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Question Number 99193 by bemath last updated on 19/Jun/20

Commented by I want to learn more last updated on 19/Jun/20

$$\mathrm{Sir},\mathrm{how}\mathrm{can}\mathrm{i}\mathrm{type}\mathrm{like}\mathrm{this}$$

Commented by Ar Brandon last updated on 19/Jun/20

You can do that on the home page.

Commented by bemath last updated on 19/Jun/20

Enter the editor menu, select font and color and then set the background color

Commented by I want to learn more last updated on 19/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

Answered by Ar Brandon last updated on 19/Jun/20

$$\mathrm{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\sqrt[3]{\frac{\mathrm{k}}{{\mathrm{n}}^4}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\sqrt[3]{\frac{\mathrm{k}}{\mathrm{n}}}$$$$\Rightarrow \mathrm{L}={\int }_0^1\sqrt[3]{\mathrm{x}}\mathrm{dx}={\left[\frac{3}{4}{\mathrm{x}}^{\frac{4}{3}}\right]}_0^1=\frac{3}{4}$$

Commented by bemath last updated on 19/Jun/20

$$\mathrm{yes}..\mathrm{Riemann}\mathrm{integral}.\mathrm{thank}\mathrm{you}$$

Commented by Ar Brandon last updated on 19/Jun/20

Cool��

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