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Question Number 128834 by bounhome last updated on 10/Jan/21

$${\int }_0^3\mid x^2-2x\mid dx=...?isthereanyoneto$$$$explainthewaytosolvethis$$

Answered by bobhans last updated on 10/Jan/21

$${\int }_0^3\mid x^2-2x\mid dx={\int }_0^2(2x-x^2)dx+{\int }_2^3(x^2-2x)dx$$

Answered by mathmax by abdo last updated on 10/Jan/21

$${\int }_0^3\mid {\mathrm{x}}^2-2\mathrm{x}\mid \mathrm{dx}={\int }_0^3\mathrm{x}\mid \mathrm{x}-2\mid \mathrm{dx}={\int }_0^2\mathrm{x}(2-\mathrm{x})\mathrm{dx}+{\int }_2^3\mathrm{x}(\mathrm{x}-2)\mathrm{dx}$$$$={\int }_0^2(2\mathrm{x}-{\mathrm{x}}^2)\mathrm{dx}+{\int }_2^3({\mathrm{x}}^2-2\mathrm{x})\mathrm{dx}={[{\mathrm{x}}^2-\frac{{\mathrm{x}}^3}{3}]}_0^2+{[\frac{{\mathrm{x}}^3}{3}-{\mathrm{x}}^2]}_2^3$$$$=2^2-\frac{2^3}{3}+(\frac{3^3}{3}-3^2-\frac{2^3}{3}+2^2)=4-\frac{8}{3}-\frac{8}{3}+4=8-\frac{16}{3}=\frac{24-16}{3}$$$$=\frac{8}{3}$$

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