Let
𝑓
be a CONTINUOUS, POSITIVE, DECREASING function on
[
0
,
∞
)
.

Which of the following statements must be true?

Select all the correct answers.

Question

Let  
𝑓
  be a CONTINUOUS, POSITIVE, DECREASING function on  
[
0
,
∞
)
.

Which of the following statements must be true?

Select all the correct answers.

BlackTom AI · Accepted Answer

Question restatement: - We have a function f that is continuous, positive, and decreasing on [0, ∞). - We consider the following two statements and must determine which must be true (select all that apply): A) ∑_{k=1}^{n} f(k) < ∫_0^{n} f(x) dx B) ∑_{k=0}^{n-1} f(k) > ∫_0^{n} f(x) dx Now, let’s analyze each option carefully, using the basic area-under-the-curve comparison for decreasing functions. Option A: ∑_{k=1}^{n} f(k) < ∫_0^{n} f(x) dx - Key idea: For each integer k = 1,2,...,n, because f is decreasing, we have f(x) ≥ f(k) for all x in [k-1, k]. Therefore, on each subinterval [k-1, k], the area under the curve is at least f(k) times the interval length 1, i.e., ∫_{k-1}^{k} f(x) dx ≥ f(k). - Summing these inequalities from k = 1 to n yields ∫_{0}^{n} f(x) dx ≥ ∑_{k=1}^{n} f(k). - If f is strictly decreasing (the common textbook interpretation of “decreasing” in a calculus context is strictly decreasing), then on each interval (k-1, k) we have f(x) > f(k), so the inequality is strict: ∫_{k-1}^{k} f(x) dx > f(k), and hence ∫_{0}^{n} f(x) dx > ∑_{k=1}^{n} f(k). - Conclusion for A: Under the usual interpretation of “decreasing” as strictly decreasing, A holds with a strict inequality. If one only assumes nonincreasing (not strictly), equality can occur when f is constant on each [k-1, k], making the strict inequality not guaranteed. Thus A is true in the strictly-decreasing case, but not necessarily in the non-strict case. Option B: ∑_{k=0}^{n-1} f(k) > ∫_0^{n} f(x) dx - Compare the left sum to the integral by breaking the integral into n unit subintervals: ∫_0^{n} f(x) dx = ∑_{k=0}^{n-1} ∫_{k}^{k+1} f(x) dx. - On each subinterval [k, k+1], since f is decreasing, we have f(x) ≤ f(k) for all x in [k, k+1]. Therefore ∫_{k}^{k+1} f(x) dx ≤ f(k)·(1) = f(k). - Summing over k = 0 to n-1 gives ∫_0^{n} f(x) dx ≤ ∑_{k=0}^{n-1} f(k). - Again, if f is strictly decreasing, the inequality is strict on at least one subinterval, yielding ∑_{k=0}^{n-1} f(k) > ∫_0^{n} f(x) dx. If f is only nonincreasing (not strictly decreasing), you could have equality when f is constant on each [k, k+1], so the strict inequality is not guaranteed in that broader interpretation. - Conclusion for B: Similar to A, B holds with a strict inequality under strict decrease, but is not guaranteed to be strict (and thus not guaranteed at all) if we only assume nonincreasing. Summary cross-check: - Both statements align with the standard integral-test-type inequalities for positive, decreasing functions when we interpret decreasing as strict. In that common interpretation, both A and B hold with strict inequalities for all n ≥ 1. If the problem intends only nonincreasing (not strictly), both inequalities could be non-strict in some cases, and the statements would not be guaranteed to be strict. Additional note on interpretation: - If the problem’s wording uses “decreasing” to mean strictly decreasing, then both options A and B must be true for all n, making them the correct selections. If the wording allows f to be constant on intervals (i.e., nonincreasing but not strictly decreasing everywhere), then neither statement can be asserted as always true with strict inequality, though the non-strict versions (≤) would still hold.

Let 𝑓 be a CONTINUOUS, POSITIVE, DECREASING function on [ 0 , ∞ ) . Which of the following statements must be true? Select all the correct answers.多项选择题

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