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Question Number 30858 by gopikrishnan005@gmail.com last updated on 27/Feb/18

$${\int }_0^1\mathrm{x}\mid \mathrm{x}-4\mid \mathrm{dx}$$

Answered by mrW2 last updated on 27/Feb/18

$${\int }_0^1\mathrm{x}\mid \mathrm{x}-4\mid \mathrm{dx}$$$$={\int }_0^1\mathrm{x}(4-\mathrm{x})\mathrm{dx}$$$$={[2x^2-\frac{x^3}{3}]}_0^1$$$$=2-\frac{1}{3}$$$$=\frac{5}{3}$$

Answered by drishtantsingh315 last updated on 27/Feb/18

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