The area bounded by the curves y=(x-1)^2,y=(x | vedclass

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(A) The curves are $y=(x-1)^2$,$y=(x+1)^2$,and the line $y=\frac{1}{4}$.By symmetry,the area is twice the area in the first quadrant bounded by $y=(x-

Vedclass · Accepted Answer

$\frac{1}{3}$ sq. units.

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(A) The curves are $y=(x-1)^2$,$y=(x+1)^2$,and the line $y=\frac{1}{4}$.By symmetry,the area is twice the area in the first quadrant bounded by $y=(x-1)^2$ and $y=\frac{1}{4}$ from $x=0$ to $x=\frac{1}{2}$.$	ext{Required Area} = 2 \int_0^{\frac{1}{2}} \left[ \frac{1}{4} - (x-1)^2 ight] dx$ is incorrect based on the graph; the region is bounded above by the curves and below by $y=\frac{1}{4}$.Correct setup: The region is bounded by $y=(x+1)^2$ and $y=(x-1)^2$ meeting at $x=0$ and the line $y=\frac{1}{4}$.Area $= 2 \int_0^{1/2} [y_{upper} - y_{lower}] dx = 2 \int_0^{1/2} [(x+1)^2 - (x-1)^2] dx$ is not correct. Looking at the graph,the area is bounded by $y=(x-1)^2$ and $y=(x+1)^2$ and $y=1/4$.Area $= 2 \int_0^{1/2} [y_{upper} - y_{lower}] dx = 2 \int_0^{1/2} [	ext{curve} - 1/4] dx$.Specifically,for $x \in [0, 1/2]$,the curve is $y=(x-1)^2$.Area $= 2 \int_0^{1/2} [(x-1)^2 - 1/4] dx = 2 \left[ \frac{(x-1)^3}{3} - \frac{x}{4} ight]_0^{1/2} = 2 \left[ (\frac{(-1/2)^3}{3} - \frac{1/2}{4}) - (\frac{(-1)^3}{3} - 0) ight] = 2 \left[ -\frac{1}{24} - \frac{1}{8} + \frac{1}{3} ight] = 2 \left[ \frac{-1-3+8}{24} ight] = 2 \left[ \frac{4}{24} ight] = \frac{1}{3} 	ext{ sq. units.}$

The area bounded by the curves $y=(x-1)^2$,$y=(x+1)^2$ and $y=\frac{1}{4}$ is

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