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

$${}_0{\int }^2\mid x^2-2x+1\mid dx=?$$

Answered by TheSupreme last updated on 07/Feb/22

$${\int }_0^2\mid {(x-1)}^2\mid dx={\int }_0^2{(x-1)}^{2}dx=\frac{{(x-1)}^3}{3}{\mid }_0^2=\frac{2}{3}$$

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

$$wrong!$$$${\int }_0^2\ne {\int }_0^1$$

Answered by alephzero last updated on 06/Feb/22

$$\underset{0}{\stackrel{2}{\int }}\mid x^2-2x+1\mid dx=$$$$=\underset{0}{\stackrel{2}{\int }}\mid {(x-1)}^2\mid dx=$$$$=\underset{0}{\stackrel{2}{\int }}{(x-1)}^{2}dx=$$$$=\underset{0}{\stackrel{2}{\int }}{x}^2-2x+1dx=$$$$={[\int x^{2}dx-\int 2xdx+\int dx]}_0^2=$$$$={[\frac{x^3}{3}-\frac{2x^2}{2}+x]}_0^2=$$$$=\frac{8}{3}-\frac{8}{2}+2=\frac{16-24}{6}+2=-\frac{8}{6}+$$$$+2=-\frac{4}{3}+2=2-\frac{4}{3}=\frac{6-4}{3}=$$$$=\frac{2}{3}$$

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