Question Number 150941 by rexford last updated on 16/Aug/21
$$\int_{\mathrm{1}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid{dx} \\ $$$${please}\:{help}\:{me}\:{out} \\ $$
Commented by puissant last updated on 25/Aug/21
![=∫_1 ^4 ∣x−2∣dx =∫_1 ^2 ∣x−2∣dx+∫_2 ^4 ∣x−2∣dx =∫_1 ^2 (2−x)dx+∫_2 ^4 (x−2)dx =[2x−(x^2 /2)]_1 ^2 +[(x^2 /2)−2x]_2 ^4 =[(4−2)−(2−(1/2))] + [(8−8)−(2−4)] ⇒ I=(1/2)+2=(5/2)..](https://www.tinkutara.com/question/Q152027.png)
$$=\int_{\mathrm{1}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \mid{x}−\mathrm{2}\mid{dx}+\int_{\mathrm{2}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right){dx}+\int_{\mathrm{2}} ^{\mathrm{4}} \left({x}−\mathrm{2}\right){dx} \\ $$$$=\left[\mathrm{2}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} +\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$=\left[\left(\mathrm{4}−\mathrm{2}\right)−\left(\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right]\:+\:\left[\left(\mathrm{8}−\mathrm{8}\right)−\left(\mathrm{2}−\mathrm{4}\right)\right] \\ $$$$\Rightarrow\:{I}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{5}}{\mathrm{2}}.. \\ $$
Commented by rexford last updated on 25/Aug/21
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by Olaf_Thorendsen last updated on 16/Aug/21
![I = ∫_1 ^4 ∣x−2∣ dx I = ∫_1 ^2 ∣x−2∣ dx+ ∫_2 ^4 ∣x−2∣ dx I = ∫_1 ^2 (2−x) dx+ ∫_2 ^4 (x−2) dx I = [2x−(x^2 /2)]_1 ^2 + [(x^2 /2)−2x]_2 ^4 I = (2−(3/2))+ (0−(−2)) I = (5/2)](https://www.tinkutara.com/question/Q150942.png)
$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \mid{x}−\mathrm{2}\mid\:{dx}+\:\int_{\mathrm{2}} ^{\mathrm{4}} \mid{x}−\mathrm{2}\mid\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{2}−{x}\right)\:{dx}+\:\int_{\mathrm{2}} ^{\mathrm{4}} \left({x}−\mathrm{2}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\left[\mathrm{2}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{2}} +\:\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{x}\right]_{\mathrm{2}} ^{\mathrm{4}} \\ $$$$\mathrm{I}\:=\:\left(\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\:\left(\mathrm{0}−\left(−\mathrm{2}\right)\right) \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 22/Aug/21
$$\mathrm{nice} \\ $$
Commented by rexford last updated on 25/Aug/21
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by ajfour last updated on 16/Aug/21
