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If $f(x)=\int_{-1}^{x}|t| d t$, then for any $x \geq 0, f(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2}\left(1+\mathrm{x}^{2}\right)$
Given, $\mathrm{f}(\mathrm{x})=\int_{-1}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}$
$$
\begin{aligned}
&=\int_{-1}^{0}|\mathrm{t}| \mathrm{dt}+\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt} \\
&=\int_{-1}^{0}-\mathrm{t} \mathrm{dt}+\int_{0}^{\mathrm{x}} \mathrm{t} \mathrm{dt} \\
&=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{x}}=-\left[0-\frac{1}{2}\right]+\left[\frac{\mathrm{x}^{2}}{2}-0\right] \\
&=\frac{1}{2}\left(1+\mathrm{x}^{2}\right)
\end{aligned}
$$
$$
\begin{aligned}
&=\int_{-1}^{0}|\mathrm{t}| \mathrm{dt}+\int_{0}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt} \\
&=\int_{-1}^{0}-\mathrm{t} \mathrm{dt}+\int_{0}^{\mathrm{x}} \mathrm{t} \mathrm{dt} \\
&=-\left[\frac{\mathrm{t}^{2}}{2}\right]_{-1}^{0}+\left[\frac{\mathrm{t}^{2}}{2}\right]_{0}^{\mathrm{x}}=-\left[0-\frac{1}{2}\right]+\left[\frac{\mathrm{x}^{2}}{2}-0\right] \\
&=\frac{1}{2}\left(1+\mathrm{x}^{2}\right)
\end{aligned}
$$
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