Question Number 227146 by Spillover last updated on 03/Jan/26
Answered by Kassista last updated on 03/Jan/26
![I = ∫_( −2) ^( 2 ) (x^2 /(1+5^x )) dx ⇒^(x=−u) ∫_( 2) ^( −2) (u^2 /(1+5^(−u) )) −du = ∫_(−2) ^( 2) (u^2 /(1+5^(−u) )) .(5^u /5^u )du = ∫_( −2) ^( 2) ((u^2 5^u )/(5^u +1)) du ∴ 2I = ∫_(−2) ^( 2) (x^2 /(1+5^x ))+((x^2 5^x )/(1+5^x )) dx = ∫_(−2) ^( 2) ((x^2 (1+5^x ))/((1+5^x )))dx ∴ 2I=2∫_0 ^( 2) x^2 dx =[(x^3 /3)]_0 ^2 =(8/3) ⇒I=(8/3)](https://www.tinkutara.com/question/Q227150.png)
$$ \\ $$$${I}\:=\:\int_{\:−\mathrm{2}} ^{\:\mathrm{2}\:} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{{x}} }\:{dx}\:\overset{{x}=−{u}} {\Rightarrow}\int_{\:\mathrm{2}} ^{\:−\mathrm{2}} \frac{{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{−{u}} }\:−{du}\: \\ $$$$=\:\int_{−\mathrm{2}} ^{\:\:\mathrm{2}} \frac{{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{−{u}} }\:.\frac{\mathrm{5}^{{u}} }{\mathrm{5}^{{u}} }{du}\:=\:\int_{\:−\mathrm{2}} ^{\:\:\mathrm{2}} \frac{{u}^{\mathrm{2}} \mathrm{5}^{{u}} }{\mathrm{5}^{{u}} +\mathrm{1}}\:{du} \\ $$$$\therefore\:\mathrm{2}{I}\:=\:\int_{−\mathrm{2}} ^{\:\:\mathrm{2}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{5}^{{x}} }+\frac{{x}^{\mathrm{2}} \mathrm{5}^{{x}} }{\mathrm{1}+\mathrm{5}^{{x}} }\:{dx}\:=\:\int_{−\mathrm{2}} ^{\:\:\mathrm{2}} \frac{{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{5}^{{x}} \right)}{\left(\mathrm{1}+\mathrm{5}^{{x}} \right)}{dx} \\ $$$$\therefore\:\mathrm{2}{I}=\mathrm{2}\int_{\mathrm{0}} ^{\:\:\mathrm{2}} {x}^{\mathrm{2}} {dx}\:=\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{0}} ^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{3}}\:\Rightarrow{I}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$
Commented by Spillover last updated on 03/Jan/26
$${thanks} \\ $$
Answered by Spillover last updated on 03/Jan/26
