Science talk:Math Exam Resources/Courses/MATH103/April 2010/Question 4 (c)

In the last part of the solution, how did you get from the second line to the third line? I don't understand what was done there.

$\int _{0}^{1}{\frac {\frac {2}{5}}{2-x}}\,dx={\frac {2}{5}}\int _{0}^{1}{\frac {1}{2-x}}\,dx={\frac {2}{5}}\left.\left(\ln |2-x|\left({\frac {1}{-1}}\right)\right)\right|_{0}^{1}={\frac {-2}{5}}(\ln |2-1|-\ln |2-0|).$ Don't forget the extra factor 1/(-1) which comes by taking the derivative of the inner function (2-x). Same with the second integral which needs an extra factor of 1/2 since (1+2x)' = 2. Does this make it more clear?

Ooooh! Okay, thanks. I didn't even know that was a thing that needed to be done!

I cleaned up the formatting of the solution only because it was too long to fit the standard width of the box and was moving it to the bottom of the page.