# Question and Answers Forum

Integration Questions

Question Number 165685 by daus last updated on 06/Feb/22

$$\:_{\mathrm{0}} \int^{\mathrm{2}} \mid\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:\mid\:{dx}\:=?\: \\$$

Answered by TheSupreme last updated on 07/Feb/22

$$\int_{\mathrm{0}} ^{\mathrm{2}} \mid\left({x}−\mathrm{1}\right)^{\mathrm{2}} \mid{dx}=\int_{\mathrm{0}} ^{\mathrm{2}} \left({x}−\mathrm{1}\right)^{\mathrm{2}} {dx}=\frac{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{3}}\mid_{\mathrm{0}} ^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}} \\$$

Commented by mr W last updated on 06/Feb/22

$${wrong}! \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}} \neq\int_{\mathrm{0}} ^{\mathrm{1}} \\$$

Answered by alephzero last updated on 06/Feb/22

$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\mid{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\mid\:{dx}\:= \\$$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\mid\left({x}−\mathrm{1}\right)^{\mathrm{2}} \mid\:{dx}\:= \\$$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:{dx}\:=\: \\$$$$=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:{dx}\:= \\$$$$=\:\left[\int{x}^{\mathrm{2}} {dx}−\int\mathrm{2}{xdx}+\int{dx}\right]_{\mathrm{0}} ^{\mathrm{2}} \:= \\$$$$=\:\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}}+{x}\right]_{\mathrm{0}} ^{\mathrm{2}} \:= \\$$$$=\:\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{2}}+\mathrm{2}\:=\:\frac{\mathrm{16}−\mathrm{24}}{\mathrm{6}}+\mathrm{2}\:=\:−\frac{\mathrm{8}}{\mathrm{6}}+ \\$$$$+\mathrm{2}\:=\:−\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{2}\:=\:\mathrm{2}−\frac{\mathrm{4}}{\mathrm{3}}\:=\:\frac{\mathrm{6}−\mathrm{4}}{\mathrm{3}}\:= \\$$$$=\:\frac{\mathrm{2}}{\mathrm{3}} \\$$

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