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Question Number 202543 by mr W last updated on 29/Dec/23
Commented by mr W last updated on 29/Dec/23
unsolved old question Q201806
$${unsolved}\:{old}\:{question}\:{Q}\mathrm{201806} \\ $$
Answered by mr W last updated on 30/Dec/23
y=⌊x^2 +1⌋ for x∈[k, k+1): x=k+t with 0≤t<1 x^2 +1=(k+t)^2 +1=k^2 +1+2kt+t^2 y=⌊x^2 +1⌋=k^2 +m+1 with m=⌊2kt+t^2 ⌋ 0≤m=⌊2kt+t^2 ⌋<2k+1≤2k m≤2kt+t^2 <m+1 (√(k^2 +m))−k≤t<(√(k^2 +m+1))−k Δx=(√(k^2 +m+1))−(√(k^2 +m)) ∫_k ^(k+1) ⌊x^2 +1⌋dx =Σ_(m=0) ^(2k) yΔx =Σ_(m=0) ^(2k) (k^2 +m+1)((√(k^2 +m+1))−(√(k^2 +m))) ∫_0 ^n ⌊x^2 +1⌋dx=Σ_(k=0) ^(n−1) Σ_(m=0) ^(2k) (k^2 +m+1)((√(k^2 +m+1))−(√(k^2 +m))) =n^3 −Σ_(k=1) ^(n^2 −1) (√k) example: ∫_(−4) ^4 ⌊x^2 +1⌋dx =2∫_0 ^4 ⌊x^2 +1⌋dx =2(4^3 −(√1)−(√2)−(√3)−...−(√(15))) ≈47.0616 0679 9715 ✓ an other example: ∫_(−10) ^(10) ⌊x^2 +1⌋dx =2(10^3 −(√1)−(√2)−(√3)−...−(√(99))) ≈677.0741 0579 3705
$${y}=\lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor \\ $$$${for}\:{x}\in\left[{k},\:{k}+\mathrm{1}\right): \\ $$$${x}={k}+{t}\:{with}\:\mathrm{0}\leqslant{t}<\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=\left({k}+{t}\right)^{\mathrm{2}} +\mathrm{1}={k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{kt}+{t}^{\mathrm{2}} \\ $$$${y}=\lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor={k}^{\mathrm{2}} +{m}+\mathrm{1} \\ $$$${with}\:{m}=\lfloor\mathrm{2}{kt}+{t}^{\mathrm{2}} \rfloor \\ $$$$\mathrm{0}\leqslant{m}=\lfloor\mathrm{2}{kt}+{t}^{\mathrm{2}} \rfloor<\mathrm{2}{k}+\mathrm{1}\leqslant\mathrm{2}{k} \\ $$$${m}\leqslant\mathrm{2}{kt}+{t}^{\mathrm{2}} <{m}+\mathrm{1} \\ $$$$\sqrt{{k}^{\mathrm{2}} +{m}}−{k}\leqslant{t}<\sqrt{{k}^{\mathrm{2}} +{m}+\mathrm{1}}−{k} \\ $$$$\Delta{x}=\sqrt{{k}^{\mathrm{2}} +{m}+\mathrm{1}}−\sqrt{{k}^{\mathrm{2}} +{m}} \\ $$$$\int_{{k}} ^{{k}+\mathrm{1}} \lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor{dx} \\ $$$$=\underset{{m}=\mathrm{0}} {\overset{\mathrm{2}{k}} {\sum}}{y}\Delta{x} \\ $$$$=\underset{{m}=\mathrm{0}} {\overset{\mathrm{2}{k}} {\sum}}\left({k}^{\mathrm{2}} +{m}+\mathrm{1}\right)\left(\sqrt{{k}^{\mathrm{2}} +{m}+\mathrm{1}}−\sqrt{{k}^{\mathrm{2}} +{m}}\right) \\ $$$$\int_{\mathrm{0}} ^{{n}} \lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor{dx}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\mathrm{2}{k}} {\sum}}\left({k}^{\mathrm{2}} +{m}+\mathrm{1}\right)\left(\sqrt{{k}^{\mathrm{2}} +{m}+\mathrm{1}}−\sqrt{{k}^{\mathrm{2}} +{m}}\right) \\ $$$$\:\:\:\:\:\:={n}^{\mathrm{3}} −\underset{{k}=\mathrm{1}} {\overset{{n}^{\mathrm{2}} −\mathrm{1}} {\sum}}\sqrt{{k}} \\ $$$${example}: \\ $$$$\int_{−\mathrm{4}} ^{\mathrm{4}} \lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{4}} \lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor{dx} \\ $$$$=\mathrm{2}\left(\mathrm{4}^{\mathrm{3}} −\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−…−\sqrt{\mathrm{15}}\right) \\ $$$$\approx\mathrm{47}.\mathrm{0616}\:\mathrm{0679}\:\mathrm{9715}\:\checkmark \\ $$$$ \\ $$$${an}\:{other}\:{example}: \\ $$$$\int_{−\mathrm{10}} ^{\mathrm{10}} \lfloor{x}^{\mathrm{2}} +\mathrm{1}\rfloor{dx} \\ $$$$=\mathrm{2}\left(\mathrm{10}^{\mathrm{3}} −\sqrt{\mathrm{1}}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−…−\sqrt{\mathrm{99}}\right) \\ $$$$\approx\mathrm{677}.\mathrm{0741}\:\mathrm{0579}\:\mathrm{3705} \\ $$
Commented by mr W last updated on 29/Dec/23

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