certificate: lim_(n→∞) ∫_0 ^1 (n/(1+n^2 x^2 ))e^x^3 dx=(π/2).


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 212600 by MrGaster last updated on 18/Oct/24

$certificate :$ $\lim_{n \to \infty} \int_{0}^{1} \frac{n}{1 + n^{2} x^{2}} e^{x^{3}} d x = \frac{π}{2} .$
	Answered by Berbere last updated on 21/Oct/24
	$=lim_(n→∞) ∫^1 _0 ((d(nx))/(1+(nx)^2 ))e^x^3 A=lim_(n→∞) {[^1 _0 tan^(−1) (nx)e^x^3 ]−∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)dx} =(π/2)e−lim_(n→∞) ∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)dx (π/2)−(1/x)≤tan^(−1) (x)=(π/2)−tan^(−1) ((1/x))≤(π/2) ((π/2)−(1/(nx)))3x^2 e^x^3 ≤3x^2 e^x^3 tan^(−1) (nx)≤(π/2).3x^2 e^x^3 lim_(n→∞) (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3 ≤u=lim_(n→∞) ∫_0 ^1 3x^2 e^x^3 tan^(−1) (nx)≤(π/2)e−(π/2) (π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^2 dx≤(π/2)e−(π/2)−(1/n)∫_0 ^1 3xe^x^3 ≤u≤(π/2)e−(π/2) lim_(n→∞) (π/2)e−(π/2)−(3/(2n))(e−1)≤u≤(π/2)e−(π/2) u=(π/2)e−(π/2);A=(π/2)e−u=(π/2)$
	$Extra \left or missing \right$ $= \frac{π}{2} e - \lim_{n \to \infty} \int_{0}^{1} 3 x^{2} e^{x^{3}} \tan^{- 1} (n x) d x$ $\frac{π}{2} - \frac{1}{x} ⩽ \tan^{- 1} (x) = \frac{π}{2} - \tan^{- 1} (\frac{1}{x}) ⩽ \frac{π}{2}$ $(\frac{π}{2} - \frac{1}{n x}) 3 x^{2} e^{x^{3}} ⩽ 3 x^{2} e^{x^{3}} \tan^{- 1} (n x) ⩽ \frac{π}{2} . 3 x^{2} e^{x^{3}}$ $\lim_{n \to \infty} \frac{π}{2} e - \frac{π}{2} - \frac{1}{n} \int_{0}^{1} 3 {x e}^{x^{3}} ⩽ u = \lim_{n \to \infty} \int_{0}^{1} 3 x^{2} e^{x^{3}} \tan^{- 1} (n x) ⩽ \frac{π}{2} e - \frac{π}{2}$ $\frac{π}{2} e - \frac{π}{2} - \frac{1}{n} \int_{0}^{1} 3 {x e}^{x^{2}} d x ⩽ \frac{π}{2} e - \frac{π}{2} - \frac{1}{n} \int_{0}^{1} 3 {x e}^{x^{3}} ⩽ u ⩽ \frac{π}{2} e - \frac{π}{2}$ $\lim_{n \to \infty} \frac{π}{2} e - \frac{π}{2} - \frac{3}{2 n} (e - 1) ⩽ u ⩽ \frac{π}{2} e - \frac{π}{2}$ $u = \frac{π}{2} e - \frac{π}{2}; A = \frac{π}{2} e - u = \frac{π}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum