Answer: D) 101
Explanation:
By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out
![\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](https://img.qammunity.org/2022/formulas/mathematics/college/j98fc1y7ab8y61gs3o8j371w2xrucx83mt.png)
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?