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Question-212762




Question Number 212762 by universe last updated on 23/Oct/24
Commented by MrGaster last updated on 23/Oct/24
Rewrite the integrand function as:  (x−1)(x−2)…(x−n)=(((−1)^n )/(n!))x(x−1)(x−2)…(x−n)  Computational integral:  ∫_0 ^n (x−1)−(x−2)…(x−n)dx=∫_0 ^n (((−1)^n )/(n!))x(x−1)(x−2)…(x−n)dx  take notice of,x(x−1)(x−2)…(x−n)  Is a descending factorial polynomial.Tayi  x=0,1,2,…n The value is 0.So the integral is divided into two parts.:  ∫_0 ^n (x−1)(x−2)…(x−n)dx=∫_0 ^1 (x−1)(x−2)…(x−n)dx∫_1 ^2 (x−1)(x−2)…(x−n)dx+…+∫_(n+1) ^n (x−1)(x−2)…(x−n)dx  Between each cell[k,k+1],function  (x−1)(x−2)…(x−k)(x−k)(x−k)(−(k+1))…(x−n)  The sign of is the same so the absolutel  vaue can be extracted and the integralt  beween each cell can be calculated.  because[k,k+1]Symbol invariance can be written as:  ∫_k ^(k+1) ∣x−k∣x−(k+1)∣…∣x−n∣dx  Because the integration result betweenc  eah cell will contain a factor k!   When divided by n! when   These integral results will be eliminatedo  s the whole integral result is 0.  lim_(n→∞) (((−1)^n )/(n!))×0=0
$$\mathrm{Rewrite}\:\mathrm{the}\:\mathrm{integrand}\:\mathrm{function}\:\mathrm{as}: \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{Computational}\:\mathrm{integral}: \\ $$$$\int_{\mathrm{0}} ^{{n}} \left({x}−\mathrm{1}\right)−\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx}=\int_{\mathrm{0}} ^{{n}} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx} \\ $$$$\mathrm{take}\:\mathrm{notice}\:\mathrm{of},{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{Is}\:\mathrm{a}\:\mathrm{descending}\:\mathrm{factorial}\:\mathrm{polynomial}.\mathrm{Tayi} \\ $$$${x}=\mathrm{0},\mathrm{1},\mathrm{2},\ldots{n}\:\mathrm{The}\:\mathrm{value}\:\mathrm{is}\:\mathrm{0}.\mathrm{So}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts}.: \\ $$$$\int_{\mathrm{0}} ^{{n}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx}\int_{\mathrm{1}} ^{\mathrm{2}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx}+\ldots+\int_{{n}+\mathrm{1}} ^{{n}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{n}\right){dx} \\ $$$$\mathrm{Between}\:\mathrm{each}\:\mathrm{cell}\left[{k},{k}+\mathrm{1}\right],\mathrm{function} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\ldots\left({x}−{k}\right)\left({x}−{k}\right)\left({x}−{k}\right)\left(−\left({k}+\mathrm{1}\right)\right)\ldots\left({x}−{n}\right) \\ $$$$\mathrm{The}\:\mathrm{sign}\:\mathrm{of}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{so}\:\mathrm{the}\:\mathrm{absolutel} \\ $$$$\mathrm{vaue}\:\mathrm{can}\:\mathrm{be}\:\mathrm{extracted}\:\mathrm{and}\:\mathrm{the}\:\mathrm{integralt} \\ $$$$\mathrm{beween}\:\mathrm{each}\:\mathrm{cell}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated}. \\ $$$$\mathrm{because}\left[{k},{k}+\mathrm{1}\right]\mathrm{Symbol}\:\mathrm{invariance}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}: \\ $$$$\int_{{k}} ^{{k}+\mathrm{1}} \mid{x}−{k}\mid{x}−\left({k}+\mathrm{1}\right)\mid\ldots\mid{x}−{n}\mid{dx} \\ $$$$\mathrm{Because}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{result}\:\mathrm{betweenc} \\ $$$$\mathrm{eah}\:\mathrm{cell}\:\mathrm{will}\:\mathrm{contain}\:\mathrm{a}\:\mathrm{factor}\:{k}!\: \\ $$$$\mathrm{When}\:\mathrm{divided}\:\mathrm{by}\:{n}!\:\mathrm{when}\: \\ $$$$\mathrm{These}\:\mathrm{integral}\:\mathrm{results}\:\mathrm{will}\:\mathrm{be}\:\mathrm{eliminatedo} \\ $$$$\mathrm{s}\:\mathrm{the}\:\mathrm{whole}\:\mathrm{integral}\:\mathrm{result}\:\mathrm{is}\:\mathrm{0}. \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}×\mathrm{0}=\mathrm{0} \\ $$
Answered by mathmax last updated on 23/Oct/24
p(x)=(x−1)(x−2).....(x−n)est un polynome de degren  ⇒p(x)=x^n +ax^(n−1) +bx^(n−2) +......+l ⇒  ∫_0 ^n p(x)dx=[(x^(n+1) /(n+1))+((ax^n )/n)+.....+lx]_0 ^n   =(n^(n+1) /(n+1))+((an^n )/n)+.....+ln ⇒  (((−1)^n )/(n!))∫_0 ^n p(x)dx=(((−1)^n )/(n!)){(n^(n+1) /(n+1))+((an^n )/n)+....+ln}  et ∣(((−1)^n )/(n!))∫_0 ^n p(x)dx∣≤(1/(n!))(n^(n+1) /(n+1))+∣a∣(n^n /(n!n))  +.....∣l∣(n/(n!))→0 (n→∞)  c est la factorielle qui commmande ⇒  lim_(n→∞) (((−1)^n )/(n!))∫_0 ^n (x−1)(x−2)...(x−n)dx =0
$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…..\left({x}−{n}\right){est}\:{un}\:{polynome}\:{de}\:{degren} \\ $$$$\Rightarrow{p}\left({x}\right)={x}^{{n}} +{ax}^{{n}−\mathrm{1}} +{bx}^{{n}−\mathrm{2}} +……+{l}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{{n}} {p}\left({x}\right){dx}=\left[\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{{ax}^{{n}} }{{n}}+…..+{lx}\right]_{\mathrm{0}} ^{{n}} \\ $$$$=\frac{{n}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{{an}^{{n}} }{{n}}+…..+{ln}\:\Rightarrow \\ $$$$\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{n}} {p}\left({x}\right){dx}=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\left\{\frac{{n}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\frac{{an}^{{n}} }{{n}}+….+{ln}\right\} \\ $$$${et}\:\mid\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{n}} {p}\left({x}\right){dx}\mid\leqslant\frac{\mathrm{1}}{{n}!}\frac{{n}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+\mid{a}\mid\frac{{n}^{{n}} }{{n}!{n}} \\ $$$$+…..\mid{l}\mid\frac{{n}}{{n}!}\rightarrow\mathrm{0}\:\left({n}\rightarrow\infty\right) \\ $$$${c}\:{est}\:{la}\:{factorielle}\:{qui}\:{commmande}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{n}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\left({x}−{n}\right){dx}\:=\mathrm{0} \\ $$

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