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Question Number 92799 by Ar Brandon last updated on 09/May/20

$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{function}\mathrm{x}\to {\mathrm{x}}^3\mathrm{is}$$$$\mathrm{of}\mathrm{Riemann}\mathrm{within}\mathrm{the}\mathrm{interval}[-1,2]$$$$\mathrm{then}\mathrm{calculate}{\int }_{-1}^2{\mathrm{x}}^2\mathrm{dx}$$

Commented by mathmax by abdo last updated on 09/May/20

$${\int }_{-1}^2{x}^{2}dx={\left[\frac{1}{3}{x}^3\right]}_{-1}^2=\frac{1}{3}(2^3-{(-1)}^3)=\frac{1}{3}(8+1)=3$$$$weknow$$$${\int }_a^{b}f\left(x\right)dx={lim}_{n\to \mathrm{\infty }}\frac{b-a}{n}\sum _{k=1}^{n}f(a+\frac{k(b-a)}{n})\Rightarrow$$$${\int }_{-1}^2{x}^{2}dx{=}_{x=t-1}{\int }_0^3{(t-1)}^{2}dt={\int }_0^3(t^2-2t+1)dt$$$$={\int }_0^3{t}^{2}dt-2{\int }_0^{3}tdt+3$$$${\int }_0^3{t}^{2}dt={lim}_{n\to +\mathrm{\infty }}\frac{3}{n}\sum _{k=1}^n{\left(\frac{3k}{n}\right)}^2={lim}_{n\to +\mathrm{\infty }}\frac{27}{n^3}\times \frac{n(n+1)(2n+1)}{6}$$$$={lim}_{n\to +\mathrm{\infty }}\frac{27}{6}\times \frac{2n^3}{n^3}=9$$$${\int }_0^{3}tdt={lim}_{n\to +\mathrm{\infty }}\frac{3}{n}\sum _{k=1}^n\frac{3k}{n}={lim}_{n\to +\mathrm{\infty }}\frac{9}{n^2}\times \frac{n(n+1)}{2}$$$$=\frac{9}{2}\Rightarrow {\int }_{-1}^2{x}^{2}dx=9-2\times \frac{9}{2}+3=3$$

Commented by Ar Brandon last updated on 10/May/20

thanks bro

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