The area of the region bounded by the curves \(y=x^2+2\), \(y=-x\), \(x=0\) and \(x=1\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{17}{6}\)
For \(0 \le x \le 1\), the parabola \(y = x^2 + 2\) is at least 2, and the line \(y = -x\) is at most 0. So the parabola is the upper curve and the line the lower curve on the whole interval:
\( \displaystyle \text{Area} = \int_{0}^{1}\left[(x^2 + 2) - (-x)\right]dx \)
\( \displaystyle = \left[\frac{x^3}{3} + 2x + \frac{x^2}{2}\right]_{0}^{1} \)
\( \displaystyle = \frac{1}{3} + 2 + \frac{1}{2} = \frac{17}{6} \)