Question Number 92799 by Ar Brandon last updated on 09/May/20
![Show that the function x→x^3 is of Riemann within the interval [−1,2] then calculate ∫_(−1) ^2 x^2 dx](https://www.tinkutara.com/question/Q92799.png)
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{x}\rightarrow\mathrm{x}^{\mathrm{3}} \:\mathrm{is} \\ $$$$\mathrm{of}\:\mathrm{Riemann}\:\mathrm{within}\:\mathrm{the}\:\mathrm{interval}\:\left[−\mathrm{1},\mathrm{2}\right] \\ $$$$\mathrm{then}\:\mathrm{calculate}\:\int_{−\mathrm{1}} ^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \mathrm{dx} \\ $$
Commented by mathmax by abdo last updated on 09/May/20
![∫_(−1) ^2 x^2 dx =[(1/3)x^3 ]_(−1) ^2 =(1/3)(2^3 −(−1)^3 ) =(1/3)(8+1) =3 we know ∫_a ^b f(x)dx =lim_(n→∞) ((b−a)/n)Σ_(k=1) ^n f(a+((k(b−a))/n)) ⇒ ∫_(−1) ^2 x^2 dx =_(x=t−1) ∫_0 ^3 (t−1)^2 dt =∫_0 ^3 (t^2 −2t+1)dt =∫_0 ^3 t^2 dt −2∫_0 ^3 t dt +3 ∫_0 ^3 t^2 dt =lim_(n→+∞) (3/n)Σ_(k=1) ^n (((3k)/n))^2 =lim_(n→+∞) ((27)/n^3 )×((n(n+1)(2n+1))/6) =lim_(n→+∞) ((27)/6)×((2n^3 )/n^3 ) =9 ∫_0 ^3 tdt =lim_(n→+∞) (3/n)Σ_(k=1) ^n ((3k)/n) =lim_(n→+∞) (9/n^2 )×((n(n+1))/2) =(9/2) ⇒∫_(−1) ^2 x^2 dx =9−2×(9/2) +3 =3](https://www.tinkutara.com/question/Q92907.png)
$$\int_{−\mathrm{1}} ^{\mathrm{2}} \:{x}^{\mathrm{2}} \:{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} \right]_{−\mathrm{1}} ^{\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{2}^{\mathrm{3}} −\left(−\mathrm{1}\right)^{\mathrm{3}} \right)\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{8}+\mathrm{1}\right)\:=\mathrm{3} \\ $$$${we}\:{know} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}\:={lim}_{{n}\rightarrow\infty} \:\:\frac{{b}−{a}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{f}\left({a}+\frac{{k}\left({b}−{a}\right)}{{n}}\right)\:\Rightarrow \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{2}} \:{x}^{\mathrm{2}} \:{dx}\:=_{{x}={t}−\mathrm{1}} \:\:\int_{\mathrm{0}} ^{\mathrm{3}} \left({t}−\mathrm{1}\right)^{\mathrm{2}} \:{dt}\:=\int_{\mathrm{0}} ^{\mathrm{3}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{3}} \:{t}^{\mathrm{2}} \:{dt}\:−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{3}} \:{t}\:{dt}\:+\mathrm{3} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \:{t}^{\mathrm{2}} {dt}\:={lim}_{{n}\rightarrow+\infty} \:\frac{\mathrm{3}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \left(\frac{\mathrm{3}{k}}{{n}}\right)^{\mathrm{2}} \:={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{27}}{{n}^{\mathrm{3}} }×\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$$$={lim}_{{n}\rightarrow+\infty} \:\:\frac{\mathrm{27}}{\mathrm{6}}×\frac{\mathrm{2}{n}^{\mathrm{3}} }{{n}^{\mathrm{3}} }\:=\mathrm{9} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{3}} \:{tdt}\:={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{3}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{3}{k}}{{n}}\:={lim}_{{n}\rightarrow+\infty} \frac{\mathrm{9}}{{n}^{\mathrm{2}} }×\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$=\frac{\mathrm{9}}{\mathrm{2}}\:\Rightarrow\int_{−\mathrm{1}} ^{\mathrm{2}} \:{x}^{\mathrm{2}} \:{dx}\:=\mathrm{9}−\mathrm{2}×\frac{\mathrm{9}}{\mathrm{2}}\:+\mathrm{3}\:=\mathrm{3} \\ $$
Commented by Ar Brandon last updated on 10/May/20
thanks bro