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$\int_{-1}^{1}|1-x| d x$ is equal to
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Verified Answer
The correct answer is:
$2$
Let $I=\int_{-1}^{1}|1-x| d x$
$$
\begin{aligned}
&=\int_{-1}^{1}(1-x) d x \quad\left[\because|x|=\left\{\begin{array}{cc}
-x, & x < 0 \\
x, & x \geq 0
\end{array}\right]\right. \\
&=\left[x-\frac{x^{2}}{2}\right]_{-1}^{1}=\left(1-\frac{1}{2}\right)-\left(-1-\frac{1}{2}\right)=\frac{1}{2}+\frac{3}{2}=2
\end{aligned}
$$
$$
\begin{aligned}
&=\int_{-1}^{1}(1-x) d x \quad\left[\because|x|=\left\{\begin{array}{cc}
-x, & x < 0 \\
x, & x \geq 0
\end{array}\right]\right. \\
&=\left[x-\frac{x^{2}}{2}\right]_{-1}^{1}=\left(1-\frac{1}{2}\right)-\left(-1-\frac{1}{2}\right)=\frac{1}{2}+\frac{3}{2}=2
\end{aligned}
$$
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