Question:

The area bounded by the parabolas $y = (x+1)^2$ and $y = (x-1)^2$ and the line $y = 1/4$ is determined by finding the region enclosed by these curves. First, find the intersection points of the parabolas with the line $y=1/4$: For $y=(x+1)^2$: $$ (x+1)^2 = \frac{1}{4} \implies x+1 = \pm\frac{1}{2} $$ $$ x = -\frac{1}{2} \quad \text{or} \quad x = -\frac{3}{2} $$ For $y=(x-1)^2$: $$ (x-1)^2 = \frac{1}{4} \implies x-1 = \pm\frac{1}{2} $$ $$ x = \frac{1}{2} \quad \text{or} \quad x = \frac{3}{2} $$ The two parabolas intersect each other at: $$ (x+1)^2 = (x-1)^2 \implies x^2+2x+1 = x^2-2x+1 \implies 4x=0 \implies x=0 $$ At $x=0$, $y=(0+1)^2=1$. So, the parabolas intersect at $(0,1)$. The problem asks for the area bounded by "the parabolas" and "the line $y=1/4$". Given the options, this typically refers to the area of one of the parabolic segments formed by the intersection of a parabola and the line $y=1/4$. The two parabolas are symmetric with respect to the y-axis, so the area of the segment formed by $y=(x+1)^2$ and $y=1/4$ is equal to the area of the segment formed by $y=(x-1)^2$ and $y=1/4$. Consider the region bounded by $y=(x-1)^2$ and $y=1/4$. For this region, the line $y=1/4$ is the upper boundary, and the parabola $y=(x-1)^2$ is the lower boundary. This region exists for $x$ values where $(x-1)^2 \le 1/4$, which means $1/2 \le x \le 3/2$. The area $A$ is given by the integral: $$ A = \int_{1/2}^{3/2} \left(\frac{1}{4} - (x-1)^2\right) dx $$ To simplify the integration, let $u = x-1$, so $du = dx$. When $x=1/2$, $u = 1/2 - 1 = -1/2$. When $x=3/2$, $u = 3/2 - 1 = 1/2$. The integral becomes: $$ A = \int_{-1/2}^{1/2} \left(\frac{1}{4} - u^2\right) du $$ Since the integrand $\left(\frac{1}{4} - u^2\right)$ is an even function, we can write: $$ A = 2 \int_{0}^{1/2} \left(\frac{1}{4} - u^2\right) du $$ Now, evaluate the integral: $$ A = 2 \left[\frac{1}{4}u - \frac{u^3}{3}\right]_{0}^{1/2} $$ $$ A = 2 \left[\left(\frac{1}{4}\cdot\frac{1}{2} - \frac{(1/2)^3}{3}\right) - \left(\frac{1}{4}\cdot 0 - \frac{0^3}{3}\right)\right] $$ $$ A = 2 \left[\frac{1}{8} - \frac{1/8}{3}\right] $$ $$ A = 2 \left[\frac{1}{8} - \frac{1}{24}\right] $$ $$ A = 2 \left[\frac{3}{24} - \frac{1}{24}\right] $$ $$ A = 2 \left[\frac{2}{24}\right] $$ $$ A = 2 \left[\frac{1}{12}\right] $$ $$ A = \frac{1}{6} $$