(Image removed from quote.)
My working out for qn 8
The attached image is the graph you want to find the area of. So you find the integral of the function from -2 to -1. I don't know why there's a negative sign in front of everything. Also, note that when you take the integral of the fraction part, it becomes log
eIx-1I, where when x-1 is negative, you take the positive value. Hopefully that fixes the problem you've been having.

Edit: Realised that you're really close the the answer. If you drop that negative in the front, then you get the answer. Also you can combine the logs to make log
e2/3. Tell me if you don't get it.
I'm going to show my working for the start of the other question since I'm not going to be online for a while. I don't know if it's right, but the method should be legit I hope.
Let y=(log
ex)
2

