Answer: D) 101
Step-by-step explanation:
By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out
[tex]\int\limits^{10}_{-1} {(2x+0.5f'(x))} \, dx =\int\limits^{10}_{-1} {2x} \, dx +0.5\int\limits^{10}_{-1} {f'(x)} \, dx =x^2|^{^{10}}_{_{-1}}+0.5f(x)|^{^{10}}_{_{-1}}\\=(100-1)+0.5(f(10)-f(-1))=99+0.5(8-4))=101[/tex]
I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?
