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Question Number 101057 by bemath last updated on 30/Jun/20

$$\mathrm{Find}\mathrm{the}\mathrm{area}\mathrm{bounded}\mathrm{the}$$$$\mathrm{curves}\mathrm{f}\left(\mathrm{x}\right)=\mid x^3-4x^2+3x\mid \mathrm{and}$$$$\mathrm{x}-\mathrm{axis}$$

Answered by john santu last updated on 30/Jun/20

Commented by john santu last updated on 30/Jun/20

$$\mathrm{Area}=\underset{0}{\stackrel{1}{\int }}(x^3-4x^2+3x)dx-\underset{1}{\stackrel{3}{\int }}(x^3-4x^2+3x)dx$$$$={[\frac{1}{4}{x}^4-\frac{4}{3}{x}^3+\frac{3}{2}{x}^2]}_0^1-$$$${[\frac{1}{4}{x}^4-\frac{4}{3}{x}^3+\frac{3}{2}{x}^2]}_1^3$$$$=\frac{5}{12}-(\frac{1}{4}\left(80\right)-\frac{4}{3}\left(26\right)+\frac{3}{2}\left(8\right))$$$$=\frac{5}{12}+\frac{104}{3}-32=\frac{5}{12}+\frac{8}{3}$$$$=\frac{37}{12}★$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{thank}$$

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