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Question Number 31499 by abdo imad last updated on 09/Mar/18

$$findthepolynialpwichverifyp\left(x\right)-p^{{}^{\prime }}\left(x\right)=x^{n}then$$$$calculate{\int }_0^{1}p\left(x\right)dx.$$

Commented by abdo imad last updated on 14/Mar/18

$$letputp\left(x\right)=\sum _{k=0}^n{a}_k{x}^{k}wehave{p}^{{}^{\prime }}\left(x\right)=\sum _{k=1}^{n}k{a}_k{x}^{k-1}$$$$p\left(x\right)-p^{{}^{\prime }}\left(x\right)=x^n\iff \sum _{k=0}^n{a}_k{x}^k-\sum _{k=1}^{n}k{a}_k{x}^{k-1}=x^n$$$$\iff \sum _{k=0}^n{a}_k{x}^k-\sum _{k=0}^{n-1}(k+1)a_{k+1}{x}^k=x^n\iff$$$$\sum _{k=0}^{n-1}(a_k-(k+1)a_{k+1})x^k+a_n{x}^n=x^n\iff$$$$a_k-(k+1)a_{k+1}=0\mathrm{\forall }k\in \left[[0,n-1]\right]and{a}_n=1\iff a_n=1and$$$$a_{k+1}=\frac{a_k}{k+1}\Rightarrow \prod _{k=0}^{n-1}{a}_{k+1}=\frac{\prod _{k=0}^{n-1}{a}_k}{\prod _{k=0}^{n-1}(k+1)}\Rightarrow$$$$a_1.a_2...a_n=\frac{1}{n!}{a}_0.a_1.a_2...a_{n-1}\Rightarrow a_n=\frac{a_0}{n!}\Rightarrow$$$$p\left(x\right)=\sum _{k=0}^{n-1}\frac{a_0}{k!}{x}^k+x^n$$$$p\left(o\right)-p^{{}^{\prime }}\left(0\right)=0\Rightarrow a_0=0\Rightarrow p\left(x\right)=\sum _{k=1}^{n-1}\frac{x^k}{k!}+x^n$$$$2){\int }_0^{1}p\left(x\right)dx=\sum _{k=1}^{n-1}\frac{1}{k!}{\int }_0^1{x}^{k}dx+\frac{1}{n+1}$$$$=\sum _{k=1}^{n-1}\frac{1}{(k+1)!}+\frac{1}{n+1}$$$$$$

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