The area of the region bounded by \(y=-1, y=2, x=y^3\) and \(x = 0\) is \(\displaystyle\frac{m}{n}\) sq. units, where gcd \((m, n) =1\), then \(m -n\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 13
The region lies between the curve \(x = y^3\) and the y-axis, for y from −1 to 2. For \(y \lt 0\), \(x = y^3\) is negative, so that part lies to the left of the y-axis. The area is the integral of |x| with respect to y:
\(\displaystyle \text{Area} = \int_{-1}^{0} (-y^3)\,dy + \int_{0}^{2} y^3\,dy\)
\(\displaystyle = \left[-\frac{y^4}{4}\right]_{-1}^{0} + \left[\frac{y^4}{4}\right]_{0}^{2}\)
\(\displaystyle = \frac{1}{4} + 4 = \frac{17}{4}\)
So \(m = 17\), \(n = 4\) (gcd 1), and
\(m - n = 13\)
Tip: \(\int_{-1}^{2} y^3\,dy = \frac{15}{4}\) would give 11 (Option (2)), because the part left of the y-axis comes out negative and cancels. For an area, take each part with a positive sign.