Question Number 79903 by Pratah last updated on 29/Jan/20
Commented by jagoll last updated on 29/Jan/20
![what definition of [x]? it same with⌊x⌋ ?](Q79907.png)
$$\mathrm{what}\:\mathrm{definition}\:\mathrm{of}\:\left[\mathrm{x}\right]? \\ $$$$\mathrm{it}\:\mathrm{same}\:\mathrm{with}\lfloor\mathrm{x}\rfloor\:? \\ $$
Commented by mr W last updated on 29/Jan/20
Commented by jagoll last updated on 29/Jan/20
![{3.47}=3.47−[3.47] = 3.47−3 = 0.47 if {5.75} = 5.75−[5.75] = 5.75−5=0.75](Q79916.png)
$$\left\{\mathrm{3}.\mathrm{47}\right\}=\mathrm{3}.\mathrm{47}−\left[\mathrm{3}.\mathrm{47}\right] \\ $$$$=\:\mathrm{3}.\mathrm{47}−\mathrm{3}\:=\:\mathrm{0}.\mathrm{47} \\ $$$$\mathrm{if}\:\left\{\mathrm{5}.\mathrm{75}\right\}\:=\:\mathrm{5}.\mathrm{75}−\left[\mathrm{5}.\mathrm{75}\right] \\ $$$$=\:\mathrm{5}.\mathrm{75}−\mathrm{5}=\mathrm{0}.\mathrm{75} \\ $$
Commented by jagoll last updated on 29/Jan/20
![{−7.02}=−7−[−7.02] =−7−(−7)=0 that right sir?](Q79917.png)
$$\left\{−\mathrm{7}.\mathrm{02}\right\}=−\mathrm{7}−\left[−\mathrm{7}.\mathrm{02}\right] \\ $$$$=−\mathrm{7}−\left(−\mathrm{7}\right)=\mathrm{0} \\ $$$$\mathrm{that}\:\mathrm{right}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 29/Jan/20
$$\mathrm{or}\:\mathrm{wrong}? \\ $$
Commented by mr W last updated on 29/Jan/20
![wrong! {−7.02}=−7.02−[−7.02] =−7.02−(−8)=0.98](Q79919.png)
$${wrong}! \\ $$$$\left\{−\mathrm{7}.\mathrm{02}\right\}=−\mathrm{7}.\mathrm{02}−\left[−\mathrm{7}.\mathrm{02}\right] \\ $$$$=−\mathrm{7}.\mathrm{02}−\left(−\mathrm{8}\right)=\mathrm{0}.\mathrm{98} \\ $$
Commented by jagoll last updated on 29/Jan/20
![oo i′m wrong in [−7.02] so if [−4.001] = −5 sir](Q79920.png)
$$\mathrm{oo}\:\mathrm{i}'\mathrm{m}\:\mathrm{wrong}\:\mathrm{in}\:\left[−\mathrm{7}.\mathrm{02}\right] \\ $$$$\mathrm{so}\:\mathrm{if}\:\left[−\mathrm{4}.\mathrm{001}\right]\:=\:−\mathrm{5}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 29/Jan/20
$${yes} \\ $$
Commented by jagoll last updated on 29/Jan/20
$$\mathrm{thank}\:\mathrm{you}. \\ $$$$\mathrm{now}\:\lfloor\mathrm{x}\rfloor\:\mathrm{what}\:\mathrm{is}\:\mathrm{definition}\:\mathrm{sir}? \\ $$
Commented by jagoll last updated on 29/Jan/20
$$\mathrm{floor}\:\mathrm{function}\:\mathrm{namely}? \\ $$
Commented by mr W last updated on 29/Jan/20
![what you saw above is about ⌊x⌋. i.e. [x]=⌊x⌋. for non−negative x, [x] always =⌊x⌋. but be careful with negative x, because there are different opinions about the definition of {x} and [x]. what you saw above is just one of those opinions, maybe the most popular one. if you want to know more, go to google!](Q79926.png)
$${what}\:{you}\:{saw}\:{above}\:{is}\:{about}\:\lfloor{x}\rfloor. \\ $$$${i}.{e}.\:\left[{x}\right]=\lfloor{x}\rfloor.\:{for}\:{non}−{negative}\:{x}, \\ $$$$\left[{x}\right]\:{always}\:=\lfloor{x}\rfloor.\:{but}\:{be}\:{careful}\:{with} \\ $$$${negative}\:{x},\:{because}\:{there}\:{are}\:{different}\: \\ $$$${opinions}\:{about}\:{the}\:{definition}\:{of}\:\left\{{x}\right\}\: \\ $$$${and}\:\left[{x}\right].\:{what}\:{you}\:{saw}\:{above}\:{is}\:{just} \\ $$$${one}\:{of}\:{those}\:{opinions},\:{maybe}\:{the} \\ $$$${most}\:{popular}\:{one}.\:{if}\:{you}\:{want}\:{to} \\ $$$${know}\:{more},\:{go}\:{to}\:{google}! \\ $$
Answered by mr W last updated on 30/Jan/20
![∫_0 ^2 [x^2 ]dx =∫_0 ^1 [x^2 ]dx+∫_1 ^(√2) [x^2 ]dx+∫_(√2) ^(√3) [x^2 ]dx+∫_(√3) ^2 [x^2 ]dx =∫_0 ^1 (0)dx+∫_1 ^(√2) (1)dx+∫_(√2) ^(√3) (2)dx+∫_(√3) ^2 (3)dx =0+((√2)−1)+2((√3)−(√2))+3(2−(√3)) =5−(√2)−(√3) ≈1.854 generally ∫_0 ^n [x^2 ]dx=n(n^2 −1)−Σ_(k=1) ^(n^2 −1) (√k) example: ∫_0 ^(100) [x^2 ]dx=100×9999−Σ_(k=1) ^(9999) (√k)≈333283.541](Q79914.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}} \left[{x}^{\mathrm{2}} \right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left[{x}^{\mathrm{2}} \right]{dx}+\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left[{x}^{\mathrm{2}} \right]{dx}+\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} \left[{x}^{\mathrm{2}} \right]{dx}+\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \left[{x}^{\mathrm{2}} \right]{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{0}\right){dx}+\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \left(\mathrm{1}\right){dx}+\int_{\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{3}}} \left(\mathrm{2}\right){dx}+\int_{\sqrt{\mathrm{3}}} ^{\mathrm{2}} \left(\mathrm{3}\right){dx} \\ $$$$=\mathrm{0}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$$$=\mathrm{5}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}} \\ $$$$\approx\mathrm{1}.\mathrm{854} \\ $$$$ \\ $$$${generally} \\ $$$$\int_{\mathrm{0}} ^{{n}} \left[{x}^{\mathrm{2}} \right]{dx}={n}\left({n}^{\mathrm{2}} −\mathrm{1}\right)−\underset{{k}=\mathrm{1}} {\overset{{n}^{\mathrm{2}} −\mathrm{1}} {\sum}}\sqrt{{k}} \\ $$$${example}: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{100}} \left[{x}^{\mathrm{2}} \right]{dx}=\mathrm{100}×\mathrm{9999}−\underset{{k}=\mathrm{1}} {\overset{\mathrm{9999}} {\sum}}\sqrt{{k}}\approx\mathrm{333283}.\mathrm{541} \\ $$
Commented by Pratah last updated on 29/Jan/20
$$\mathrm{great} \\ $$