Integration Questions

Question Number 17604 by tawa tawa last updated on 08/Jul/17

$$\int_{\:\:\mathrm{0}} ^{\:\:\mathrm{n}} \:\mathrm{x}^{\mathrm{2}} \left(\mathrm{n}\:−\:\mathrm{x}\right)^{\mathrm{p}} \:\mathrm{dx}\:\:\:\:\:\:\:\mathrm{for}\:\:\:\mathrm{p}\:>\:\mathrm{0} \\$$

Answered by sma3l2996 last updated on 08/Jul/17

$${t}={n}−{x}\Rightarrow{dt}=−{dx} \\$$$${I}=\int_{\mathrm{0}} ^{{n}} {x}^{\mathrm{2}} \left({n}−{x}\right)^{{p}} {dx}=\int_{\mathrm{0}} ^{{n}} \left({n}−{t}\right)^{\mathrm{2}} {t}^{{p}} {dt}=\int_{\mathrm{0}} ^{{n}} \left({n}^{\mathrm{2}} {t}^{{p}} −\mathrm{2}{nt}^{{p}+\mathrm{1}} +{t}^{{p}+\mathrm{2}} \right){dt} \\$$$$=\left[\frac{{n}^{\mathrm{2}} }{{p}+\mathrm{1}}{t}^{{p}+\mathrm{1}} −\frac{\mathrm{2}{n}}{{p}+\mathrm{2}}{t}^{{p}+\mathrm{2}} +\frac{{t}^{{p}+\mathrm{3}} }{{p}+\mathrm{3}}\right]_{\mathrm{0}} ^{{n}} \\$$$$=\frac{{n}^{\mathrm{2}} }{{p}+\mathrm{1}}{n}^{{p}+\mathrm{1}} −\frac{\mathrm{2}{n}}{{p}+\mathrm{2}}{n}^{{p}+\mathrm{2}} +\frac{{n}^{{p}+\mathrm{3}} }{{p}+\mathrm{3}} \\$$$$={n}^{{p}+\mathrm{3}} \left(\frac{\mathrm{1}}{{p}+\mathrm{1}}−\frac{\mathrm{2}}{{p}+\mathrm{2}}+\frac{\mathrm{1}}{{p}+\mathrm{3}}\right)={n}^{{p}+\mathrm{3}} \left(\frac{\left({p}+\mathrm{3}\right)\left({p}+\mathrm{2}\right)−\mathrm{2}\left({p}+\mathrm{1}\right)\left({p}+\mathrm{3}\right)+\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\left({p}+\mathrm{3}\right)}\right) \\$$$$={n}^{{p}+\mathrm{3}} \left(\frac{{p}^{\mathrm{2}} +\mathrm{5}{p}+\mathrm{6}−\mathrm{2}{p}^{\mathrm{2}} −\mathrm{8}{p}−\mathrm{6}+{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{2}}{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\left({p}+\mathrm{3}\right)}\right) \\$$$${I}=\frac{\mathrm{2}{n}^{{p}+\mathrm{3}} }{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\left({p}+\mathrm{3}\right)}\: \\$$

Commented by tawa tawa last updated on 08/Jul/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$

Answered by alex041103 last updated on 08/Jul/17

$$\mathrm{We}\:\mathrm{use}\:\mathrm{the}\:\mathrm{substitution}\: \\$$$${u}={n}−{x}\Rightarrow−{du}={dx}\:\mathrm{and}\:{x}^{\mathrm{2}} \:=\:\left({n}−{u}\right)^{\mathrm{2}} \\$$$$\mathrm{And}\:\mathrm{we}\:\mathrm{change}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{of}\:\mathrm{integration} \\$$$${x}=\mathrm{0}\:\:\:\:{x}={n} \\$$$${u}={n}\:\:\:\:{u}=\mathrm{0} \\$$$$\Rightarrow\underset{\mathrm{0}} {\overset{{n}} {\int}}\:{x}^{\mathrm{2}} \left({n}−{x}\right)^{{p}} \:{dx}\:=\:−\underset{{n}} {\overset{\mathrm{0}} {\int}}\left({n}−{u}\right)^{\mathrm{2}} {u}^{{p}} \:{du} \\$$$$=\underset{\mathrm{0}} {\overset{{n}} {\int}}\:\left({n}^{\mathrm{2}} −\mathrm{2}{nu}+{u}^{\mathrm{2}} \right){u}^{{p}} \:{du} \\$$$$={n}^{\mathrm{2}} \underset{\mathrm{0}} {\overset{{n}} {\int}}{u}^{{p}} {du}\:−\:\mathrm{2}{n}\underset{\mathrm{0}} {\overset{{n}} {\int}}{u}^{{p}+\mathrm{1}} \:{du}\:+\:\underset{\mathrm{0}} {\overset{{n}} {\int}}{u}^{{p}+\mathrm{2}} \:{du} \\$$$$={n}^{\mathrm{2}} \:\frac{{n}^{{p}+\mathrm{1}} −\mathrm{0}}{{p}+\mathrm{1}}\:−\:\mathrm{2}{n}\frac{{n}^{{p}+\mathrm{2}} −\mathrm{0}}{{p}+\mathrm{2}}+\frac{{n}^{{p}+\mathrm{3}} −\mathrm{0}}{{p}+\mathrm{3}} \\$$$$=\frac{{n}^{{p}+\mathrm{3}} }{{p}+\mathrm{1}}−\mathrm{2}\frac{{n}^{{p}+\mathrm{3}} }{{p}+\mathrm{2}}+\frac{{n}^{{p}+\mathrm{3}} }{{p}+\mathrm{3}} \\$$$$={n}^{{p}+\mathrm{3}} \left(\frac{\mathrm{1}}{{p}+\mathrm{1}}−\frac{\mathrm{2}}{{p}+\mathrm{2}}+\frac{\mathrm{1}}{{p}+\mathrm{3}}\right) \\$$$$\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as} \\$$$$\underset{\mathrm{0}} {\overset{{n}} {\int}}\:{x}^{\mathrm{2}} \left({n}−{x}\right)^{{p}} \:{dx}\:=\:\frac{\mathrm{2}{n}^{{p}+\mathrm{3}} }{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)\left({p}+\mathrm{3}\right)} \\$$

Commented by tawa tawa last updated on 08/Jul/17

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$