Question Number 224642 by Rogerkeyter last updated on 23/Sep/25
Answered by som(math1967) last updated on 24/Sep/25
![I=∫_0 ^1 (dx/(1+(e^(√(1−x)) /e^(√x) ))) =∫_0 ^1 ((e^(√x) dx)/(e^(√x) +e^(√(1−x)) )) =∫_0 ^1 ((e^(√(1+0−x)) dx)/(e^(√(1+0−x)) +e^(√(1−1−0+x)) )) =∫_0 ^1 ((e^(√(1−x)) dx)/(e^(√(1−x)) +e^(√x) )) ∴ 2I=∫_0 ^1 ((e^(√x) +e^(√(1−x)) dx)/(e^(√x) +e^(√(1−x)) ))=∫_0 ^1 dx 2I=[x]_0 ^1 =1−0 ∴ I=(1/2)](https://www.tinkutara.com/question/Q224647.png)
$$\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{dx}}}{\mathrm{1}+\frac{\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} }{\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} }} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} \boldsymbol{{dx}}}{\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} +\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} } \\ $$$$\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{e}}^{\sqrt{\mathrm{1}+\mathrm{0}−\boldsymbol{{x}}}} \boldsymbol{{dx}}}{\boldsymbol{{e}}^{\sqrt{\mathrm{1}+\mathrm{0}−\boldsymbol{{x}}}} +\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\mathrm{1}−\mathrm{0}+\boldsymbol{{x}}}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} \boldsymbol{{dx}}}{\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} +\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} } \\ $$$$\therefore\:\mathrm{2}\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} +\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} \boldsymbol{{dx}}}{\boldsymbol{{e}}^{\sqrt{\boldsymbol{{x}}}} +\boldsymbol{{e}}^{\sqrt{\mathrm{1}−\boldsymbol{{x}}}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{dx}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\left[\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1}−\mathrm{0} \\ $$$$\therefore\:\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Ghisom_ last updated on 24/Sep/25
$$\mathrm{yes}! \\ $$