Question Number 215458 by depressiveshrek last updated on 07/Jan/25
$$\mathrm{Evaluate}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}^{\mathrm{2}} \:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}} \mathrm{sin}\left(\pi{x}\right){dx} \\ $$
Answered by MathematicalUser2357 last updated on 11/Jan/25
![lim_(n→∞) n^2 ∫_0 ^1 x^(n+1) sin πxdx=lim_(n→∞) ∫_0 ^1 n^2 x^(n+1) sin πxdx=lim_(n→∞) [n^2 ∫x^(n+1) sin πxdx]_0 ^1 determinant ((D,I),((x^(n+1) =(((n+1)!)/((n+1)!))x^(n+1) ),(sin πx)),(((n+1)x^n =(((n+1)!)/(n!))x^n ),(−(1/π)cos πx)),(((n+1)nx^(n−1) =(((n+1)!)/((n−1)!))x^(n−1) ),(−(1/π^2 )sin πx)),(((((n+1)!)/((n−2)!))x^(n−2) ),((1/π^3 )cos πx)),((...),((1/π^4 )sin πx))) =lim_(n→∞) [Σ_(k=0) ^∞ n^2 (−1)^(k+1) {(((n+1)!)/(π^(2k+1) (n+1−2k)!))x^(n+1−2k) cos πx−(((n+1)!)/(π^(2k+2) (n−2k)!))x^(n−2k) sin πx}]_0 ^1 =0](https://www.tinkutara.com/question/Q215586.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+\mathrm{1}} \mathrm{sin}\:\pi{xdx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} {n}^{\mathrm{2}} {x}^{{n}+\mathrm{1}} \mathrm{sin}\:\pi{xdx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[{n}^{\mathrm{2}} \int{x}^{{n}+\mathrm{1}} \mathrm{sin}\:\pi{xdx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{D}}&\hline{{I}}\\{{x}^{{n}+\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}{x}^{{n}+\mathrm{1}} }&\hline{\mathrm{sin}\:\pi{x}}\\{\left({n}+\mathrm{1}\right){x}^{{n}} =\frac{\left({n}+\mathrm{1}\right)!}{{n}!}{x}^{{n}} }&\hline{−\frac{\mathrm{1}}{\pi}\mathrm{cos}\:\pi{x}}\\{\left({n}+\mathrm{1}\right){nx}^{{n}−\mathrm{1}} =\frac{\left({n}+\mathrm{1}\right)!}{\left({n}−\mathrm{1}\right)!}{x}^{{n}−\mathrm{1}} }&\hline{−\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\mathrm{sin}\:\pi{x}}\\{\frac{\left({n}+\mathrm{1}\right)!}{\left({n}−\mathrm{2}\right)!}{x}^{{n}−\mathrm{2}} }&\hline{\frac{\mathrm{1}}{\pi^{\mathrm{3}} }\mathrm{cos}\:\pi{x}}\\{…}&\hline{\frac{\mathrm{1}}{\pi^{\mathrm{4}} }\mathrm{sin}\:\pi{x}}\\\hline\end{array} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{n}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left\{\frac{\left({n}+\mathrm{1}\right)!}{\pi^{\mathrm{2}{k}+\mathrm{1}} \left({n}+\mathrm{1}−\mathrm{2}{k}\right)!}{x}^{{n}+\mathrm{1}−\mathrm{2}{k}} \mathrm{cos}\:\pi{x}−\frac{\left({n}+\mathrm{1}\right)!}{\pi^{\mathrm{2}{k}+\mathrm{2}} \left({n}−\mathrm{2}{k}\right)!}{x}^{{n}−\mathrm{2}{k}} \mathrm{sin}\:\pi{x}\right\}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{0} \\ $$