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Question Number 149516 by mathdanisur last updated on 05/Aug/21

$$\mathrm{\Omega }=\underset{0}{\stackrel{3}{\int }}\left(\sqrt{{(x+2)}^2-8x}\right)dx=?$$

Answered by Ar Brandon last updated on 05/Aug/21

$$={\int }_0^3\sqrt{x^2-4x+4}dx={\int }_0^3\sqrt{{(x-2)}^2}dx$$$$={\int }_0^3\mid x-2\mid dx={\int }_2^3(x-2)dx-{\int }_0^2(x-2)dx$$$$={[\frac{x^2}{2}-2x]}_2^3-{[\frac{x^2}{2}-2x]}_0^2=(\frac{9}{2}-6)-2(2-4)$$$$=-\frac{3}{2}+4=\frac{5}{2}$$

Commented by mathdanisur last updated on 06/Aug/21

$$\mathrm{Thankyou}\mathbf{Ser}$$

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