The area of the region bounded by the curves \(y = x^2+2,y=x,x=0\) and \(x = 2\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{14}{3}\) Sq. units
The parabola lies above the line for every x, because
\(\displaystyle x^2 + 2 - x = \left(x - \frac{1}{2}\right)^2 + \frac{7}{4} \gt 0\)
So the area between x = 0 and x = 2 is the integral of upper curve minus lower curve:
\(\displaystyle A = \int_{0}^{2}\left[(x^2 + 2) - x\right]dx\)
\(\displaystyle = \left[\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_{0}^{2}\)
\(\displaystyle = \frac{8}{3} - 2 + 4 = \frac{14}{3}\) sq. units
Why the other options are incorrect
Option (1) \(\frac{11}{3}\) Sq. units — Incorrect. The area is \(\frac{14}{3}\), not \(\frac{11}{3}\).
Option (2) \(\frac{2}{3}\) Sq. units — Incorrect. This is \(\int_{0}^{2}(x^2 - x)\,dx = \frac{8}{3} - 2\), with the constant 2 of the parabola left out.
Option (4) \(\frac{26}{3}\) Sq. units — Incorrect. This adds the two areas under the curves, \(\frac{20}{3} + 2\), instead of subtracting them.