Answer: D.
Step-by-step explanation:
Use screenshot below to help you.
I. [tex]\int\limits^4_ {-1} \, sinxdx[/tex] works since (b-a)/n = [tex]5/n[/tex] & the -1 in [tex]f(a+(b-a)/n(k))[/tex] works aswell
II. [tex]\int\limits^5_0 {sin(-1+x)} \, dx[/tex] also works since (b-a)/n = [tex](5/n)[/tex] and the -1 is already inside the function making [tex]f(a+(b-a)/n(k))[/tex] work aswell.
III. [tex]5\int\limits^1_0 {sin(-1+5x)} \, dx[/tex] works since you are taking a coeffienct out and multiplying it by 5 on the inside function. It is a proven rule, and since condition II works, this does aswell.
That's it!
