find-0-1-1-x-4-dx-

Question Number 51550 by maxmathsup by imad last updated on 28/Dec/18

$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx} \\ $$

Commented by Abdo msup. last updated on 29/Dec/18

let I =∫_0 ^1 (√(1+x^4 ))dx no elementary function so we give I at form of serie we have (1+u)^α =1+((αu)/(1!)) +((α(α−1))/(2!)) u^2 + ...((α(α−1)...(α−n+1))/(n!)) u^n +...⇒ (1+x^4 )^(1/4) =1+(1/4)x^4 +(1/2)(1/4)((1/4)−1)x^8 +... (((1/4)((1/4)−1)...((1/4)−n+1))/(n!)) x^(4n) +... =1+(x^4 /4) −(3/(32)) x^8 +...⇒ I ∼ ∫_0 ^1 (1+(x^4 /4) −(3/(32)) x^8 )dx =[x +(x^5 /(20)) −(3/(9.32)) x^9 ]_0 ^1 =1+(1/(20)) −(1/(96)) =((21)/(20)) −(1/(96)) .

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:\:{no}\:{elementary}\:{function}\:{so} \\ $$$${we}\:{give}\:{I}\:{at}\:{form}\:{of}\:{serie} \\ $$$${we}\:{have}\:\left(\mathrm{1}+{u}\right)^{\alpha} \:=\mathrm{1}+\frac{\alpha{u}}{\mathrm{1}!}\:+\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}\:{u}^{\mathrm{2}} \:+ \\ $$$$…\frac{\alpha\left(\alpha−\mathrm{1}\right)…\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\:{u}^{{n}} \:+…\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}{x}^{\mathrm{4}} \:+\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right){x}^{\mathrm{8}} \:+… \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)…\left(\frac{\mathrm{1}}{\mathrm{4}}−{n}+\mathrm{1}\right)}{{n}!}\:{x}^{\mathrm{4}{n}} \:+… \\ $$$$=\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{\mathrm{3}}{\mathrm{32}}\:{x}^{\mathrm{8}} \:\:+…\Rightarrow \\ $$$${I}\:\sim\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}}\:−\frac{\mathrm{3}}{\mathrm{32}}\:{x}^{\mathrm{8}} \right){dx} \\ $$$$=\left[{x}\:+\frac{{x}^{\mathrm{5}} }{\mathrm{20}}\:−\frac{\mathrm{3}}{\mathrm{9}.\mathrm{32}}\:{x}^{\mathrm{9}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{20}}\:−\frac{\mathrm{1}}{\mathrm{96}} \\ $$$$=\frac{\mathrm{21}}{\mathrm{20}}\:−\frac{\mathrm{1}}{\mathrm{96}}\:. \\ $$$$ \\ $$

Commented by Abdo msup. last updated on 29/Dec/18