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) :}`