The area of the region bounded by $y=-1,y=2,x=y^3$ and $x = 0$ is $\frac{m}{n}$ 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
Given: Curves and lines: $y = -1$, $y = 2$, $x = y^3$, and $x = 0$
Required Area:
$\int_{y = -1}^{2} (y^3 - 0)\,dy = \int_{-1}^{2} y^3\,dy$
$= \left[\frac{y^4}{4}\right]_{-1}^{2} = \left(\frac{2^4}{4} - \frac{(-1)^4}{4}\right)$
$= \left(\frac{16}{4} - \frac{1}{4}\right) = \frac{15}{4}$
Since $x = y^3$ lies on both sides of the y-axis in this interval, and $x = 0$ is the left boundary, area is computed as:
$\int_{-1}^{0} |y^3| dy + \int_{0}^{2} y^3 dy$
First integral: $\int_{-1}^{0} (-y^3)\,dy = \left[ -\frac{y^4}{4} \right]_{-1}^{0} = 0 - \left( -\frac{1}{4} \right) = \frac{1}{4}$
Second integral: $\int_{0}^{2} y^3\,dy = \left[ \frac{y^4}{4} \right]_0^2 = \frac{16}{4} = 4$
Total Area = $\frac{1}{4} + 4 = \frac{17}{4}$
So, $m = 17$, $n = 4$, and $m - n = 13$