Final answer:
Only option (D), which presents f(x) values that ascend from 0 at x=4 to positive values without any negatives, meets the criteria of an always increasing function and the integral from 4 to 1 being 0.
Step-by-step explanation:
If f'(x) > 0 for all real numbers x, this indicates that the function f(x) is increasing over its domain. Given the condition that the integral of f(t) from 4 to 1 is 0, we infer that the net area under the curve from x=1 to x=4 is zero. This suggests that f(x) must take on negative values at some point(s) between 1 and 4 to cancel out the positive areas, given the function is continuous and always increasing.
Reviewing the provided options, we can eliminate any table of values that includes negative values for f(x) at x > 4 because f(x) should be increasing. Similarly, we eliminate any tables with f(x) > 0 for x < 4, as the integral from 4 to 1 should cancel out to zero. Only option (D) presents values for f(x) that are consistent with an integral of 0 from 4 to 1 and an always increasing function, as the values go from 0 to positive without any negative values.