Question:
The area bounded by the lines \(y=\big|\,|x-1|-2\,\big|\) is ___________.
The area bounded by the lines \(y=\big|\,|x-1|-2\,\big|\) is ___________.
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Sketch the W-shaped graph; the only closed region is the triangle between its two zeros and the x-axis.
Updated On: Nov 4, 2026
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Correct Answer: 4
Solution and Explanation
Let \(t=x-1\). Then \(y=\big||t|-2\big|\). For \(|t|\le2\): \(y=2-|t|\); for \(|t|\gt2\): \(y=|t|-2\), which rises without bound.
The curve meets the x-axis at \(t=\pm2\) (i.e. \(x=-1,\,3\)) and has a peak \(y=2\) at \(t=0\) (\(x=1\)). The only bounded region is between this peak and the x-axis for \(-1\le x\le3\).
Area \(=\int_{-1}^{3}\big(2-|x-1|\big)dx = 2\int_{0}^{2}(2-t)\,dt = 2\left[2t-\frac{t^2}{2}\right]_0^2=2\times2=4\).
Answer: 4
The curve meets the x-axis at \(t=\pm2\) (i.e. \(x=-1,\,3\)) and has a peak \(y=2\) at \(t=0\) (\(x=1\)). The only bounded region is between this peak and the x-axis for \(-1\le x\le3\).
Area \(=\int_{-1}^{3}\big(2-|x-1|\big)dx = 2\int_{0}^{2}(2-t)\,dt = 2\left[2t-\frac{t^2}{2}\right]_0^2=2\times2=4\).
Answer: 4
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