}}, & 0 \lt x \le 2 \end{cases}\). The area (in sq. units) of the region bounded by the curve y = f(x) and the x-axis is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2+\pi\)
Both pieces are on or above the x-axis and the curve meets the x-axis at x = −2 and x = 2, so the area is the integral of f from −2 to 2, split at x = 0.
The line piece:
\( \displaystyle \int_{-2}^{0}(x+2)\,dx = \left[\frac{x^{2}}{2}+2x\right]_{-2}^{0} = 0-(2-4) = 2 \)
The circle piece is the first-quadrant quarter of the disc \(x^{2}+y^{2} \le 4\):
\( \displaystyle \int_{0}^{2}\sqrt{4-x^{2}}\,dx = \frac{1}{4}\pi(2)^{2} = \pi \)
\(A = 2+\pi\)