Let \(f(x) = \begin{cases} x^{2}, & x \lt 0 \\ 2x, & x \ge 0 \end{cases}\). The area (in sq. units) of the region bounded by the curve y = f(x), the x-axis and the lines x = −2 and x = 3 is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{35}{3}\)
The function has a different formula on each side of x = 0, so the area is split at x = 0. On the left \(x^{2} \ge 0\), and on the right \(2x \ge 0\), so both parts lie above the x-axis:
\( \displaystyle A = \int_{-2}^{0}x^{2}\,dx+\int_{0}^{3}2x\,dx \)
\( \displaystyle = \left[\frac{x^{3}}{3}\right]_{-2}^{0}+\left[x^{2}\right]_{0}^{3} \)
\( \displaystyle = \frac{8}{3}+9 = \frac{35}{3} \)