Question Number 37895 by abdo mathsup 649 cc last updated on 19/Jun/18
![calculate A_n =∫_0 ^n (x−[(√x)])dx and lim_(n→+∞) A_n](Q37895.png)
$${calculate}\:\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left({x}−\left[\sqrt{{x}}\right]\right){dx}\:{and} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
![∫_0 ^1 (x−[(√x) ])dx+∫_1 ^4 (x−[(√x) ]dx+∫_4 ^9 (x−[(√x) ] dx +... ∫_0 ^1 (x−0)dx+∫_1 ^4 (x−1)dx+∫_4 ^9 (x−2)dx+.. +−−∫_(((√(n−1))^ )^2 ) ^(((√n) )^2 ) (x−(√(n−1)) )dx ={((1^2 −o^2 )/2)+((4^2 −1^2 )/2)+((9^2 −4^2 )/2)+...+((((√n) )^2 −((√(n−1)) )^2 )/2) +{0.(1−0)+(4−1)+2(9−4)+3(16−9)+...+( ((√(n−1)) (n−n+1)} =(n/2)+(3+10+21+](Q37956.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\left[\sqrt{{x}}\:\right]\right){dx}+\int_{\mathrm{1}} ^{\mathrm{4}} \left({x}−\left[\sqrt{{x}}\:\right]{dx}+\int_{\mathrm{4}} ^{\mathrm{9}} \left({x}−\left[\sqrt{{x}}\:\right]\:{dx}\right.\right. \\ $$$$+... \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\mathrm{0}\right){dx}+\int_{\mathrm{1}} ^{\mathrm{4}} \left({x}−\mathrm{1}\right){dx}+\int_{\mathrm{4}} ^{\mathrm{9}} \left({x}−\mathrm{2}\right){dx}+.. \\ $$$$+−−\int_{\left(\sqrt{\boldsymbol{{n}}−\mathrm{1}}\:^{} \right)^{\mathrm{2}} } ^{\left(\sqrt{{n}}\:\right)^{\mathrm{2}} } \left(\boldsymbol{{x}}−\sqrt{\boldsymbol{{n}}−\mathrm{1}}\:\right)\boldsymbol{{dx}} \\ $$$$=\left\{\frac{\mathrm{1}^{\mathrm{2}} −{o}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{4}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{9}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}+...+\frac{\left(\sqrt{{n}}\:\right)^{\mathrm{2}} −\left(\sqrt{{n}−\mathrm{1}}\:\right)^{\mathrm{2}} }{\mathrm{2}}\right. \\ $$$$+\left\{\mathrm{0}.\left(\mathrm{1}−\mathrm{0}\right)+\left(\mathrm{4}−\mathrm{1}\right)+\mathrm{2}\left(\mathrm{9}−\mathrm{4}\right)+\mathrm{3}\left(\mathrm{16}−\mathrm{9}\right)+...+\left(\right.\right. \\ $$$$\left(\sqrt{{n}−\mathrm{1}}\:\left({n}−{n}+\mathrm{1}\right)\right\} \\ $$$$=\frac{{n}}{\mathrm{2}}+\left(\mathrm{3}+\mathrm{10}+\mathrm{21}+\right. \\ $$