For a, let u= 1- x. Then du= -dx and x= 1- u. When x= 0, u= 1 and when x= 1, u= 0.
So \(\displaystyle \int_0^1 xf'(1- x)dx= -\int_1^0 (1- u)f'(u)du= \int_0^1 f'(u)du- \int_0^1 uf'(u)du\).
Of course, \(\displaystyle \int_0^1 f'(u)du= f(1)- f(0)= -3- 8= -11\).
To integrate \(\displaystyle \int_0^1 uf'(u)du\) use integration by parts.
