C.1 · 단변수 미적분 및 응용
GRE Subject Test · GRE Mathematics 주제 1 3:27 영어 내레이션 · 영어 + 중국어 자막 burned-in
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This unit uses limits, continuity and differentiability; derivatives, integrals and the fundamental theorem; and Riemann sums, kept apart from infinite series.
本单元使用极限、连续性和可导性；导数、积分和微积分基本定理；以及黎曼和，并把它与无穷级数区分开。
Every one of these tools is a limit.
这些工具每一个都是极限。
You need exact algebra, radians and inequalities first.
先要掌握精确代数、弧度和不等式。
A limit describes values near a point; the function need not even be defined there.
极限描述的是一点附近的取值；函数在该点甚至可以没有定义。
Continuity adds two things: the value exists, and it equals the limit.
连续性多了两个要求：函数值存在，并且等于极限。
Cancelling a factor is valid away from its zero, and can reveal a removable limit.
约去一个因式在其零点以外成立，可以揭示可去的极限。
A derivative implies continuity, but not the other way round.
可导推出连续，反过来不成立。
The absolute value of x is continuous at zero, yet its one-sided slopes are minus 1 on the left and plus 1 on the right.
|x| 在零点连续，但左侧斜率是 −1，右侧是 +1。
They disagree, so there is no derivative at zero.
两者不相等，所以零点处没有导数。
The fundamental theorem: the derivative of the integral from a to x of a continuous f is f of x.
微积分基本定理：连续函数 f 从 a 到 x 的积分，其导数是 f(x)。
With a variable upper bound h of x, multiply by h dash.
上限是 h(x) 时，乘以 h′。
With two variable bounds, subtract the lower bound's contribution.
上下限都变化时，减去下限的贡献。
Divide top and bottom by n squared: the sum of n over n squared plus k squared is 1 over n times the sum of 1 over 1 plus k over n squared.
分子分母同除以 n²：n/(n² + k²) 的和等于 (1/n)·Σ 1/(1 + (k/n)²)。
Each term is a rectangle of width 1 over n under 1 over 1 plus t squared, from 0 to 1, so the limit is pi over 4.
每一项都是 1/(1 + t²) 下面一个宽为 1/n 的矩形，范围从 0 到 1，所以极限是 π/4。
A series is a different limit.
级数是另一种极限。
The terms 1 over k tend to zero, but the partial sums keep growing: each block from 2 to the j minus 1, plus 1, up to 2 to the j adds at least a half, so they grow without bound.
1/k 趋于零，但部分和一直增长：从 2^{j−1} + 1 到 2^j 的每一块至少增加 1/2，所以部分和无界增长。
For F of x, the integral from 0 to x squared of e to the t: the theorem puts the bound into the integrand, e to the x squared, and the chain rule multiplies by 2x.
F(x) 是 e^t 从 0 到 x² 的积分：定理把上限代入被积函数，得 e^{x²}，链式法则再乘以 2x。
At x equals 1 that is 2e.
x = 1 时是 2e。
Omitting 2x misses the upper limit's rate of change.
漏掉 2x 就忽略了上限的变化率。
A-level calculus is useful prerequisite material, but it does not cover the undergraduate analysis and applications tested here.
A-level 微积分是有用的预备知识，但它没有涵盖这里考查的本科分析和应用。
F of x is the integral from x to x squared of 1 plus t squared.
F(x) 是 1 + t² 从 x 到 x² 的积分。
The integrand is continuous, so the theorem applies at both bounds: 1 plus x to the fourth times 2x, minus 1 plus x squared.
被积函数连续，所以定理在两个限上都适用：(1 + x⁴)·2x − (1 + x²)。
At 2 that is 68 minus 5, which is 63.
x = 2 时是 68 − 5，即 63。
The antiderivative t plus t cubed over 3 gives the same derivative.
用原函数 t + t³/3 得到同样的导数。
Rewrite each term as 1 over n times k over n, over 1 plus k over n squared: a Riemann sum for t over 1 plus t squared on 0 to 1, whose integral is a half ln 2.
把每一项写成 (1/n)·(k/n)/(1 + (k/n)²)：这是 t/(1 + t²) 在 上的黎曼和，积分为 (1/2)ln 2。
The harmonic series is different: each block adds at least a half, so it diverges.
调和级数不同：每一块至少增加 1/2，所以发散。
Check yourself.
自我检测。
The absolute value is continuous at zero without a derivative.
|x| 在零点连续但没有导数。
Differentiating to x cubed gives 3 x squared f of x cubed.
积分上限为 x³ 时，导数是 3x²f(x³)。
Zero over zero is a form, never an answer.
0/0 是一种形式，从来不是答案。
Terms tending to zero prove nothing.
项趋于零什么也证明不了。
And 1 over n times the sum of f of k over n tends to the integral from 0 to 1.
(1/n)Σf(k/n) 趋于 f 从 0 到 1 的积分。