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Question Number 97218 by bobhans last updated on 07/Jun/20

$$\underset{-2}{\stackrel{3}{\int }}\mid x-2\mid \lfloor \frac{x}{2}\rfloor \mathrm{sgn}(x-1)dx$$

Answered by john santu last updated on 07/Jun/20

$$\mathrm{sgn}(x-1)=1,\mathrm{when}x>1$$$$\mathrm{sgn}(x-1)=-1,\mathrm{when}x<1$$$$\mathrm{J}=-\underset{-2}{\stackrel{1}{\int }}\mid x-2\mid \lfloor \frac{x}{2}\rfloor dx+$$$$\underset{1}{\stackrel{3}{\int }}\mid x-2\mid \lfloor \frac{x}{2}\rfloor dx$$$$\mathrm{J}=\underset{-2}{\stackrel{1}{\int }}(x-2)\lfloor \frac{x}{2}\rfloor dx-$$$$\underset{1}{\stackrel{2}{\int }}(x-2)\lfloor \frac{x}{2}\rfloor dx+\underset{2}{\stackrel{3}{\int }}(x-2)\lfloor \frac{x}{2}\rfloor dx$$$$\mathrm{J}=\underset{-2}{\stackrel{3}{\int }}(x-2)\lfloor \frac{x}{2}\rfloor dx-2\underset{1}{\stackrel{2}{\int }}(x-2)\lfloor \frac{x}{2}\rfloor dx$$$$\mathrm{let}x=2z$$$$\mathrm{J}=\underset{-1}{\stackrel{\frac{3}{2}}{\int }}(2\mathrm{z}-2)\lfloor \mathrm{z}\rfloor .2\mathrm{dz}-2\underset{\frac{1}{2}}{\stackrel{1}{\int }}(2\mathrm{z}-2)\lfloor \mathrm{z}\rfloor .2\mathrm{dz}$$$$\mathrm{J}=\underset{-1}{\stackrel{\frac{3}{2}}{\int }}(4\mathrm{z}-4)\lfloor \mathrm{z}\rfloor \mathrm{dz}-\underset{\frac{1}{2}}{\stackrel{1}{\int }}(8\mathrm{z}-8)\lfloor \mathrm{z}\rfloor \mathrm{dz}$$$$\mathrm{J}=\underset{-1}{\stackrel{0}{\int }}(4-4\mathrm{z})\mathrm{dz}+\underset{1}{\stackrel{\frac{3}{2}}{\int }}(4\mathrm{z}-4)\mathrm{dz}$$$$\mathrm{J}={\left[(4\mathrm{z}-{2\mathrm{z}}^2)\right]}_{-1}^0+{\left[({2\mathrm{z}}^2-4\mathrm{z})\right]}_1^{\frac{3}{2}}$$$$\mathrm{J}=0-(-4-2)+2(\frac{9}{4}-1)-4(\frac{3}{2}-1)$$$$\mathrm{J}=6+2\left(\frac{5}{4}\right)-4\left(\frac{1}{2}\right)$$$$\mathrm{J}=6+\frac{5}{2}-2=4+\frac{5}{2}=\frac{13}{2}$$

Answered by mathmax by abdo last updated on 07/Jun/20

$$\mathrm{A}={\int }_{-2}^3\mid \mathrm{x}-2\mid \left[\frac{\mathrm{x}}{2}\right]\delta (\mathrm{x}-1)\mathrm{dx}\mathrm{with}\delta (\mathrm{x}-1)=1\mathrm{if}\mathrm{x}>1\mathrm{and}\delta (\mathrm{x}-1)=-1\mathrm{if}\mathrm{x}<1$$$$\mathrm{changement}\frac{\mathrm{x}}{2}=\mathrm{t}\mathrm{give}$$$$\mathrm{A}={\int }_{-1}^{\frac{3}{2}}\mid 2\mathrm{t}-2\mid \left[\mathrm{t}\right]\delta (2\mathrm{t}-1)2\mathrm{dt}=4{\int }_{-1}^{\frac{3}{2}}\mid \mathrm{t}-1\mid \left[\mathrm{t}\right]\delta (2\mathrm{t}-1)\mathrm{dt}$$$$=4({\int }_{-1}^0(1-\mathrm{t})(-1)(-1)\mathrm{dt}+{\int }_0^{\frac{1}{2}}(1-\mathrm{t})0(-1)\mathrm{dt}+{\int }_{\frac{1}{2}}^1(1-\mathrm{t})0\left(1\right)\mathrm{dt}$$$$+{\int }_1^{\frac{3}{2}}(\mathrm{t}-1)1\mathrm{dt})=4({\int }_{-1}^0(1-\mathrm{t})\mathrm{dt}+{\int }_1^{\frac{3}{2}}(\mathrm{t}-1)\mathrm{dt})$$$$=4\{{[\mathrm{t}-\frac{{\mathrm{t}}^2}{2}]}_{-1}^0+{[\frac{{\mathrm{t}}^2}{2}-\mathrm{t}]}_1^{\frac{3}{2}}\}=4\{-(-1-\frac{1}{2})+\frac{9}{8}-\frac{3}{2}-(-\frac{1}{2})\}$$$$=4\{\frac{9}{8}+\frac{1}{2}\}=\frac{9}{2}+2=\frac{13}{2}$$

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