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What is the area under the curve $\mathrm{y}=|\mathrm{x}|+|\mathrm{x}-1|$ between $\mathrm{x}=0$ and $\mathrm{x}=1 ?$
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The correct answer is:
1
$|\mathrm{x}|$ for $\mathrm{x} \geq 0$
$=\mathrm{x}$ and $|\mathrm{x}-1|$ for $\mathrm{x} \leq 1$
$=-(\mathrm{x}-1)$
so, $\int_{0}^{1}(|\mathrm{x}|+|\mathrm{x}-| 1 \mid)=$ required area
$\mathrm{a}=\int_{0}^{1} \mathrm{x} \mathrm{dx}-\int_{0}^{1}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}-\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{0}^{1}=\frac{1}{2}-\left(\frac{1}{2}-1\right)=1 \mathrm{sq}$ units
$=\mathrm{x}$ and $|\mathrm{x}-1|$ for $\mathrm{x} \leq 1$
$=-(\mathrm{x}-1)$
so, $\int_{0}^{1}(|\mathrm{x}|+|\mathrm{x}-| 1 \mid)=$ required area
$\mathrm{a}=\int_{0}^{1} \mathrm{x} \mathrm{dx}-\int_{0}^{1}(\mathrm{x}-1) \mathrm{d} \mathrm{x}$
$=\left[\frac{\mathrm{x}^{2}}{2}\right]_{0}^{1}-\left[\frac{\mathrm{x}^{2}}{2}-\mathrm{x}\right]_{0}^{1}=\frac{1}{2}-\left(\frac{1}{2}-1\right)=1 \mathrm{sq}$ units
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