Interpreting Limits of Riemann Sums as Integrals
Read this topicExplanations 1
Problems 6
- Find the limit limn→∞∑k=1n9k2n3\lim_{n\to\infty} \sum_{k=1}^{n}\frac{9k^2}{n^3}limn→∞∑k=1nn39k2 (as the positive integer nnn tends to infinity).
- Let L=limn→∞∑i=0n−1(n2−i2)−1/2L = \lim_{n\to\infty} \sum_{i=0}^{n-1} (n^2 - i^2)^{-1/2}L=limn→∞∑i=0n−1(n2−i2)−1/2. Find π/L\pi/Lπ/L. Source: HMMT February 2003
- Evaluate limn→∞(1n+1+1n+2+⋯+1n+n)\lim_{n\rightarrow\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n+n}\right)n→∞lim(n+11+n+21+⋯+n+n1)
- Rewrite the following as a definite integral: limn→∞∑i=0∞11+(in)2Δx\lim_{n\to\infty} \sum_{i=0}^{\infty} \frac{1}{1+(\frac{i}{n})^2} \Delta xlimn→∞∑i=0∞1+(ni)21Δx where Δx=…
- Which of the choices below is closest to the optimal number of pockets (irregular ones included) and the common side length of the square pockets?
- Evaluate the sum-limit limn→∞∑k=0n−14n(2−4kn)4.\lim_{n \rightarrow \infty}\sum^{n - 1}_{k = 0}\frac{4}{n}\left(2 - \frac{4k}{n}\right)^4.n→∞limk=0∑n−1n4(2−n4k)4.