advanced calculus of several variables

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1. Evaluate the integral , f where A is a subset in Rº defined by A = {(1, 12, 13, 14): 21+x2+23+24
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Explanation & Answer

The solutions are ready.I'll use these results for questions 4.6 and 4.7:)

1. Express this four-dimension integral as a chain of one-dimensional integrals.
It is evident that 𝑥1 ∈ [0,1], 𝑥2 ∈ [0,1 − 𝑥1 ] for each 𝑥1 , 𝑥3 ∈ [0,1 − 𝑥2 − 𝑥3 ] for each 𝑥1 , 𝑥2
and 𝑥4 ∈ [0,1 − 𝑥1 − 𝑥2 − 𝑥3 ]. The chain integral becomes
1−𝑥1

1

1−𝑥1 −𝑥2

∫ ( ∫ ( ∫
𝑥1 =0

𝑥2 =0

1−𝑥1 −𝑥2 −𝑥3

(

𝑥3 =0

(𝑥1 𝑥2 𝑥3 𝑥4 )𝑑𝑥4 ) 𝑑𝑥3 ) 𝑑𝑥2 ) 𝑑𝑥1 .


𝑥4 =0

The inner integral is equal to
1−𝑥1 −𝑥2 −𝑥3



1−𝑥1 −𝑥2 −𝑥3

𝑑𝑥4 (𝑥1 𝑥2 𝑥3 𝑥4 ) = (𝑥1 𝑥2 𝑥3 )

𝑥4 =0


𝑥4 =0

1
𝑑𝑥4 (𝑥4 ) = (𝑥1 𝑥2 𝑥3 )(1 − 𝑥1 − 𝑥2 − 𝑥3 )2 =
2

1
= 𝑥1 𝑥2 (𝑥3 (1 − 𝑥1 − 𝑥2 )2 − 2𝑥32 (1 − 𝑥1 − 𝑥2 ) + 𝑥33 ).
2
The next integral is
1
𝑥 𝑥
2 1 2

1−𝑥1 −𝑥2



𝑑𝑥3 (𝑥3 (1 − 𝑥1 − 𝑥2 )2 − 2𝑥32 (1 − 𝑥1 − 𝑥2 ) + 𝑥33 ) =

𝑥3 =0

1
1
2
1
= 𝑥1 𝑥2 ( (1 − 𝑥1 − 𝑥2 )2 (1 − 𝑥1 − 𝑥2 )2 − (1 − 𝑥1 − 𝑥2 )(1 − 𝑥1 − 𝑥2 )3 + (1 − 𝑥1 − 𝑥2 )4 ) =
2
2
3
4
1
1 2 1
1
= 𝑥1 𝑥2 (1 − 𝑥1 − 𝑥2 )4 ( − + ) =
𝑥 𝑥 (1 − 𝑥1 − 𝑥2 )4 =
2
2 3 4
24 1 2
=

1
𝑥 ((1 − 𝑥1 )4 𝑥2 − 4(1 − 𝑥1 )3 𝑥22 + 6(1 − 𝑥1 )2 𝑥23 − 4(1 − 𝑥1 )𝑥24 + 𝑥25 ).
24 1

Then
1−𝑥1

1
𝑥 ∫ ((1 − 𝑥1 )4 𝑥2 − 4(1 − 𝑥1 )3 𝑥22 + 6(1 − 𝑥1 )2 𝑥23 − 4(1 − 𝑥1 )𝑥24 + 𝑥25 )𝑑𝑥2 =
24 1
𝑥2 =0

=

1
1 4 6 4 1
1
𝑥1 (1 − 𝑥1 )6 ( − + − + ) =
𝑥 (1 − 𝑥1 )6 .
24
2 3 4 5 6
720 1

And finally
1

1

1
1
∫(
𝑥1 (1 − 𝑥1 )6 ) 𝑑𝑥1 =
∫(𝑥1 − 6𝑥12 + 15𝑥13 − 20𝑥14 + 15𝑥15 − 6𝑥16 + 𝑥17 )𝑑𝑥1 =
720
720
0

0

=

1 1 6 ...


Anonymous
Nice! Really impressed with the quality.

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