Question Number 198845 by sonukgindia last updated on 25/Oct/23
Answered by som(math1967) last updated on 25/Oct/23
![I=∫_(−1) ^1 (((1−1−x)^2 dx)/(1+e^(1−1−x) )) =∫_(−1) ^1 ((x^2 dx)/(1+e^(−x) )) =∫_(−1) ^1 ((e^x x^2 dx)/(1+e^x )) ∴2I=∫_(−1) ^1 ((x^2 (1+e^x )dx)/((1+e^x ))) =[(x^3 /3)]_(−1) ^1 =(2/3) ∴I=(1/3)](https://www.tinkutara.com/question/Q198847.png)
$${I}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−\mathrm{1}−{x}\right)^{\mathrm{2}} {dx}}{\mathrm{1}+{e}^{\mathrm{1}−\mathrm{1}−{x}} } \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {dx}}{\mathrm{1}+{e}^{−{x}} } \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{e}^{{x}} {x}^{\mathrm{2}} {dx}}{\mathrm{1}+{e}^{{x}} } \\ $$$$\therefore\mathrm{2}{I}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} \left(\mathrm{1}+{e}^{{x}} \right){dx}}{\left(\mathrm{1}+{e}^{{x}} \right)} \\ $$$$\:=\left[\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{1}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\therefore{I}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$
Commented by Frix last updated on 25/Oct/23
$$\mathrm{Nice}! \\ $$
Commented by som(math1967) last updated on 25/Oct/23
$${Thank}\:{you}\:{sir} \\ $$