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The value of $\int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \frac{x_{1}+x_{2}+x_{3}-x_{4}}{x_{1}+x_{2}+x_{3}+x_{4}} d x_{1} d x_{2} d x_{3} d x_{4}$

In the given solution:

$\int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \frac{x_{i} d x_{1} d x_{2} d x_{3} d x_{4}}{x_{1}+x_{2}+x_{3}+x_{4}}=\frac{1}{4}; as \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \frac{x_{1}+x_{2}+x_{3}-x_{4}}{x_{1}+x_{2}+x_{3}+x_{4}} d x_{1} d x_{2} d x_{3} d x_{4}=1$ $\therefore \mathrm{I}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$

How it is coming?

1 Answers1

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You need to prove this :-

$\int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \frac{x_{i} d x_{1} d x_{2} d x_{3} d x_{4}}{x_{1}+x_{2}+x_{3}+x_{4}}=\frac{1}{4};$

Let $I_i=\int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \int_{1}^{2} \frac{x_{i} d x_{1} d x_{2} d x_{3} d x_{4}}{x_{1}+x_{2}+x_{3}+x_{4}}$

It is to be noted that $I_i=I_j$ for every $1\le i,j \le 4$ since for $i\ne j$ , interchange $i$ and $j$ in $I_j$ (say) and the order of integration to show the equality.

Then $\sum_{i=1}^4 I_i =1$

which gives $I_i=1/4$

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