Question Number 150489 by mathdanisur last updated on 12/Aug/21
![∫_( 0) ^( 2) ∣x∣ x^([x+1]) sgn(x) dx = ?](Q150489.png)
$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{2}} {\int}}\:\mid\mathrm{x}\mid\:\mathrm{x}^{\left[\boldsymbol{\mathrm{x}}+\mathrm{1}\right]} \:\mathrm{sgn}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 12/Aug/21
![I = ∫_0 ^2 ∣x∣.x^([x+1]) sgn(x) dx I = ∫_0 ^2 x.x^([x+1]) (+1) dx I = ∫_0 ^1 x.x^([x+1]) dx+∫_1 ^2 x.x^([x+1]) dx I = ∫_0 ^1 x.x^1 dx+∫_1 ^2 x.x^2 dx I = ∫_0 ^1 x^2 dx+∫_1 ^2 x^3 dx I = [(x^3 /3)]_0 ^1 +[(x^4 /4)]_1 ^2 I = (1/3)+((15)/4) = ((49)/(12))](Q150491.png)
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \mid{x}\mid.{x}^{\left[{x}+\mathrm{1}\right]} \mathrm{sgn}\left({x}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} {x}.{x}^{\left[{x}+\mathrm{1}\right]} \left(+\mathrm{1}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}.{x}^{\left[{x}+\mathrm{1}\right]} \:{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {x}.{x}^{\left[{x}+\mathrm{1}\right]} \:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}.{x}^{\mathrm{1}} \:{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {x}.{x}^{\mathrm{2}} \:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \:{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{3}} \:{dx} \\ $$$$\mathrm{I}\:=\:\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} +\left[\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{15}}{\mathrm{4}}\:=\:\frac{\mathrm{49}}{\mathrm{12}} \\ $$
Commented by mathdanisur last updated on 12/Aug/21
![Cool See thankyou If ∫_0 ^2 [x] x^([x+1] ) sgn(x)dx](Q150493.png)
$$\mathrm{Cool}\:\mathrm{See}\:\mathrm{thankyou} \\ $$$$\mathrm{If}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\left[\mathrm{x}\right]\:\mathrm{x}^{\left[\mathrm{x}+\mathrm{1}\right]\:} \mathrm{sgn}\left(\mathrm{x}\right)\mathrm{dx} \\ $$
Commented by Olaf_Thorendsen last updated on 13/Aug/21
![J = ∫_0 ^2 [x].x^([x+1]) sgn(x) dx J = ∫_0 ^2 [x].x^([x+1]) (+1) dx J = ∫_0 ^1 [x].x^([x+1]) dx+∫_1 ^2 [x].x^([x+1]) dx J = ∫_0 ^1 0.x^1 dx+∫_1 ^2 1.x^2 dx J = ∫_1 ^2 x^2 dx J = [(x^3 /3)]_1 ^2 J = (8/3)−(1/3) = (7/3)](Q150499.png)
$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \left[{x}\right].{x}^{\left[{x}+\mathrm{1}\right]} \mathrm{sgn}\left({x}\right)\:{dx} \\ $$$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}} \left[{x}\right].{x}^{\left[{x}+\mathrm{1}\right]} \left(+\mathrm{1}\right)\:{dx} \\ $$$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left[{x}\right].{x}^{\left[{x}+\mathrm{1}\right]} \:{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \left[{x}\right].{x}^{\left[{x}+\mathrm{1}\right]} \:{dx} \\ $$$$\mathrm{J}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{0}.{x}^{\mathrm{1}} \:{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{1}.{x}^{\mathrm{2}} \:{dx} \\ $$$$\mathrm{J}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} \:{dx} \\ $$$$\mathrm{J}\:=\:\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{J}\:=\:\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:=\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 13/Aug/21
$$\mathrm{Cool}\:\mathrm{Ser}\:\mathrm{thankyou} \\ $$