I was trying to solve this given problem,

When $f(x)$ is continuous on $[a, b]$, there exists infinitely many reals $p_1, p_2, p_3$ and $q_1, q_2, q_3, q_4$, which satisfies the following equations. $$(1) \int_a^bf(x)dx=\lim_{n \to \infty}\sum_{k=1}^n(p_1f(x_{2k})+p_2f(x_{2k-1})+p_3f(x_{2k-2}))\Delta x, \left(\Delta x=\frac{b-a}{2n}, x_k=a+k\Delta x\right)$$ $$(2) \int_a^bf(x)dx=\lim_{n \to \infty}\sum_{k=1}^n(q_1f(x_{3k})+q_2f(x_{3k-1})+q_3f(x_{3k-2})+q_4f(x_{3k-3}))\Delta x, \left(\Delta x=\frac{b-a}{3n}, x_k=a+k \Delta x\right)$$

The problem was to determine the condition when $(1), (2)$ would each hold.

Obviously, the answer was $$p_1+p_2+p_3=2, q_1+q_2+q_3+q_4=3$$ respectively.

My question is, how does this relate to the definition of definite integrals? I understand that sample points can be chosen arbitrarily on each interval when setting a Riemann Sum. But I couldn't understand why $(1), (2)$ would hold. It kind of reminded me of Midpoint Rule, and Trapezoidal Rule, but I had no clue how to prove the answer.

Thank you very much if you can give me a full answer, or can anyone at least explain the geometric (intuitive) explanation for this problem? Thanks.

It would be great to have a analytic solution.


1 Answer 1


When you invoke continuity, you can show that (1) and (2) will be $\epsilon$ close to the definition of the definite integral, for any $\epsilon > 0$.

For example, in (1), if you fix one $x_{2k}$, for a mesh that is sufficiently small, $f(x_{2k}) \approx f(x_{2k-1}) \approx f(x_{2k-2})$.

  • $\begingroup$ I am not very familiar with proving things with $\epsilon-\delta$ but I understand what you meant. Then why would the sum of $p_i$s and $q_i$s be 2 and 3? $\endgroup$
    – zxcvber
    Commented Jun 22, 2016 at 15:35
  • $\begingroup$ If you agree to the approximations, then $p_1 f(x_{2k}) + p_2f(x_{2k-1}) + p_3f(x_{2k-2}) \approx (p_1 + p_2 + p_3) f(x_{2k})$, so, intuitively, for the Riemann integral you want $p_1 + p_2 + p_3$ to roughly be the 'width' of the rectangles under consideration. $\endgroup$
    – Roy D.
    Commented Jun 22, 2016 at 15:38
  • $\begingroup$ Isn't the width supposed to be $\Delta x$? But I see what you mean. So $p_1+p_2+p_3$ would have to be 2 so that $(p_1+p_2+p_3)\Delta x$ is roughly the 'width'? $\endgroup$
    – zxcvber
    Commented Jun 22, 2016 at 15:55
  • $\begingroup$ Yeah, exactly. Since $\Delta x$ as formulated above has an extra $1/2$ term in it. $\endgroup$
    – Roy D.
    Commented Jun 22, 2016 at 18:26

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