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What is the area of the region bounded by the curve
$f(x)=1-\frac{x^{2}}{4}, x \in[-2,2]$, and the $x$ -axis? $\quad$
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$f(x)=1-\frac{x^{2}}{4}, x \in[-2,2]$, and the $x$ -axis? $\quad$
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Verified Answer
The correct answer is:
$\frac{8}{3} \mathrm{sq}$ unit
Required area $=\int_{-2}^{2}\left(1-\frac{x^{2}}{4}\right) \mathrm{dx}$
Since, $\left(1-\frac{\mathrm{x}^{2}}{4}\right)$ is an even function therefore
$\int_{-2}^{2}\left(1-\frac{x^{2}}{4}\right) d x=2 \int_{0}^{2}\left(1-\frac{x^{2}}{4}\right) d x$
$=2\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=2\left[2-\frac{2^{3}}{12}\right]$
$=2\left(2-\frac{2}{3}\right)=\frac{8}{3} \mathrm{sq}$ unit
Since, $\left(1-\frac{\mathrm{x}^{2}}{4}\right)$ is an even function therefore
$\int_{-2}^{2}\left(1-\frac{x^{2}}{4}\right) d x=2 \int_{0}^{2}\left(1-\frac{x^{2}}{4}\right) d x$
$=2\left[\mathrm{x}-\frac{\mathrm{x}^{3}}{12}\right]_{0}^{2}=2\left[2-\frac{2^{3}}{12}\right]$
$=2\left(2-\frac{2}{3}\right)=\frac{8}{3} \mathrm{sq}$ unit
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