Question Number 151860 by maged last updated on 23/Aug/21
$$\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}\mid\mathrm{2}−\mid\mathrm{3}−\boldsymbol{\mathrm{x}}\mid\mid\mathrm{dx} \\ $$
Commented by lyubita last updated on 23/Aug/21
$${It}\:{means}\:{area}\:{bounded}\:{by}\:{the}\:{graph} \\ $$$${and}\:{x}\:{axis}\:{between}\:{x}\:=\:\mathrm{1}\:{and}\:{x}\:=\:\mathrm{5} \\ $$
Commented by lyubita last updated on 23/Aug/21
Commented by maged last updated on 23/Aug/21
$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by puissant last updated on 23/Aug/21
![Q=∫_1 ^5 ∣2−∣3−x∣∣dx =∫_1 ^3 ∣2−3+x∣dx+∫_3 ^5 ∣2+3−x∣dx =∫_1 ^3 ∣x−1∣dx+∫_3 ^5 ∣5−x∣dx =[(x^2 /2)−x]_1 ^3 +[5x−(x^2 /2)]_3 ^5 =[((9/2)−3)−((1/2)−1)]+[(25−((25)/2))−(15−(9/2))] =2+2 =4 ∴∵ Q= ∫_1 ^5 ∣2−∣3−x∣∣dx = 4..](Q151868.png)
$${Q}=\int_{\mathrm{1}} ^{\mathrm{5}} \mid\mathrm{2}−\mid\mathrm{3}−{x}\mid\mid{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \mid\mathrm{2}−\mathrm{3}+{x}\mid{dx}+\int_{\mathrm{3}} ^{\mathrm{5}} \mid\mathrm{2}+\mathrm{3}−{x}\mid{dx} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{3}} \mid{x}−\mathrm{1}\mid{dx}+\int_{\mathrm{3}} ^{\mathrm{5}} \mid\mathrm{5}−{x}\mid{dx} \\ $$$$=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right]_{\mathrm{1}} ^{\mathrm{3}} +\left[\mathrm{5}{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{3}} ^{\mathrm{5}} \\ $$$$=\left[\left(\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{3}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)\right]+\left[\left(\mathrm{25}−\frac{\mathrm{25}}{\mathrm{2}}\right)−\left(\mathrm{15}−\frac{\mathrm{9}}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{2}+\mathrm{2} \\ $$$$=\mathrm{4} \\ $$$$\therefore\because\:\:{Q}=\:\int_{\mathrm{1}} ^{\mathrm{5}} \mid\mathrm{2}−\mid\mathrm{3}−{x}\mid\mid{dx}\:=\:\mathrm{4}.. \\ $$
Commented by maged last updated on 23/Aug/21
$$\mathrm{thank}\:\mathrm{you} \\ $$