Integration Questions

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

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

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

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

$$\mathrm{Area}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}\right)\:{dx}\:−\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\left({x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{3}{x}\right){dx} \\$$$$=\:\left[\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \:\right]_{\mathrm{0}} ^{\mathrm{1}} − \\$$$$\left[\:\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} \: \\$$$$=\:\frac{\mathrm{5}}{\mathrm{12}}−\left(\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{80}\right)−\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{26}\right)+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{8}\right)\right) \\$$$$=\:\frac{\mathrm{5}}{\mathrm{12}}+\frac{\mathrm{104}}{\mathrm{3}}−\mathrm{32}\:=\:\frac{\mathrm{5}}{\mathrm{12}}\:+\:\frac{\mathrm{8}}{\mathrm{3}} \\$$$$=\:\frac{\mathrm{37}}{\mathrm{12}}\bigstar\: \\$$

Commented by bemath last updated on 30/Jun/20

$$\mathrm{thank} \\$$