Question Number 194548 by horsebrand11 last updated on 09/Jul/23
Commented by horsebrand11 last updated on 09/Jul/23
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Answered by AST last updated on 09/Jul/23
$$\frac{{y}−\mathrm{1}}{{x}−\mathrm{1}}=\frac{\mathrm{1}−\mathrm{0}}{\mathrm{1}−\mathrm{0}}\Rightarrow{y}={x}\left({equation}\:{of}\:{line}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {xdx}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} ={x}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)={x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}\right){dx}=\frac{{x}^{\mathrm{4}} }{\mathrm{4}}−\frac{\mathrm{4}{x}^{\mathrm{3}} }{\mathrm{3}}+\mathrm{2}{x}^{\mathrm{2}} \mid_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\Rightarrow{R}_{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{12}} \\ $$$${Similarly},{we}\:{get}\:\int_{\mathrm{0}} ^{\mathrm{2}} {x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{3}}\Rightarrow{R}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{11}}{\mathrm{12}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$
Answered by MM42 last updated on 09/Jul/23
$$ \\ $$$${R}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} {x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} {dx}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${R}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{2}} {x}\left({x}−\mathrm{2}\right)^{\mathrm{2}} {dx}−{R}_{\mathrm{1}} =\frac{\mathrm{11}}{\mathrm{12}} \\ $$$$ \\ $$