Integration Questions

Question Number 58222 by salahahmed last updated on 20/Apr/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{x}} {dx} \\$$

Commented bymaxmathsup by imad last updated on 21/Apr/19

$${we}\:{have}\:{x}^{{x}} \:={e}^{{xln}\left({x}\right)} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{x}} {dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} \left({ln}\left({x}\right)\right)^{{n}} }{{n}!}\right){dx} \\$$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \left({ln}\left({x}\right)\right)^{{n}} \:{dx}\:\:{let}\:{A}_{{n},{p}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \left({ln}\left({x}\right)\right)^{{p}} {dx}\:\:\:{by}\:{parts}\: \\$$ $${u}^{'} \:={x}^{{n}} \:{and}\:{v}\:=\left({ln}\left({x}\right)\right)^{{p}} \:\Rightarrow{u}\:=\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:{and}\:{v}^{'} \:=\frac{{p}}{{x}}\left({ln}\left({x}\right)\right)^{{p}−\mathrm{1}} \:\Rightarrow \\$$ $${A}_{{n},{p}} \:=\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:\left({ln}\left({x}\right)\right)^{{p}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{x}^{{n}+\mathrm{1}} \:\frac{{p}}{{x}}\:\left({ln}\left({x}\right)\right)^{{p}−\mathrm{1}} \:{dx} \\$$ $$=−\frac{{p}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:\:\left({ln}\left({x}\right)\right)^{{p}−\mathrm{1}} \:=−\frac{{p}}{{n}+\mathrm{1}}\:{A}_{{n},{p}−\mathrm{1}} \:=\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} {p}\left({p}−\mathrm{1}\right)}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:{A}_{{n},{p}−\mathrm{2}} \\$$ $$=\frac{\left(−\mathrm{1}\right)^{{p}} \:{p}!}{\left({n}+\mathrm{1}\right)^{{p}} }\:{A}_{{n},\mathrm{0}} \:\:\:\:\:{but}\:\:{A}_{{n},{o}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \:=\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\Rightarrow \\$$ $${A}_{{n},{p}} =\frac{\left(−\mathrm{1}\right)^{{p}} {p}!}{\left({n}+\mathrm{1}\right)^{{p}+\mathrm{1}} }\:\Rightarrow{A}_{{n},{n}} =\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\Rightarrow \\$$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{x}} \:{dx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }\:−\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }\:+.... \\$$

Commented bymaxmathsup by imad last updated on 21/Apr/19

$${if}\:{we}\:{want}\:{a}\:{best}\:{approximation}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{x}} \:{dx}\:\:{we}\:{can}\:{take}\:\mathrm{10}\:{terms}\:{of}\:{the}\: \\$$ $${serie}\:. \\$$

Commented bysalahahmed last updated on 21/Apr/19

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\$$

Commented bymaxmathsup by imad last updated on 21/Apr/19

$${you}\:{are}\:{welcome}\:. \\$$

Answered by Kunal12588 last updated on 20/Apr/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{x}} \:{dx}=\left[\frac{{x}^{{x}+\mathrm{1}} }{{x}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$ $$=\frac{\mathrm{1}^{\mathrm{1}+\mathrm{1}} }{\mathrm{1}+\mathrm{1}}−\frac{\mathrm{0}^{\mathrm{0}+\mathrm{1}} }{\mathrm{0}+\mathrm{1}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0}=\frac{\mathrm{1}}{\mathrm{2}} \\$$ $${is}\:{this}\:{correct}? \\$$

Commented bymr W last updated on 20/Apr/19

$${that}'{s}\:{wrong}\:{sir},\:{because} \\$$ $$\left(\frac{{x}^{{x}+\mathrm{1}} }{{x}+\mathrm{1}}\right)^{'} \neq{x}^{{x}} . \\$$ $${in}\:\:\int{x}^{{n}} {dx}=\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}+{C}\:\:\:{n}\:{must}\:{be}\:{constant}\:{w}.{r}.{t}.\:{x}. \\$$

Commented byKunal12588 last updated on 20/Apr/19

Ohh.. Thanks sir. So how can we solve that.

Answered by MJS last updated on 20/Apr/19

$$\approx.\mathrm{7834305109} \\$$

Commented bysalahahmed last updated on 20/Apr/19

$$\mathrm{how}?\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{way}? \\$$

Commented byMJS last updated on 20/Apr/19

$$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate}\:\mathrm{by}\:\mathrm{lower}\:\mathrm{sums} \\$$ $$\mathrm{and}\:\mathrm{upper}\:\mathrm{sums} \\$$