You need to solve for y the inner integral, considering x as constant, such that:

`int_(x-1)^(x+1) e^(x+y) dy = e^(x+y)|_(x-1)^(x+1)`

`int_(x-1)^(x+1) e^(x+y) dy = e^(x+x+1) - e^(x+x-1)`

`int_(x-1)^(x+1) e^(x+y) dy = e^(2x+1) - e^(2x-1)`

You need to solve for x the outer integral such that:

`int_(-1-y)^(1-y) e^(2x+1) dx - int_(-1-y)^(1-y)e^(2x-1) dx = (1/2)(e^(2x+1) - e^(2x-1))|_(-1-y)^(1-y)`

`int_(-1-y)^(1-y) e^(2x+1) dx - int_(-1-y)^(1-y)e^(2x-1) dx = (1/2)(e^(2-2y+1) - e^(-2 - 2y+1) - e^(2 - 2y -1) + e^(-2 - 2y - 1))`

`int_(-1-y)^(1-y) e^(2x+1) dx - int_(-1-y)^(1-y)e^(2x-1) dx = (1/2)(e^(3-2y) - e^(-1 - 2y) - e^(1 - 2y) + e^(-3 - 2y))`

** Hence, evaluating the double integral under given conditions yields**

`int_(-1-y)^(1-y) int_(x-1)^(x+1) e^(x+y) dy dx = (1/2)(e^(3-2y) - e^(-1 - 2y) - e^(1 - 2y) + e^(-3 - 2y)).`

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now