Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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.

$$\:{Given}\:{f}\left({x}\right)=\:\begin{cases}{\mathrm{1}−\mid{x}\mid\:;\:{x}\leqslant\mathrm{1}}\\{\mid{x}\mid−\mathrm{1}\:;\:{x}>\mathrm{1}}\end{cases} \\ $$ $$\:{find}\:\int_{−\mathrm{3}} ^{\:\mathrm{8}} \left[{f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)\right]\:{dx}.\: \\ $$

Answered by mr W last updated on 14/Dec/21

Commented bymr 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

$${red}:\:{f}\left({x}\right) \\ $$ $${blue}:\:{f}\left({x}−\mathrm{1}\right) \\ $$ $${green}:\:{f}\left({x}+\mathrm{1}\right) \\ $$ $$\int_{−\mathrm{3}} ^{\mathrm{8}} {f}\left({x}−\mathrm{1}\right){dx}=−\frac{\mathrm{3}×\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{6}×\mathrm{6}}{\mathrm{2}}=\frac{\mathrm{29}}{\mathrm{2}} \\ $$ $$\int_{−\mathrm{3}} ^{\mathrm{8}} {f}\left({x}+\mathrm{1}\right){dx}=−\frac{\mathrm{1}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}×\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{8}×\mathrm{8}}{\mathrm{2}}=\frac{\mathrm{65}}{\mathrm{2}} \\ $$ $$\int_{−\mathrm{3}} ^{\mathrm{8}} \left[{f}\left({x}−\mathrm{1}\right)+{f}\left({x}+\mathrm{1}\right)\right]{dx}=\frac{\mathrm{29}+\mathrm{65}}{\mathrm{2}}=\mathrm{47} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com