📚 10 Definite Integrals Having Variable Limits | 含变量上下限的定积分
Definite integrals with variable limits are fundamental in calculus, appearing whenever we need to define a function as an accumulation of another function. The Fundamental Theorem of Calculus links differentiation and integration, allowing us to differentiate such integrals and solve related problems. In this article, we examine ten examples that illustrate how to handle integrals with upper limits, lower limits, or both being functions of the variable. Each example is written in the style expected in IB Mathematics and A Level examinations, with the key rule, the working, and the common pitfalls made explicit.
具有可变上下限的定积分是微积分的基础，当我们需要将函数定义为另一个函数的累积时就会出现。微积分基本定理将微分与积分联系起来，使我们可以对此类积分求导并解决相关问题。在本文中，我们研究十个例题，展示如何处理上限、下限或两者均为变量的函数的积分。每个例题都按照 IB 数学与 A Level 考试常见的形式书写，并明确给出核心法则、解题过程以及易错点。
1. The Fundamental Theorem with Variable Limits | 变量上下限与微积分基本定理
is continuous on an interval containing a and x, then the function F defined by F(x) = ∫ₐˣ le that disappears after integration. This theorem is the foundation of every example in this article.
微积分基本定理第一部分指出，若 f 在包含 a 与 x 的区间上连续，则由 F(x) = ∫ₐˣ f(t) dt 定义的函数 F 在该区间上可导，且导数为 F′(x) = f(x)。变量 x 只出现在上限，而 t 是积分后消失的哑变量。这个定理是本文每个例题的基础。
When a limit is itself a function of x, the chain rule must be applied. If the upper limit is v(x), then the derivative is f(v(x))·v′(x). If the lower limit is u(x), then the derivative includes a negative sign, because reversing the limits changes the sign of the integral. When both limits are functions of x, we combine the two contributions into the Leibniz rule.
当上下限本身是 x 的函数时，必须使用链式法则。若上限为 v(x)，则导数为 f(v(x))·v′(x)。若下限为 u(x)，则导数带负号，因为交换上下限会改变积分的符号。当上下限都是 x 的函数时，我们把两部分贡献合并为莱布尼茨公式。
d/dx ∫[u(x), v(x)] f(t) dt = f(v(x))·v′(x) − f(u(x))·u′(x)
This single formula covers the upper-limit case, the lower-limit case, and the constant-limit case. If a limit is constant, its derivative is zero, so that term vanishes automatically. This is why a constant lower limit does not affect the derivative of an integral with variable upper limit.
这一个公式涵盖了上限情形、下限情形以及常数上下限情形。如果某个上下限是常数，其导数为零，因此该项自动消失。这就是为什么常数下限不会影响上限为变量的积分之导数。
2. Example 1: Derivative of a Simple Integral with Upper Limit x | 例题1：上限为 x 的简单积分求导
Problem: Let F(x) = ∫₀ˣ (t² + 1) dt. Find F′(x).
问题：设 F(x) = ∫₀ˣ (t² + 1) dt，求 F′(x)。
he upper limit. The lower limit contributes nothing because its derivative is zero. Therefore, F′(x) = x² + 1.
这里上限正好是 x，下限是常数 0。根据微积分基本定理第一部分，在积分上限处求被积函数的值即得导数。下限不产生贡献，因为它的导数为零。因此，F′(x) = x² + 1。
This simple example confirms that differentiation 'undoes' the integration of a continuous function. It also shows the essential idea behind all later examples: the derivative of an accumulation function is the rate at which the accumulation is increasing, namely the integrand evaluated at the moving boundary.
这个简单例子验证了微分对连续函数积分具有'撤销'效果。它也展示了后续所有例题的核心思想：累积函数的导数就是累积增加的速率，即被积函数在移动边界处的取值。
F′(x) = x² + 1
3. Example 2: Upper Limit x with a Square Root Integrand | 例题2：上限为 x 且被积函数含根式
Problem: Let F(x) = ∫₁ˣ √t³ + 2 dt. Find F′(x).
问题：设 F(x) = ∫₁ˣ √t³ + 2 dt，求 F′(x)。
plying the Fundamental Theorem directly, we replace t by x in the integrand. Hence F′(x) = √x³ + 2. There is no extra factor because the upper limit is x itself, not a more complicated function of x.
下限是常数 1，因此不会出现在导数中。上限是 x，其导数为 1。直接应用微积分基本定理，把被积函数中的 t 换成 x，因此 F′(x) = √x³ + 2。由于上限就是 x 本身，而不是更复杂的 x 的函数，所以不需要额外的因子。
A common exam error is to try to integrate √t³ + 2 first. That is unnecessary and often impossible in elementary terms. The variable-limit theorem allows us to differentiate without ever finding an antiderivative explicitly.
常见考试错误是先去求 √t³ + 2 的原函数。这不仅没有必要，而且通常无法用初等函数表示。变量限定理使我们无需显式求出原函数就可以求导。
F′(x) = √x³ + 2
4. Example 3: Variable Lower Limit and the Negative Sign | 例题3：变量下限与负号
Problem: Let F(x) = ∫ₓ⁵ cos(t²) dt. Find F′(x).
问题：设 F(x) = ∫ₓ⁵ cos(t²) dt，求 F′(x)。
e first rewrite the integral so that the variable limit becomes an upper limit: F(x) = −∫₅ˣ swapping the limits. Now the upper limit is x, so differentiating gives F′(x) = −cos(x²).
此时变量位于下限。我们先把积分改写成变量限在上限的形式：F(x) = −∫₅ˣ cos(t²) dt。负号来自交换上下限。此时上限为 x，求导得到 F′(x) = −cos(x²)。
Students often forget this sign. A useful memory aid is: if the variable is at the bottom, put a minus sign in front. Equivalently, in the Leibniz rule the lower-limit term is subtracted, which produces exactly the same negative sign.
学生经常忘记这个负号。一个有用的记忆方法是：变量在下方，就在前面加负号。等价地，在莱布尼茨公式中下限项被减去，正好产生同样的负号。
F′(x) = −cos(x²)
5. Example 4: Both Limits Are Functions of x | 例题4：上下限都是 x 的函数
Problem: Let F(x) = ∫[x, x²] e−t² dt. Find F′(x).
问题：设 F(x) = ∫[x, x²] e−t² dt，求 F′(x)。
= x, with derivative u′(x) = 1. Substituting into the rule gives F′(x) = e^(−(x²)²)·2x − e−x²·1.
这里上下限都依赖 x，因此使用莱布尼茨公式。上限为 v(x) = x²，其导数 v′(x) = 2x。下限为 u(x) = x，其导数 u′(x) = 1。代入公式得到 F′(x) = e^(−(x²)²)·2x − e−x²·1。
Simplifying the first exponential, (x²)² = x⁴, so the derivative becomes 2x·e−x⁴ − e−x² eping track of which limit produces which factor is the key skill in this type of problem.
化简第一个指数，因为 (x²)² = x⁴，所以导数变为 2x·e−x⁴ − e−x²。下限项被减去，而且它没有 2x 因子，因为下限是 x，不是 x²。分清哪个上下限产生哪个因子，是这类题目的关键技能。
F′(x) = 2x·e−x⁴ − e−x²
6. Example 5: Trigonometric Upper Limit | 例题5：三角上限
Problem: Let F(x) = ∫₀^(sin x) (1 + t²) dt. Find F′(x).
问题：设 F(x) = ∫₀^(sin x) (1 + t²) dt，求 F′(x)。
Fundamental Theorem, F′(x) = (1 + (sin x)²)·cos x. We can write this as cos x·(1 + sin²x).
上限为 v(x) = sin x，因此 v′(x) = cos x。下限为常数 0，贡献为零。根据微积分基本定理的链式法则形式，F′(x) = (1 + (sin x)²)·cos x。可以写成 cos x·(1 + sin²x)。
This example shows that the upper limit does not have to be a polynomial. Any differentiable function of x can appear as a limit, and its derivative becomes the chain-rule factor. The integrand is evaluated at the limiting function, not at x itself.
这个例子说明上限不一定是多项式。任何可导的 x 的函数都可以作为上下限，其导数成为链式法则因子。被积函数要在上下限函数处求值，而不是在 x 处求值。
F′(x
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