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Question Number 85020 by M±th+et£s last updated on 18/Mar/20

$$solveintegration$$$${\int }_1^{2}xd\lfloor x^2\rfloor$$

Commented by mr W last updated on 18/Mar/20

$$1⩽x<\sqrt{2}:\lfloor x^2\rfloor =1,d(\lfloor x^2\rfloor )=0dx$$$$\sqrt{2}⩽x<\sqrt{3}:\lfloor x^2\rfloor =2,d(\lfloor x^2\rfloor )=0dx$$$$\sqrt{3}⩽x<2:\lfloor x^2\rfloor =3,d(\lfloor x^2\rfloor )=0dx$$$${\int }_1^{2}xd\lfloor x^2\rfloor ={\int }_1^{\sqrt{2}}xd\lfloor x^2\rfloor +{\int }_{\sqrt{2}}^{\sqrt{3}}xd\lfloor x^2\rfloor +{\int }_{\sqrt{3}}^{2}xd\lfloor x^2\rfloor$$$$={\int }_1^{\sqrt{2}}x\times 0dx+{\int }_{\sqrt{2}}^{\sqrt{3}}x\times 0dx+{\int }_{\sqrt{3}}^{2}x\times 0dx$$$$=0$$

Commented by M±th+et£s last updated on 18/Mar/20

$$ithinkthesolutionis2+\sqrt{3}+\sqrt{2}$$$$u=x\to \to du=dx$$$$v=d\lfloor x^2\rfloor \to \to v=\lfloor x^2\rfloor$$$$I={[\lfloor x^2\rfloor ]}_1^2-{\int }_1^2\lfloor x^2\rfloor dx$$$$(8-1)-\left[({\int }_1^{\sqrt{2}}\lfloor x^2\rfloor dx+{\int }_{\sqrt{2}}^{\sqrt{3}}\lfloor x^2\rfloor dx+{\int }_{\sqrt{3}}^2\lfloor x^2\rfloor dx)\right]$$$$7-[{\left[1\right]}_1^{\sqrt{2}}+{\left[2\right]}_{\sqrt{2}}^{\sqrt{3}}+{\left[3\right]}_{\sqrt{3}}^{\sqrt{2}}]$$$$7-[\sqrt{2}-1+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})]$$$$7-[-\sqrt{2}-\sqrt{3}+5]$$$$2+\sqrt{3}+\sqrt{2}$$$$$$

Commented by mr W last updated on 18/Mar/20

$$iff\left(x\right)=\lfloor x^2\rfloor ,say1⩽x<\sqrt{2},$$$$whatis\frac{df\left(x\right)}{dx}=?$$

Commented by M±th+et£s last updated on 18/Mar/20

$$idontthinkicanderivativethis$$$$butbecusedv=d\lfloor x^2\rfloor sov=\lfloor x^2\rfloor$$

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