Yes, integration by parts should do it. Let u= f(x) and dv= g'' dx so that du= f' dx and v= g'. \(\displaystyle \int_2^4 fg''dx= \left[f(x)g'(x)\right]_2^4- \int_2^4 f'g' dx\).
The first part is \(\displaystyle \left[f(x)g'(x)\right]_2^4= f(4)g'(4)- f(2)g'(2)= (13)(7)- (7)(1)= 84\). Now we need to calculate \(\displaystyle \int_2^4 f'g'dx\). But we are told that "f'(x)= 3 for all x". Since f' is a constant, we can take it outside the integral: \(\displaystyle 3\int_2^4 g'(x)dx\).
