# Solve this.

dear student

${x}^{2}-1=-1\phantom{\rule{0ex}{0ex}}x=0\phantom{\rule{0ex}{0ex}}{x}^{2}-1=0\phantom{\rule{0ex}{0ex}}x=1\phantom{\rule{0ex}{0ex}}{x}^{2}-1=1\phantom{\rule{0ex}{0ex}}x=\sqrt{2}\phantom{\rule{0ex}{0ex}}{x}^{2}-1=2\phantom{\rule{0ex}{0ex}}x=\sqrt{3}\phantom{\rule{0ex}{0ex}}{x}^{2}-1=3\phantom{\rule{0ex}{0ex}}x=2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}I={\int }_{0}^{2}x\left[{x}^{2}-1\right]dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}x\left[{x}^{2}-1\right]dx+{\int }_{1}^{\sqrt{2}}x\left[{x}^{2}-1\right]dx+{\int }_{\sqrt{2}}^{\sqrt{3}}x\left[{x}^{2}-1\right]dx+{\int }_{\sqrt{3}}^{2}x\left[{x}^{2}-1\right]dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{1}x\left(-1\right)dx+{\int }_{1}^{\sqrt{2}}x\left(0\right)dx+{\int }_{\sqrt{2}}^{\sqrt{3}}x\left(1\right)dx+{\int }_{\sqrt{3}}^{2}x\left(2\right)dx\phantom{\rule{0ex}{0ex}}=-1{\left[\frac{{x}^{2}}{2}\right]}_{0}^{1}+1{\left[\frac{{x}^{2}}{2}\right]}_{\sqrt{2}}^{\sqrt{3}}+2{\left[\frac{{x}^{2}}{2}\right]}_{\sqrt{3}}^{2}\phantom{\rule{0ex}{0ex}}=-\frac{1}{2}+\frac{1}{2}+4-3\phantom{\rule{0ex}{0ex}}=1$

regards

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