The area bounded by the curves $y = \left(x - 1\right)^2$, $y = \left(x + 1\right)^2$ and $y = \frac{1}{8}$ is
Step-by-Step Solution
Key Concept: Find intersection points of curves and integrate the difference between upper and lower bounds.
The curves $y = (x-1)^2$, $y = (x+1)^2$, and $y = \frac{1}{4}$ are analyzed to find intersection points. Setting $(x-1)^2 = \frac{1}{4}$ gives $x = \frac{1}{2}$, and setting $(x+1)^2 = \frac{1}{4}$ gives $x = -\frac{3}{2}$. The required area is calculated as $2\int_{-3/2}^{1/2}[(x-1)^2 - \frac{1}{4}]dx = 2[\frac{(x-1)^3}{3} - \frac{x}{4}]_{-3/2}^{1/2} = 2[-\frac{3}{8} - (-\frac{1}{4})] = \frac{3}{4}$ sq unit.
Correct Answer: 2