Question Number 200603 by Calculusboy last updated on 20/Nov/23
Answered by witcher3 last updated on 20/Nov/23
![x^2 =(n−x)^2 −2n(n−x)+n^2 ∫_0 ^n (n−x)^(p+2) dx−2n∫_0 ^n (n−x)^(p+1) +n^2 ∫_0 ^n (n−x)^p dx −(((n−x)^(p+3) )/(p+3))]_0 ^n +2n(((n−x)^(p+2) )/(p+2))]_0 ^n −(n^2 /(p+1))(n−x)^(p+1) ]_0 ^n =(n^(p+3) /(p+3))−((2n.n^(p+2) )/(p+2))+(n^(p+3) /(p+1)) =n^(p+3) ((1/(p+3))−(2/(p+2))+(1/(p+1))) =((n^(p+3) ((p^2 +3p+2)−2(p^2 +4p+3)+p^2 +5p+6))/((p+1)(p+2)(p+3))) =((2n^(p+3) )/((p+1)(p+2)(p+3)))](https://www.tinkutara.com/question/Q200609.png)
$$\mathrm{x}^{\mathrm{2}} =\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{2n}\left(\mathrm{n}−\mathrm{x}\right)+\mathrm{n}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}+\mathrm{2}} \mathrm{dx}−\mathrm{2n}\int_{\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}+\mathrm{1}} +\mathrm{n}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{n}} \left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}} \mathrm{dx} \\ $$$$\left.−\left.\frac{\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}+\mathrm{3}} }{\mathrm{p}+\mathrm{3}}\left.\right]_{\mathrm{0}} ^{\mathrm{n}} +\mathrm{2n}\frac{\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}+\mathrm{2}} }{\mathrm{p}+\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{n}} −\frac{\mathrm{n}^{\mathrm{2}} }{\mathrm{p}+\mathrm{1}}\left(\mathrm{n}−\mathrm{x}\right)^{\mathrm{p}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{n}} \\ $$$$=\frac{\mathrm{n}^{\mathrm{p}+\mathrm{3}} }{\mathrm{p}+\mathrm{3}}−\frac{\mathrm{2n}.\mathrm{n}^{\mathrm{p}+\mathrm{2}} }{\mathrm{p}+\mathrm{2}}+\frac{\mathrm{n}^{\mathrm{p}+\mathrm{3}} }{\mathrm{p}+\mathrm{1}} \\ $$$$=\mathrm{n}^{\mathrm{p}+\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{p}+\mathrm{3}}−\frac{\mathrm{2}}{\mathrm{p}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{p}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{n}^{\mathrm{p}+\mathrm{3}} \left(\left(\mathrm{p}^{\mathrm{2}} +\mathrm{3p}+\mathrm{2}\right)−\mathrm{2}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{4p}+\mathrm{3}\right)+\mathrm{p}^{\mathrm{2}} +\mathrm{5p}+\mathrm{6}\right)}{\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{p}+\mathrm{2}\right)\left(\mathrm{p}+\mathrm{3}\right)} \\ $$$$=\frac{\mathrm{2n}^{\mathrm{p}+\mathrm{3}} }{\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{p}+\mathrm{2}\right)\left(\mathrm{p}+\mathrm{3}\right)} \\ $$
Commented by Calculusboy last updated on 21/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$