Integration Questions

Question Number 81549 by jagoll last updated on 13/Feb/20

$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\lfloor\mathrm{x}\rfloor^{\mathrm{2}} \:\mathrm{dx}\:=\: \\$$$$\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\lfloor\mathrm{x}^{\mathrm{2}} \rfloor\mathrm{dx}= \\$$

Commented by jagoll last updated on 14/Feb/20

$$\mathrm{formula}\:\underset{\mathrm{1}} {\overset{\mathrm{n}+\mathrm{1}} {\int}}\lfloor\mathrm{x}\rfloor^{\mathrm{2}} \mathrm{dx}\:=\:\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{n}^{\mathrm{2}} \\$$$$=\:\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{2n}+\mathrm{1}\right)}{\mathrm{6}} \\$$

Commented by jagoll last updated on 14/Feb/20

$$\underset{\mathrm{1}} {\overset{\mathrm{4}} {\int}}\lfloor\mathrm{x}^{\mathrm{2}} \rfloor\mathrm{dx}=\:\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\mathrm{2}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}\right)+ \\$$$$...+\mathrm{15}\left(\sqrt{\mathrm{16}}−\sqrt{\mathrm{15}}\right) \\$$$$=\:\mathrm{60}\:−\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+...+\sqrt{\mathrm{15}}\right) \\$$

Commented by mathmax by abdo last updated on 14/Feb/20

$$\int_{\mathrm{0}} ^{\mathrm{4}} \left[{x}\right]^{\mathrm{2}} \:{dx}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:\int_{{k}} ^{{k}+\mathrm{1}} {k}^{\mathrm{2}} \:{dx}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{3}} \:{k}^{\mathrm{2}} \:=\mathrm{1}^{\mathrm{2}} \:+\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} =\mathrm{1}+\mathrm{4}\:+\mathrm{9}=\mathrm{14} \\$$

Commented by mathmax by abdo last updated on 14/Feb/20

$$\int_{\mathrm{0}} ^{\mathrm{4}} \left[{x}^{\mathrm{2}} \right]{dx}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{15}} \:\int_{\sqrt{{k}}} ^{\sqrt{{k}+\mathrm{1}}} \:\:\left[{x}^{\mathrm{2}} \right]{dx}\:=\sum_{{k}=\mathrm{1}} ^{\mathrm{15}} \:\int_{\sqrt{{k}}} ^{\sqrt{{k}+\mathrm{1}}} {kdx} \\$$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{15}} {k}\left(\sqrt{{k}+\mathrm{1}}−\sqrt{{k}}\right)\:=\sqrt{\mathrm{2}}−\mathrm{1}\:+\mathrm{2}\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)+\mathrm{3}\left(\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}}\right) \\$$$$+....+\mathrm{15}\left(\sqrt{\mathrm{16}}−\sqrt{\mathrm{15}}\right) \\$$

Commented by mathmax by abdo last updated on 14/Feb/20

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