Evaluate the double integral (a) int^(3_-1) int^(2_0)  x^(3) y dx dy

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to identify the inner integral such that:

`int_0^2 x^3 y dx`

Hence, you should evaluate this integral considering y as a constant such that:

`int_0^2 x^3 y dx = (y*x^4)/4|_0^2`

`int_0^2 x^3 y dx = y(2^4/4 - 0^4/4)`

`int_0^2 x^3 y dx = 4y`

You need to evaluate the outer integral, hence, you need to substitute 4y for inner integral such that:

`int_(-1)^3 int_0^2 x^3 y dx dy = int_(-1)^3 4y dy`

`int_(-1)^3 4y dy = 4y^2/2|_(-1)^3`

`int_(-1)^3 4y dy = 2(3^2 - (-1)^2)`

`int_(-1)^3 4y dy = 2(9 - 1) => int_(-1)^3 4y dy = 16`

Hence, evaluating the double integral `int_(-1)^3 int_0^2 x^3 y dx dy`  yields `int_(-1)^3 int_0^2 x^3 y dx dy = 16` .

lambert86's profile pic

lambert86 | Student, Undergraduate | (Level 2) eNoter

Posted on

This is a question on double integral.

First do the integral w.r.t y assuming x is a constant.

so;

int^(3_-1)int^(2_0)x(^3)ydxdy

=int^(2_0)x^(3)[y^2/2](3_-1)dx

=int^(2_0)x^(3)[(3^2-(-1)^2)/2]dx

=int^(2_0)x^(3)(8)dx

Now integrate the function w.r.t x;

=int^(2_0)8x^(3)dx

=8[X^4/4](2_0)

=2x^4 (2_0)

=2[(2)^4-(0)^4]

= 2*16

=32

Therefore the answer is 32.

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