Question

$\tiny \displaystyle \bf \red{\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ \alpha + \beta + \gamma + \theta - \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$
Cʜᴀʟʟᴇɴɢᴇ❤️❤️❤️❤️❤️​

Answer 1

Step-by-step explanation:

$\displaystyle \sf \red{I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ \alpha + \beta + \gamma + \theta - \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

According to symmetry

$\displaystyle \sf \red{I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ \alpha + \beta + \gamma - \theta + \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

$\displaystyle \sf \red{I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ \alpha + \beta - \gamma + \theta + \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

$\displaystyle \sf \red{I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ \alpha - \beta + \gamma + \theta + \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

$\displaystyle \sf \red{I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ - \alpha + \beta + \gamma + \theta + \eta}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

Take sum of 5 integrals

$\displaystyle \sf \green{5I=\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \frac{ 3(\alpha + \beta + \gamma + \theta + \eta)}{\alpha + \beta + \gamma + \theta + \eta} \: d \alpha d \beta d \gamma d \theta d \eta}$

$\displaystyle \sf \pink{5I=3\int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \int^{3}_{1} \: d \alpha d \beta d \gamma d \theta d \eta}

$

$\large \sf \green{5I=3( {2)}^{5} }$

$\large \sf \pink{I={ \frac{96}{5} }^{} }$

Answer 2

Answer:

I = 96/10 = 96:10 is your correct answer.