Question Number 206391 by hardmath last updated on 13/Apr/24
$$\mathrm{Find}: \\ $$$$\int_{−\mathrm{3}} ^{\:−\mathrm{2}} \:\left(\mid\mathrm{x}\mid\:+\:\mid\mathrm{x}\:−\:\mathrm{4}\mid\right)\:\mathrm{dx}\:=\:? \\ $$
Answered by MM42 last updated on 13/Apr/24
![I=∫_(−3) ^(−2) (−2x+4)dx =(−x^2 +4x)]_(−3) ^(−2) =−12+21=9✓](https://www.tinkutara.com/question/Q206410.png)
$${I}=\int_{−\mathrm{3}} ^{−\mathrm{2}} \left(−\mathrm{2}{x}+\mathrm{4}\right){dx} \\ $$$$\left.=\left(−{x}^{\mathrm{2}} +\mathrm{4}{x}\right)\right]_{−\mathrm{3}} ^{−\mathrm{2}} =−\mathrm{12}+\mathrm{21}=\mathrm{9}\checkmark \\ $$$$ \\ $$
Answered by TonyCWX08 last updated on 13/Apr/24
![∫_(−3) ^(−2) (∣x∣+∣x−4∣)dx =∫_(−3) ^(−2) (∣x∣)dx +∫_(−3) ^(−2) (∣x−4∣)dx =∫_(−3) ^(−2) (−x)dx +∫_(−3) ^(−2) −(x−4)dx =[−(x^2 /2)]_(−3) ^(−2) +[−(x^2 /2)+4x]_(−3) ^(−2) =(−2+(9/2))+(−10+((33)/2)) =(5/2)+((13)/2) =9](https://www.tinkutara.com/question/Q206411.png)
$$\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\int}}\left(\mid{x}\mid+\mid{x}−\mathrm{4}\mid\right){dx} \\ $$$$=\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\int}}\left(\mid{x}\mid\right){dx}\:+\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\int}}\left(\mid{x}−\mathrm{4}\mid\right){dx} \\ $$$$=\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\int}}\left(−{x}\right){dx}\:+\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\int}}−\left({x}−\mathrm{4}\right){dx} \\ $$$$=\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\right]}}+\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}{x}\underset{−\mathrm{3}} {\overset{−\mathrm{2}} {\right]}} \\ $$$$=\left(−\mathrm{2}+\frac{\mathrm{9}}{\mathrm{2}}\right)+\left(−\mathrm{10}+\frac{\mathrm{33}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$=\mathrm{9} \\ $$
Commented by hardmath last updated on 13/Apr/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors}\:\mathrm{cool} \\ $$