Question Number 4900 by prakash jain last updated on 19/Mar/16
$$\int_{{n}} ^{\:{n}+\mathrm{1}} {f}\left({x}\right)\mathrm{d}{x},\:{n}\in\mathbb{N} \\ $$$${where}\:{f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\prod}}\left({x}−{i}\right) \\ $$$$\mathrm{Can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{integrated}\:\mathrm{with}\:\mathrm{evaluating}\:\mathrm{the} \\ $$$$\mathrm{product}? \\ $$
Commented by Yozzii last updated on 20/Mar/16
![It certainly is possible, but with direct expansion of f(x), evaluating will be terribly difficult. A Riemann series representation is possible for the value of the integral. f(x)=x(x−1)(x−2)...(x−n)(x−n−1) ⇒f(1)=0. f(n)=0,f(n+1)=0. Use right end−points on the curve of y=f(x) on the interval [n,n+1]. Let the area between the curve and y=f(x) be regularly partitioned into strips each of width q=((n+1−n)/N)=(1/N) where N is the number of strips (N∈N). The height of each strip is given by f(n+(j/N))=Π_(i=0) ^(n+1) (n+(j/N)−i) where 1≤j≤N So ∫_n ^(n+1) f(x)dx=lim_(N→∞) [Σ_(j=1) ^N {f((j/N)+n)×(1/N)}] ∫_n ^(n+1) f(x)dx=lim_(N→∞) {(1/N)(f((1/N)+n)+f((2/N)+n)+f((3/N)+n)+f((4/N)+n)+...+f(((N−1)/N)+n)+f(1+n))} ∫_n ^(n+1) f(x)dx=lim_(N→∞) {(1/N)(f((1/N)+n)+f((2/N)+n)+f((3/N)+n)+f((4/N)+n)+...+f(((N−1)/N)+n))}](https://www.tinkutara.com/question/Q4903.png)
$${It}\:{certainly}\:{is}\:{possible},\:{but}\:{with} \\ $$$${direct}\:{expansion}\:{of}\:{f}\left({x}\right),\:{evaluating} \\ $$$${will}\:{be}\:{terribly}\:{difficult}.\:{A}\:{Riemann}\:{series} \\ $$$${representation}\:{is}\:{possible}\:{for} \\ $$$${the}\:{value}\:{of}\:{the}\:{integral}. \\ $$$${f}\left({x}\right)={x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)…\left({x}−{n}\right)\left({x}−{n}−\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{0}.\:{f}\left({n}\right)=\mathrm{0},{f}\left({n}+\mathrm{1}\right)=\mathrm{0}. \\ $$$${Use}\:{right}\:{end}−{points}\:{on}\:{the}\:{curve}\:{of} \\ $$$${y}={f}\left({x}\right)\:{on}\:{the}\:{interval}\:\left[{n},{n}+\mathrm{1}\right]. \\ $$$${Let}\:{the}\:{area}\:{between}\:{the}\:{curve}\:{and} \\ $$$${y}={f}\left({x}\right)\:{be}\:{regularly}\:{partitioned}\:{into} \\ $$$${strips}\:{each}\:{of}\:{width}\:{q}=\frac{{n}+\mathrm{1}−{n}}{{N}}=\frac{\mathrm{1}}{{N}} \\ $$$${where}\:{N}\:{is}\:{the}\:{number}\:{of}\:{strips}\:\left({N}\in\mathbb{N}\right). \\ $$$${The}\:{height}\:{of}\:{each}\:{strip}\:{is}\:{given}\:{by} \\ $$$${f}\left({n}+\frac{{j}}{{N}}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}+\mathrm{1}} {\prod}}\left({n}+\frac{{j}}{{N}}−{i}\right)\:{where}\:\mathrm{1}\leqslant{j}\leqslant{N} \\ $$$${So}\:\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{f}\left({x}\right){dx}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\underset{{j}=\mathrm{1}} {\overset{{N}} {\sum}}\left\{{f}\left(\frac{{j}}{{N}}+{n}\right)×\frac{\mathrm{1}}{{N}}\right\}\right] \\ $$$$\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{f}\left({x}\right){dx}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{N}}\left({f}\left(\frac{\mathrm{1}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{2}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{3}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{4}}{{N}}+{n}\right)+…+{f}\left(\frac{{N}−\mathrm{1}}{{N}}+{n}\right)+{f}\left(\mathrm{1}+{n}\right)\right)\right\} \\ $$$$\underset{{n}} {\overset{{n}+\mathrm{1}} {\int}}{f}\left({x}\right){dx}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{{N}}\left({f}\left(\frac{\mathrm{1}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{2}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{3}}{{N}}+{n}\right)+{f}\left(\frac{\mathrm{4}}{{N}}+{n}\right)+…+{f}\left(\frac{{N}−\mathrm{1}}{{N}}+{n}\right)\right)\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$