Answer
Using integration by parts and the conditions $f(1)=2 , f(4)=7, f'(1)=5 , f'(4)=3$ we find that
$$\int_1^4{xf''(x)dx}=2$$
Work Step by Step
Since $f''(x)$ is continuous we can solve $\int_1^4{xf''(x)dx}$ by parts using the formula $$A=uv-\int{vdu}$$
Putting $u=x$ and $dv=f''(x)dx$
then
$$A=[xf'(x)]_{1}^{4}-\int_1^4{f'(x)dx}$$
solving the second term
$$A=[xf'(x)]_{1}^{4}-[f(x)]_{1}^{4}$$
since $f(1)=2 , f(4)=7, f'(1)=5 , f'(4)=3$
then $$A=4*3-1*5-(7-2)=2$$
