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Question Number 38716 by maxmathsup by imad last updated on 28/Jun/18

calculate ∫_2 ^5 (dx/((x +1−[x])^2 ))

$${calculate}\:\:\:\int_{\mathrm{2}} ^{\mathrm{5}} \:\:\:\:\:\frac{{dx}}{\left({x}\:+\mathrm{1}−\left[{x}\right]\right)^{\mathrm{2}} } \\ $$

Commented by abdo mathsup 649 cc last updated on 29/Jun/18

I = Σ_(k=2) ^4 ∫_k ^(k+1) (dx/((x+1−k)^2 )) =Σ_(k=2) ^4 [−(1/(x+1−k))]_k ^(k+1) =Σ_(k=2) ^4 {1−(1/2)}=(1/2)Σ_(k=2) ^4 (1) =(1/2)(4−2+1) =(3/2) .

$${I}\:=\:\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{dx}}{\left({x}+\mathrm{1}−{k}\right)^{\mathrm{2}} } \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\:\left[−\frac{\mathrm{1}}{{x}+\mathrm{1}−{k}}\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \:\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right\}=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{2}} ^{\mathrm{4}} \left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−\mathrm{2}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

∫_2 ^3 (dx/((x+1−2)^2 ))+∫_3 ^4 (dx/((x+1−3)^2 ))+∫_4 ^5 (dx/((x+1−4)^2 )) =∣(((x−1)^(−1) )/(−1))∣_2 ^3 +∣(((x−2)^(−1) )/(−1))∣_3 ^4 +∣(((x−3)^(−1) )/(−1))∣_4 ^5 =−1{((1/2)−(1/1)+(1/2)−(1/1)+(1/2)−(1/1))} =−1((3/2)−3) =(3/2)

$$\int_{\mathrm{2}} ^{\mathrm{3}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{2}\right)^{\mathrm{2}} }+\int_{\mathrm{3}} ^{\mathrm{4}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} }+\int_{\mathrm{4}} ^{\mathrm{5}} \frac{{dx}}{\left({x}+\mathrm{1}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$=\mid\frac{\left({x}−\mathrm{1}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{2}} ^{\mathrm{3}} +\mid\frac{\left({x}−\mathrm{2}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{3}} ^{\mathrm{4}} +\mid\frac{\left({x}−\mathrm{3}\right)^{−\mathrm{1}} }{−\mathrm{1}}\mid_{\mathrm{4}} ^{\mathrm{5}} \\ $$$$=−\mathrm{1}\left\{\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}}\right)\right\} \\ $$$$=−\mathrm{1}\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{3}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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