Question Number 30858 by gopikrishnan005@gmail.com last updated on 27/Feb/18
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\mid\mathrm{x}−\mathrm{4}\mid\mathrm{dx} \\ $$
Answered by mrW2 last updated on 27/Feb/18
![∫_0 ^1 x∣x−4∣dx =∫_0 ^1 x(4−x)dx =[2x^2 −(x^3 /3)]_0 ^1 =2−(1/3) =(5/3)](https://www.tinkutara.com/question/Q30863.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\mid\mathrm{x}−\mathrm{4}\mid\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}\left(\mathrm{4}−\mathrm{x}\right)\mathrm{dx} \\ $$$$=\left[\mathrm{2}{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$=\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Answered by drishtantsingh315 last updated on 27/Feb/18
