Symmetry?

Calculus Level pending

Evaluate:

∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4} \,dx_1dx_2dx_3dx_4


The answer is 0.5.

Mark Recio by Mark Recio , 21, Philippines

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1 solution

Mark Recio
Feb 27, 2017

Let I = ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 I=\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1+x_2+x_3-x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

By symmetry, I I is equal to:

I = ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 I=\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1+x_2-x_3+x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

I = ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 I=\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1-x_2+x_3+x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

I = ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 − 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 I=\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{-x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

Adding all of these we obtain:

4 I = ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 2 x 1 + 2 x 2 + 2 x 3 + 2 x 4 x 1 + x 2 + x 3 + x 4 d x 1 d x 2 d x 3 d x 4 4I=\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{2x_1+2x_2+2x_3+2x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

4 I = 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 4I=2\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4

4 I = 2 ( 1 ) 4I=2(1)

. . ˙ I = 1 2 \dot{.\hspace{.0005in}.}\hspace{0.25in} I=\dfrac{1}{2}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Why did the integral become 1?

Jess Late - 4 years, 3 months ago

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∫ 1 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 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 2 ∫ 1 2 ∫ 1 2 ∫ 1 2 ( 1 ) d x 1 d x 2 d x 3 d x 4 = ∫ 1 2 ∫ 1 2 ∫ 1 2 x 1 ∣ 1 2 d x 2 d x 3 d x 4 = ∫ 1 2 ∫ 1 2 x 2 ∣ 1 2 d x 3 d x 4 = ∫ 1 2 x 3 ∣ 1 2 d x 4 = 1 \large \displaystyle \begin{aligned} \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \dfrac{x_1+x_2+x_3+x_4}{x_1+x_2+x_3+x_4}\,dx_1dx_2dx_3dx_4 \\ & = \int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\int_{1}^{2} (1) \,dx_1dx_2dx_3dx_4 \\ & = \int_{1}^{2}\int_{1}^{2}\int_{1}^{2} \left. x _1 \right |_{1}^{2} \, dx_2dx_3dx_4 \\ & = \int_{1}^{2}\int_{1}^{2} \left. x_2 \right |_{1}^{2} \, dx_3dx_4 \\ & = \int_{1}^{2} \left. x_3 \right |_{1}^{2} \, dx_4 \\ & = \boxed{1} \end{aligned}

Christian Daang - 4 years, 3 months ago

How such a variable can be changed such a way?

Kushal Bose - 4 years, 3 months ago

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Since the lower bound and upper bound of each integral are all the same, you can interchange the integrals and hence, you can interchange the variables that way.

Christian Daang - 4 years, 3 months ago

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Well said .Thanks

Kushal Bose - 4 years, 3 months ago

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@Kushal Bose – Welcome sir. :)

Christian Daang - 4 years, 3 months ago

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