Question Number 161229 by cortano last updated on 14/Dec/21
![Given f(x)= { ((1−∣x∣ ; x≤1)),((∣x∣−1 ; x>1)) :} find ∫_(−3) ^( 8) [f(x−1)+f(x+1)] dx.](https://www.tinkutara.com/question/Q161229.png)
$$\:{Given}\:{f}\left({x}\right)=\:\begin{cases}{\mathrm{1}−\mid{x}\mid\:;\:{x}\leqslant\mathrm{1}}\\{\mid{x}\mid−\mathrm{1}\:;\:{x}>\mathrm{1}}\end{cases} \\ $$$$\:{find}\:\int_{−\mathrm{3}} ^{\:\mathrm{8}} \left[{f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)\right]\:{dx}.\: \\ $$
Answered by mr W last updated on 14/Dec/21
Commented by mr W last updated on 14/Dec/21
![red: f(x) blue: f(x−1) green: f(x+1) ∫_(−3) ^8 f(x−1)dx=−((3×3)/2)+((2×1)/2)+((6×6)/2)=((29)/2) ∫_(−3) ^8 f(x+1)dx=−((1×1)/2)+((2×1)/2)+((8×8)/2)=((65)/2) ∫_(−3) ^8 [f(x−1)+f(x+1)]dx=((29+65)/2)=47](https://www.tinkutara.com/question/Q161246.png)
$${red}:\:{f}\left({x}\right) \\ $$$${blue}:\:{f}\left({x}−\mathrm{1}\right) \\ $$$${green}:\:{f}\left({x}+\mathrm{1}\right) \\ $$$$\int_{−\mathrm{3}} ^{\mathrm{8}} {f}\left({x}−\mathrm{1}\right){dx}=−\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{6}×\mathrm{6}}{\mathrm{2}}=\frac{\mathrm{29}}{\mathrm{2}} \\ $$$$\int_{−\mathrm{3}} ^{\mathrm{8}} {f}\left({x}+\mathrm{1}\right){dx}=−\frac{\mathrm{1}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{8}×\mathrm{8}}{\mathrm{2}}=\frac{\mathrm{65}}{\mathrm{2}} \\ $$$$\int_{−\mathrm{3}} ^{\mathrm{8}} \left[{f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)\right]{dx}=\frac{\mathrm{29}+\mathrm{65}}{\mathrm{2}}=\mathrm{47} \\ $$