The area (in sq. units) bounded by the curves | vedclass

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(C) The curves are $y=(x+1)^2$ and $y=(x-1)^2$. The line is $y=\frac{1}{4}$.To find the intersection of $y=(x-1)^2$ and $y=\frac{1}{4}$,we set $(x-1)^

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$\frac{1}{3}$

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(C) The curves are $y=(x+1)^2$ and $y=(x-1)^2$. The line is $y=\frac{1}{4}$.To find the intersection of $y=(x-1)^2$ and $y=\frac{1}{4}$,we set $(x-1)^2 = \frac{1}{4}$,which gives $x-1 = \pm \frac{1}{2}$,so $x = \frac{1}{2}$ or $x = \frac{3}{2}$.Similarly,for $y=(x+1)^2$ and $y=\frac{1}{4}$,we get $x = -\frac{1}{2}$ or $x = -\frac{3}{2}$.The region is symmetric about the $y$-axis. The bounded region is between $x = -\frac{1}{2}$ and $x = \frac{1}{2}$.For $x \in [0, \frac{1}{2}]$,the upper boundary is $y = \min((x+1)^2, (x-1)^2)$ and the lower boundary is $y = \frac{1}{4}$.Specifically,for $x \in [0, \frac{1}{2}]$,the curve $y=(x-1)^2$ is the upper boundary.$	ext{Required Area} = 2 \int_0^{\frac{1}{2}} \left[ (x-1)^2 - \frac{1}{4} ight] dx$$= 2 \left[ \frac{(x-1)^3}{3} - \frac{x}{4} ight]_0^{\frac{1}{2}}$$= 2 \left[ \left( \frac{(-1/2)^3}{3} - \frac{1/2}{4} ight) - \left( \frac{(-1)^3}{3} - 0 ight) ight]$$= 2 \left[ \left( -\frac{1}{24} - \frac{1}{8} ight) - \left( -\frac{1}{3} ight) ight]$$= 2 \left[ -\frac{4}{24} + \frac{1}{3} ight] = 2 \left[ -\frac{1}{6} + \frac{2}{6} ight] = 2 \left( \frac{1}{6} ight) = \frac{1}{3} 	ext{ sq. units.}$

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

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