Evaluate lim_(n→∞) n^2 ∫_0 ^1 x^(n+1) sin(πx)dx


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Question Number 215458 by depressiveshrek last updated on 07/Jan/25

$Evaluate \lim_{n \to \infty} n^{2} \int_{0}^{1} x^{n + 1} \sin (π x) d x$
	Answered by MathematicalUser2357 last updated on 11/Jan/25
	$lim_(n→∞) n^2 ∫_0 ^1 x^(n+1) sin πxdx=lim_(n→∞) ∫_0 ^1 n^2 x^(n+1) sin πxdx=lim_(n→∞) [n^2 ∫x^(n+1) sin πxdx]_0 ^1 determinant ((D,I),((x^(n+1) =(((n+1)!)/((n+1)!))x^(n+1) ),(sin πx)),(((n+1)x^n =(((n+1)!)/(n!))x^n ),(−(1/π)cos πx)),(((n+1)nx^(n−1) =(((n+1)!)/((n−1)!))x^(n−1) ),(−(1/π^2 )sin πx)),(((((n+1)!)/((n−2)!))x^(n−2) ),((1/π^3 )cos πx)),((...),((1/π^4 )sin πx))) =lim_(n→∞) [Σ_(k=0) ^∞ n^2 (−1)^(k+1) {(((n+1)!)/(π^(2k+1) (n+1−2k)!))x^(n+1−2k) cos πx−(((n+1)!)/(π^(2k+2) (n−2k)!))x^(n−2k) sin πx}]_0 ^1 =0$
	$\lim_{n \to \infty} n^{2} \int_{0}^{1} x^{n + 1} \sin π x d x = \lim_{n \to \infty} \int_{0}^{1} n^{2} x^{n + 1} \sin π x d x = \lim_{n \to \infty} {[n^{2} \int x^{n + 1} \sin π x d x]}_{0}^{1}$ $\begin{array}{cccccc} D & I \\ x^{n + 1} = \frac{(n + 1)!}{(n + 1)!} x^{n + 1} & \sin π x \\ (n + 1) x^{n} = \frac{(n + 1)!}{n!} x^{n} & - \frac{1}{π} \cos π x \\ (n + 1) {n x}^{n - 1} = \frac{(n + 1)!}{(n - 1)!} x^{n - 1} & - \frac{1}{π^{2}} \sin π x \\ \frac{(n + 1)!}{(n - 2)!} x^{n - 2} & \frac{1}{π^{3}} \cos π x \\ . . . & \frac{1}{π^{4}} \sin π x \end{array}$ $= \lim_{n \to \infty} {[\underset{k = 0}{\sum^{\infty}} n^{2} {(- 1)}^{k + 1} {\frac{(n + 1)!}{π^{2 k + 1} (n + 1 - 2 k)!} x^{n + 1 - 2 k} \cos π x - \frac{(n + 1)!}{π^{2 k + 2} (n - 2 k)!} x^{n - 2 k} \sin π x}]}_{0}^{1}$ $= 0$
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