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 127704 by NATTAPONG4359 last updated on 01/Jan/21

  if f(x)= { ((x−n ; 2n ≤ x ≤2n+1)),((n+1 ; 2n+1≤x≤2n+2 )) :} where  n =0,1,2,3,..,9  find ∫_0 ^(20) f(x)dx

$$ \\ $$$${if}\:{f}\left({x}\right)=\begin{cases}{{x}−{n}\:;\:\mathrm{2}{n}\:\leqslant\:{x}\:\leqslant\mathrm{2}{n}+\mathrm{1}}\\{{n}+\mathrm{1}\:;\:\mathrm{2}{n}+\mathrm{1}\leqslant{x}\leqslant\mathrm{2}{n}+\mathrm{2}\:}\end{cases}\:{where}\:\:{n}\:=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},..,\mathrm{9} \\ $$$${find}\:\int_{\mathrm{0}} ^{\mathrm{20}} {f}\left({x}\right){dx} \\ $$

Answered by mahdipoor last updated on 01/Jan/21

∫_(2n) ^( 2n+1) f(x)dx=∫_(2n) ^( 2n+1) (x−n)dx=[(x^2 /2)−nx]_(2n) ^(2n+1)   =((((2n+1)^2 )/2)−n(2n+1))−((((2n)^2 )/2)−2(2n))  =n−(1/2)  ∫_(2n+1) ^( 2n+2) f(x)dx=∫_(2n+1) ^( 2n+2) (1+n)dx=[(1+n)x]_(2n+1) ^(2n+2)   =(1+n)(2n+2−(2n+1))=1+n     ∫_0 ^(20) f(x)dx=(∫_0 ^1 f(x)dx+∫_2 ^( 3) f(x)dx+...+∫_(18) ^( 19) f(x)dx)   +(∫_1 ^( 2) f(x)dx+∫_3 ^( 4) f(x)dx+...+∫_(19) ^( 20) f(x)dx)=  Σ_(n=0) ^9 (n−(1/2))+Σ_(n=0) ^9 (1+n)=Σ_(n=0) ^9 (2n+(1/2))=  2×(((9×10)/2))+10×(1/2)=95

$$\int_{\mathrm{2}{n}} ^{\:\mathrm{2}{n}+\mathrm{1}} {f}\left({x}\right){dx}=\int_{\mathrm{2}{n}} ^{\:\mathrm{2}{n}+\mathrm{1}} \left({x}−{n}\right){dx}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{nx}\right]_{\mathrm{2}{n}} ^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\left(\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}−{n}\left(\mathrm{2}{n}+\mathrm{1}\right)\right)−\left(\frac{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\left(\mathrm{2}{n}\right)\right) \\ $$$$={n}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\mathrm{2}{n}+\mathrm{1}} ^{\:\mathrm{2}{n}+\mathrm{2}} {f}\left({x}\right){dx}=\int_{\mathrm{2}{n}+\mathrm{1}} ^{\:\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{1}+{n}\right){dx}=\left[\left(\mathrm{1}+{n}\right){x}\right]_{\mathrm{2}{n}+\mathrm{1}} ^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$=\left(\mathrm{1}+{n}\right)\left(\mathrm{2}{n}+\mathrm{2}−\left(\mathrm{2}{n}+\mathrm{1}\right)\right)=\mathrm{1}+{n} \\ $$$$ \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{20}} {f}\left({x}\right){dx}=\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}+\int_{\mathrm{2}} ^{\:\mathrm{3}} {f}\left({x}\right){dx}+...+\int_{\mathrm{18}} ^{\:\mathrm{19}} {f}\left({x}\right){dx}\right) \\ $$$$\:+\left(\int_{\mathrm{1}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}+\int_{\mathrm{3}} ^{\:\mathrm{4}} {f}\left({x}\right){dx}+...+\int_{\mathrm{19}} ^{\:\mathrm{20}} {f}\left({x}\right){dx}\right)= \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\left(\mathrm{1}+{n}\right)=\underset{{n}=\mathrm{0}} {\overset{\mathrm{9}} {\sum}}\left(\mathrm{2}{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)= \\ $$$$\mathrm{2}×\left(\frac{\mathrm{9}×\mathrm{10}}{\mathrm{2}}\right)+\mathrm{10}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{95} \\ $$

Commented by mehedi last updated on 22/Dec/21

4

$$\mathrm{4} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com