Calculus I, Fall 2014 (http://www.math.nthu.edu.tw/~wangwc/)

Brief answer to selected problems in Homework 11

1. Section 5.6: problem 112: use the change of variable u = 1 − x. problem 117: use the change of variable u = x − c.

2. Chapter 5: problem 4: 1 = \frac{d}{dx}x = \frac{d}{dy} ∫_{0}^{y} \frac{1}{\sqrt{1+4t^{2}}} dt × \frac{dy}{dx}, implies \frac{dy}{dx} = \sqrt{1 + 4t^{2}}. Hence

\frac{d^{2}y}{dx^{2}} = 4y

problem 20: lim_{x→∞} \frac{1}{x} ∫_{0}^{x} tan^{−1} tdt = \frac{π}{2}

problem 22: lim_{n→∞} \frac{1}{n}(e^{\frac{1}{n}} + e^{\frac{2}{n}} + ⋯ + e^{\frac{n−1}{n}} + e^{\frac{n}{n}}) = ∫_{0}^{1} e^{x}dx

problem 26:

a. lim_{n→∞} \frac{1}{n^{2}} (2 + 4 + 6 + ⋯ + 2n) = lim_{n→∞} \frac{1}{n}(2\frac{1}{n} + 2\frac{2}{n} + ⋯ + 2\frac{n}{n}) = ∫_{0}^{1} 2xdx

b. lim_{n→∞} \frac{1}{n^{16}} (1^{15} + 2^{15} + ⋯ + n^{15}) = lim_{n→∞} \frac{1}{n}((\frac{1}{n})^{15} + (\frac{2}{n})^{15} + ⋯ + (\frac{n}{n})^{15}) = ∫_{0}^{1} x^{15}dx

c. lim_{n→∞} \frac{1}{n}(sin \frac{π}{n} + sin \frac{2π}{n} + ⋯ + sin \frac{nπ}{n}) = ∫_{0}^{1} sin πxdx

By b, we can derive the following

d. lim_{n→∞} \frac{1}{n^{17}} (1^{15} + 2^{15} + ⋯ + n^{15}) = lim_{n→∞}[\frac{1}{n}((\frac{1}{n})^{15} + (\frac{2}{n})^{15} + ⋯ + (\frac{n}{n})^{15})] × \frac{1}{n} = 0

e. lim_{n→∞} \frac{1}{n^{15}} (1^{15} + 2^{15} + ⋯ + n^{15}) = lim_{n→∞}[\frac{1}{n}((\frac{1}{n})^{15} + (\frac{2}{n})^{15} + ⋯ + (\frac{n}{n})^{15})] × n = ∞