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Question Number 2629 by Filup last updated on 24/Nov/15

Can you evalate: A=∫_a ^( b) ⌊f(x)⌋dx for example: ∫_(0.5) ^( 2.5) ⌊x^2 ⌋dx

$$\mathrm{Can}\:\mathrm{you}\:\mathrm{evalate}: \\ $$$${A}=\int_{{a}} ^{\:{b}} \lfloor{f}\left({x}\right)\rfloor{dx} \\ $$$$ \\ $$$${for}\:{example}: \\ $$$$\int_{\mathrm{0}.\mathrm{5}} ^{\:\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx} \\ $$

Commented by Filup last updated on 24/Nov/15

∫_(0.5) ^( 2.5) ⌊x^2 ⌋dx=∫_(0.5) ^( 1) ⌊x^2 ⌋dx+∫_1 ^2 ⌊x^2 ⌋dx+∫_2 ^( 2.5) ⌊x^2 ⌋dx ?

$$\int_{\mathrm{0}.\mathrm{5}} ^{\:\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\int_{\mathrm{0}.\mathrm{5}} ^{\:\mathrm{1}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}+\int_{\mathrm{1}} ^{\mathrm{2}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}+\int_{\mathrm{2}} ^{\:\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx} \\ $$$$? \\ $$

Commented by Yozzi last updated on 24/Nov/15

Yes, but there is a way to solve integrals like that using the geometric interpretation of the integral.

$${Yes},\:{but}\:{there}\:{is}\:{a}\:{way}\:{to}\:{solve} \\ $$$${integrals}\:{like}\:{that}\:{using}\:{the}\: \\ $$$${geometric}\:{interpretation}\:{of}\:{the} \\ $$$${integral}. \\ $$

Answered by Yozzi last updated on 24/Nov/15

For 0≤x<1⇒⌊x^2 ⌋=0 For 1≤x<(√2)⇒⌊x^2 ⌋=1 For (√2)≤x<(√3)⇒2≤x^2 <3⇒⌊x^2 ⌋=2 For (√3)≤x<2⇒3≤x^2 <4⇒⌊x^2 ⌋=3 For 2≤x<2.1⇒4≤x^2 <4.41⇒⌊x^2 ⌋=4 For 2.1≤x<2.23⇒4.41≤x^2 <4.9729⇒⌊x^2 ⌋=4 For 2.23≤x<(√5)⇒⌊x^2 ⌋=4 For (√5)≤x<(√6)⇒⌊x^2 ⌋=5 For (√6)≤x≤2.5⇒⌊x^2 ⌋=6 ∴∫_0 ^(2.5) ⌊x^2 ⌋dx=(1−0)×0+((√2)−1)×1+2((√3)−(√2)) +3(2−(√3))+4(2.1−2) +4(2.23−2.1)+4((√5)−2.23) +5((√6)−(√5))+6(2.5−(√6)) ∫_0 ^(2.5) ⌊x^2 ⌋dx=(1)((√2)−1)+2((√3)−(√2))+3((√4)−(√3))+4((√5)−(√4))+5((√6)−(√5))+6(2.5−(√6)) I=∫_0 ^(2.5) ⌊x^2 ⌋dx=Σ_(r=1) ^5 r((√(r+1))−(√r))+6(2.5−(√6)) I=∫_(0.5) ^(2.5) ⌊x^2 ⌋dx as well since ∫_0 ^(2.5) ⌊x^2 ⌋dx=∫_(0.5) ^(2.5) ⌊x^2 ⌋dx+∫_0 ^(0.5) ⌊x^2 ⌋dx and ∫_0 ^(0.5) ⌊x^2 ⌋dx=0 (∵ for 0≤x<1⇒0≤x^2 <1⇒⌊x^2 ⌋=0). ∫_0 ^(n/2) ⌊x^2 ⌋dx=Σ_(r=1) ^n r((√(r+1))−(√r))+(n+1)((n/2)−(√(n+1))) ??? (n≥3,n odd)

$${For}\:\mathrm{0}\leqslant{x}<\mathrm{1}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{0} \\ $$$${For}\:\mathrm{1}\leqslant{x}<\sqrt{\mathrm{2}}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{1} \\ $$$${For}\:\sqrt{\mathrm{2}}\leqslant{x}<\sqrt{\mathrm{3}}\Rightarrow\mathrm{2}\leqslant{x}^{\mathrm{2}} <\mathrm{3}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{2} \\ $$$${For}\:\sqrt{\mathrm{3}}\leqslant{x}<\mathrm{2}\Rightarrow\mathrm{3}\leqslant{x}^{\mathrm{2}} <\mathrm{4}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{3} \\ $$$${For}\:\mathrm{2}\leqslant{x}<\mathrm{2}.\mathrm{1}\Rightarrow\mathrm{4}\leqslant{x}^{\mathrm{2}} <\mathrm{4}.\mathrm{41}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{4} \\ $$$${For}\:\mathrm{2}.\mathrm{1}\leqslant{x}<\mathrm{2}.\mathrm{23}\Rightarrow\mathrm{4}.\mathrm{41}\leqslant{x}^{\mathrm{2}} <\mathrm{4}.\mathrm{9729}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{4} \\ $$$${For}\:\mathrm{2}.\mathrm{23}\leqslant{x}<\sqrt{\mathrm{5}}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{4} \\ $$$${For}\:\sqrt{\mathrm{5}}\leqslant{x}<\sqrt{\mathrm{6}}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{5} \\ $$$${For}\:\sqrt{\mathrm{6}}\leqslant{x}\leqslant\mathrm{2}.\mathrm{5}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{6} \\ $$$$ \\ $$$$\therefore\int_{\mathrm{0}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\left(\mathrm{1}−\mathrm{0}\right)×\mathrm{0}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)×\mathrm{1}+\mathrm{2}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)+\mathrm{4}\left(\mathrm{2}.\mathrm{1}−\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{2}.\mathrm{23}−\mathrm{2}.\mathrm{1}\right)+\mathrm{4}\left(\sqrt{\mathrm{5}}−\mathrm{2}.\mathrm{23}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{5}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{5}}\right)+\mathrm{6}\left(\mathrm{2}.\mathrm{5}−\sqrt{\mathrm{6}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\left(\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}\right)+\mathrm{4}\left(\sqrt{\mathrm{5}}−\sqrt{\mathrm{4}}\right)+\mathrm{5}\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{5}}\right)+\mathrm{6}\left(\mathrm{2}.\mathrm{5}−\sqrt{\mathrm{6}}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\underset{{r}=\mathrm{1}} {\overset{\mathrm{5}} {\sum}}{r}\left(\sqrt{{r}+\mathrm{1}}−\sqrt{{r}}\right)+\mathrm{6}\left(\mathrm{2}.\mathrm{5}−\sqrt{\mathrm{6}}\right) \\ $$$${I}=\int_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}\:{as}\:{well}\:{since}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\int_{\mathrm{0}.\mathrm{5}} ^{\mathrm{2}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}+\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx} \\ $$$${and}\:\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\mathrm{0}\:\:\:\left(\because\:{for}\:\mathrm{0}\leqslant{x}<\mathrm{1}\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} <\mathrm{1}\Rightarrow\lfloor{x}^{\mathrm{2}} \rfloor=\mathrm{0}\right). \\ $$$$\int_{\mathrm{0}} ^{\frac{{n}}{\mathrm{2}}} \lfloor{x}^{\mathrm{2}} \rfloor{dx}=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}\left(\sqrt{{r}+\mathrm{1}}−\sqrt{{r}}\right)+\left({n}+\mathrm{1}\right)\left(\frac{{n}}{\mathrm{2}}−\sqrt{{n}+\mathrm{1}}\right)\:???\:\left({n}\geqslant\mathrm{3},{n}\:{odd}\right) \\ $$

Commented by Filup last updated on 24/Nov/15

Wow thats amazing! So crazy!

$$\mathrm{Wow}\:\mathrm{thats}\:\mathrm{amazing}!\:\mathrm{So}\:\mathrm{crazy}! \\ $$

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