Question:
$\int\limits^2_0 |1 -x| dx$ is equal to
$\int\limits^2_0 |1 -x| dx$ is equal to
Updated On: Jan 30, 2025
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- $\frac{3}{2}$
- $\frac{1}{2}$
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The Correct Option is B
Solution and Explanation
$\int\limits_{0}^{2}|1-x| d x=\int\limits_{0}^{1}(1-x) d x+\int\limits_{1}^{2}(x-1) d x$
$=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{2}$
$=1-\frac{1}{2}+\left[2-2-\left(\frac{1}{2}-1\right)\right]$
$=\frac{1}{2}+\frac{1}{2}=1$
$=\left[x-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{2}}{2}-x\right]_{1}^{2}$
$=1-\frac{1}{2}+\left[2-2-\left(\frac{1}{2}-1\right)\right]$
$=\frac{1}{2}+\frac{1}{2}=1$
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