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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Aicharya · Accepted Answer

$\begin{aligned} 	ext { Required area } & =2 \int_0^{\frac{1}{2}}\left[(x-1)^2-\frac{1}{4}ight] \mathrm{d} x \ & =2\left[\frac{(x-1)^3}{3}ight]_0^{\frac{1}{2}}-\frac{1}{2}[x]_0^{\frac{1}{2}} \ & =\frac{2}{3}\left(-\frac{1}{8}+1ight)-\frac{1}{2}\left(\frac{1}{2}-0ight) \ & =\frac{1}{3} 	ext { sq. units }\end{aligned}$

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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