Area bounded by the curves
\( y=\begin{cases}x+2 & ,x\ge -1\\-x & ,x\lt -1\end{cases},x=-2,x=3 \) and \( y=0 \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{27}{2} \) Square units
On [−2, 3] the curve lies above the x-axis (at x = −2, y = 2; at x = −1, y = 1; at x = 3, y = 5), so the area is the integral of y, split at x = −1.
\( \displaystyle \text{Area}=\int_{-2}^{-1}\displaystyle(-x)\,dx+\int_{-1}^{3}\displaystyle(x+2)\,dx \)
\( \displaystyle =\left[-\frac{x^2}{2}\right]_{-2}^{-1}+\left[\frac{x^2}{2}+2x\right]_{-1}^{3} \)
\( \displaystyle =\left(-\frac{1}{2}+2\right)+\left(\left(\frac{9}{2}+6\right)-\left(\frac{1}{2}-2\right)\right) \)
\( \displaystyle =\frac{3}{2}+12 \)
\( \displaystyle =\frac{27}{2} \)
Shortcut: Both pieces are trapeziums: from −2 to −1 the parallel sides are 2 and 1 with width 1, area \( \frac{3}{2} \); from −1 to 3 they are 1 and 5 with width 4, area 12. Total \( \frac{27}{2} \).