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Question Number 146859 by mathdanisur last updated on 16/Jul/21

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

Answered by mathmax by abdo last updated on 16/Jul/21

$$\mathrm{I}={\int }_0^3\sqrt{{(\mathrm{x}+2)}^2-8\mathrm{x}}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^3\sqrt{{\mathrm{x}}^2+4\mathrm{x}+4-8\mathrm{x}}\mathrm{dx}$$$$={\int }_0^3\sqrt{{(\mathrm{x}-2)}^2}\mathrm{dx}={\int }_0^3\mid \mathrm{x}-2\mid \mathrm{dx}={\int }_0^2\mid \mathrm{x}-2\mid \mathrm{dx}+{\int }_2^3\mid \mathrm{x}-2\mid \mathrm{dx}$$$$={\int }_0^2(2-\mathrm{x})\mathrm{dx}+{\int }_2^3(\mathrm{x}-2)\mathrm{dx}={[2\mathrm{x}-\frac{{\mathrm{x}}^2}{2}]}_0^2+{[\frac{{\mathrm{x}}^2}{2}-2\mathrm{x}]}_2^3$$$$=4-2+\frac{9}{2}-6-2+4=-2+\frac{9}{2}=\frac{5}{2}$$

Commented by mathdanisur last updated on 16/Jul/21

$$thankyouSer$$

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