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What is the area bounded by the curves $|y|=1-x^{2}$ ?
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The correct answer is:
$\frac{8}{3}$ square units
Since $|y|=\left\{\begin{array}{cc}y & y>0 \\ -y & y < 0 \\ 0 & y=0\end{array}\right.$
For $y>0 \Rightarrow y=1-x^{2}$
For $y < 0 \Rightarrow y=x^{2}-1$
For $y=0 \Rightarrow x=\pm 1$
So area under the curve $=4 \times$ Area under the region OABO (symmetry)
$\begin{aligned}=& 4 \times \int_{0}^{1} 1-x^{2} \mathrm{~d} \mathrm{x} \\ & \left.=4 \times\left[x-\frac{x^{3}}{3}\right]_{0}^{1}\right] \end{aligned}$
$\quad=4\left(1-\frac{1}{3}\right)=4 \times \frac{2}{3}=\frac{8}{3}$ sq. units
For $y>0 \Rightarrow y=1-x^{2}$
For $y < 0 \Rightarrow y=x^{2}-1$
For $y=0 \Rightarrow x=\pm 1$
So area under the curve $=4 \times$ Area under the region OABO (symmetry)
$\begin{aligned}=& 4 \times \int_{0}^{1} 1-x^{2} \mathrm{~d} \mathrm{x} \\ & \left.=4 \times\left[x-\frac{x^{3}}{3}\right]_{0}^{1}\right] \end{aligned}$
$\quad=4\left(1-\frac{1}{3}\right)=4 \times \frac{2}{3}=\frac{8}{3}$ sq. units
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