Finding the limit with definite integral is generally to find the limit of some summation formulas, and classifying the limit of summation formula as definite integral is based on the definition of definite integral and the properties of integrable function. We know that definite integral is the limit of special summation formula, in which there are functions, interval lengths and arbitrary points between cells, and the division method between these cells and the selection of arbitrary points are arbitrary and complicated. On the other hand, we know that continuous function is integrable in its continuous interval, that is, definite integral exists, which can of course be expressed by the limit of summation formula. Here we choose a special method to get the points between cells: divide the interval into n equal parts (or n- 1 equal parts) and select the end points of the interval, so the limit of the sum formula obtained should also be equal to the definite integral. We see this kind of problem, which is generally called sum limit. Turning into definite integral requires some skills and reverse thinking. In the process of transformation, the key is to turn each term in the sum into a factor of 1/n. This topic has already been found, and the remaining k-th term is sin(kπ/n), and the integrand can be sinπ x.n. Then the value range of k/n should be1/n-> 1。 Because n tends to infinity and 1/n tends to 0, we can know that the integral interval is [0, 1], and the limit can be converted into the definite integral of sinπx on [0, 1]. (You can try it the other way around, divide the interval into equal parts, express sinπx as sum limit by definition, and that gram c takes the endpoint between cells).

Note: If kπ/n is x, the integrand is sinx, and the integration interval becomes [0, π].