The area bounded by curves \( y=-\displaystyle\frac{2}{3}x+2,\ x=-1,\ x=2 \) and the \( x \)-axis is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 square units
The line meets the x-axis at x = 3, so it is above the x-axis on [−1, 2] and the area is the integral itself:
\( \displaystyle \int_{-1}^{2}\left(-\frac{2}{3}x+2\right)dx=\left[-\frac{x^2}{3}+2x\right]_{-1}^{2} \)
\( \displaystyle =\frac{8}{3}-\left(-\frac{7}{3}\right) \)
\( \displaystyle =5 \)
Shortcut: The region is a trapezium with parallel sides \( y(-1)=\frac{8}{3} \) and \( y(2)=\frac{2}{3} \) and width 3:
\( \displaystyle \text{Area}=\frac{1}{2}\times\left(\frac{8}{3}+\frac{2}{3}\right)\times 3 \)
\( \displaystyle =5 \)