Hi, I have done all the exercises I have to do for tomorrow, but I am stuck on this one...my professor said it is really easy, but I have no clue how to solve this. Any help would be very much appreciated, thanks!
Given the following:
$\displaystyle u(a)=-4, u'(b)=-4, u'(a)=-1, u(b)=-4$
$\displaystyle v'(a)=-8, v(b)=3, v(a)=4, v'(b)=2$
$\displaystyle \int^b_a u(x) v(x) dx = 4, \int^b_a u(x) dx = 3$
$\displaystyle \int^b_a v'(x) dx = -1, \int^b_a u'(x) v(x) dx = -6$
What is the value of $\displaystyle \int^b_a u(x) v'(x) dx$?