Since `−(2 − 3x) = 3x − 2`, we can bring the denominator to the top and write the question as:

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

Put `u = 3x − 2` then `du = 3\ dx`.

So

`int_-1^1e^(3x-2)dx=1/3[e^(3x-2)]_-1^1`

`=1/3[e^1-e^-5]`

`=0.9038`

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