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x-2-sgn-2x-dx-




Question Number 12171 by uni last updated on 15/Apr/17
∫x^2 ×sgn(2x)dx=?
$$\int\mathrm{x}^{\mathrm{2}} ×\mathrm{sgn}\left(\mathrm{2x}\right)\mathrm{dx}=? \\ $$
Answered by mrW1 last updated on 15/Apr/17
f(x)=x^2 ×sgn(2x)  = { ((x^2    x>0)),((0     x=0)),((−x^2     x<0)) :}  ∫f(x)dx=∫x^2 dx=(x^3 /3)+C     (x≥0)  =−∫x^2 dx=−(x^3 /3)+C    (x<0)  or   ∫f(x)dx=((∣x^3 ∣)/3)+C
$${f}\left({x}\right)={x}^{\mathrm{2}} ×{sgn}\left(\mathrm{2}{x}\right) \\ $$$$=\begin{cases}{{x}^{\mathrm{2}} \:\:\:{x}>\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:{x}=\mathrm{0}}\\{−{x}^{\mathrm{2}} \:\:\:\:{x}<\mathrm{0}}\end{cases} \\ $$$$\int{f}\left({x}\right){dx}=\int{x}^{\mathrm{2}} {dx}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{C}\:\:\:\:\:\left({x}\geqslant\mathrm{0}\right) \\ $$$$=−\int{x}^{\mathrm{2}} {dx}=−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{C}\:\:\:\:\left({x}<\mathrm{0}\right) \\ $$$${or}\: \\ $$$$\int{f}\left({x}\right){dx}=\frac{\mid{x}^{\mathrm{3}} \mid}{\mathrm{3}}+{C} \\ $$

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