一、什么是”积分上限为变量的定积分”：从面积函数到微积分基本定理 | What Is a Definite Integral with a Variable Upper Limit: From Area Functions to the Fundamental Theorem
在 IB 数学 AA HL 的微积分部分，我们最先遇到的定积分都有固定的上下限，例如 ∫₁³ x² dx，它算出的是一个具体的数：26/3。但如果我们把上限从固定的 3 改成变量 x，就得到一个全新的对象：F(x) = ∫₁ˣ t² dt。这个 F(x) 不再是某个数值，而是一个关于 x 的函数。读者可以把 F(x) 理解为”从 1 到 x 这一段的面积”，当 x 变化时，面积也跟着变化。
In the calculus section of IB Mathematics AA HL, the definite integrals we meet first have fixed limits, such as ∫₁³ x² dx, which evaluates to a single number: 26/3. But if we change the upper limit from a fixed value like 3 to a variable x, we obtain a completely new object: F(x) = ∫₁ˣ t² dt. This F(x) is no longer a number; it is a function of x. You can think of F(x) as “the area from 1 up to x”. As x moves, the area moves with it.
这个看似简单的改动，恰恰是微积分基本定理（Fundamental Theorem of Calculus，FTC）的核心。它回答了微积分中一个最深刻的问题：微分和积分到底是什么关系？答案出人意料地简洁：它们互为逆运算。对一个变上限积分求导，得到的正是被积函数本身。本文将从定义出发，逐步推导出 FTC 的两大推论，并配以 IB 真题风格的例题与易错点总结，帮助同学们彻底掌握这类题型。
This apparently small change is exactly the heart of the Fundamental Theorem of Calculus (FTC). It answers one of the deepest questions in calculus: what is the true relationship between differentiation and integration? The answer is surprisingly simple: they are inverse operations. Differentiating a definite integral with a variable upper limit returns the integrand itself. Starting from the definition, this article derives the two main consequences of the FTC step by step, and then provides IB exam-style worked examples plus a summary of common mistakes, so that students can master this question type completely.
二、微积分基本定理第一部分的严格表述 | FTC Part 1: The Exact Statement You Must Memorise
IB 考试中，微积分基本定理通常以文字题（state the fundamental theorem of calculus）的形式出现，因此同学们必须能够准确地写出它的两种形式。第一部分（FTC-1）的标准表述如下：设函数 f 在闭区间 [a, b] 上连续，定义 F(x) = ∫ₐˣ f(t) dt，其中 x 属于 [a, b]。则 F 在 [a, b] 上连续，在开区间 (a, b) 内可导，并且 F'(x) = f(x)。
In IB exams, the Fundamental Theorem of Calculus often appears as a “state the theorem” question, so you must be able to write down both of its forms accurately. The standard statement of Part 1 (FTC-1) is: let f be continuous on the closed interval [a, b], and define F(x) = ∫ₐˣ f(t) dt for x in [a, b]. Then F is continuous on [a, b], differentiable on the open interval (a, b), and F'(x) = f(x).
注意三个细节。第一，被积函数里的变量用 t 而不是 x，这叫做”哑变量”（dummy variable），它的名字不影响积分的值；第二，定理要求 f 连续，这个条件保证了 F 一定存在并且可导；第三，a 可以是任意常数，甚至比 x 大，这时 F(x) 表示的是”带符号面积”（signed area），即曲线在 x 轴下方时面积为负。
Note three details. First, the integration variable inside the integrand is t, not x; this is called a dummy variable, and its name does not affect the value of the integral. Second, the theorem requires f to be continuous, a condition that guarantees F exists and is differentiable. Third, a can be any constant, even larger than x; in that case F(x) represents the signed area, meaning the area is negative when the curve lies below the x-axis.
第二部分（FTC-2）则把积分与反导数联系起来：如果 f 在 [a, b] 上连续，且 F 是 f 的任意一个反导数（即 F’ = f），那么 ∫ₐᵇ f(x) dx = F(b) − F(a)。两个部分合在一起，构成了”微分与积分互逆”的完整图景：FTC-1 说”积分再求导等于原函数”，FTC-2 说”求导再积分等于函数值的差”。
Part 2 (FTC-2) links integration to antiderivatives: if f is continuous on [a, b] and F is any antiderivative of f (that is, F’ = f), then ∫ₐᵇ f(x) dx = F(b) − F(a). Together, the two parts form the complete picture of “differentiation and integration are inverse”: FTC-1 says “integrate then differentiate gives the original function”, while FTC-2 says “differentiate then integrate gives the difference of function values”.
三、为什么 F'(x) = f(x)：积分的”求逆”性质的直观证明 | Why F'(x) = f(x): An Intuitive Proof of the Inverse Relationship
很多同学能够背出 FTC-1，却不知道它为什么成立。理解证明过程，能帮助你在考试中自信地处理变形题。我们从一个直观的几何论证开始。设 F(x) = ∫ₐˣ f(t) dt，考虑 x 增加一个很小的量 h，则 F(x + h) − F(x) = ∫ₓˣ⁺ʰ f(t) dt，这正是从 x 到 x + h 这一小条的面积。
Many students can recite FTC-1 but do not know why it is true. Understanding the proof helps you handle variant questions confidently in exams. We begin with an intuitive geometric argument. Let F(x) = ∫ₐˣ f(t) dt and consider increasing x by a small amount h. Then F(x + h) − F(x) = ∫ₓˣ⁺ʰ f(t) dt, which is exactly the area of the thin strip from x to x + h.
当 h 很小时，这一小条近似于一个矩形，其高约为 f(x)，宽为 h，所以 F(x + h) − F(x) ≈ f(x) · h。两边除以 h 并令 h 趋于 0，就得到 F'(x) = f(x)。严谨地说，这一步用到积分中值定理（Mean Value Theorem for Integrals）：在区间 [x, x+h] 内必存在一点 c，使得 ∫ₓˣ⁺ʰ f(t) dt = f(c) · h，而当 h → 0 时 c → x，由 f 的连续性即得结论。
When h is small, this strip is approximately a rectangle with height about f(x) and width h, so F(x + h) − F(x) ≈ f(x) · h. Dividing both sides by h and letting h tend to 0 gives F'(x) = f(x). Rigorously, this step uses the Mean Value Theorem for Integrals: there exists a point c in [x, x + h] such that ∫ₓˣ⁺ʰ f(t) dt = f(c) · h, and as h → 0 we have c → x; the continuity of f then gives the result.
这个证明揭示了 FTC 的本质：F 的导数，就是”面积随右端点移动的瞬时增长速度”，而这个速度恰好等于右端点处曲线的高度 f(x)。换句话说，曲线越高，面积增长得越快。这一几何直觉在 IB Paper 3 的探究题中经常被考查。
This proof reveals the essence of the FTC: the derivative of F is the instantaneous rate at which the area grows as the right endpoint moves, and this rate equals the height f(x) of the curve at that endpoint. In other words, the higher the curve, the faster the area accumulates. This geometric intuition is frequently examined in IB Paper 3 exploration-style questions.
四、变上限积分的几何意义：面积函数的斜率 | The Geometric Meaning of a Variable-Limit Integral: The Slope of the Area Function
把 F(x) = ∫ₐˣ f(t) dt 画出来，得到的曲线称为”面积函数”（area function）。它有一个非常实用的解读方式：F(x) 在某点的斜率 = f(x) 在该点的函数值。因此，如果题目给你 f 的图像而要求你画出 F 的图像，你只需”读斜率”即可：f 为正时 F 上升，f 为负时 F 下降，f 为零的地方 F 有水平切线（极值点）。
Plotting F(x) = ∫ₐˣ f(t) dt gives a curve called the area function. It has a very practical interpretation: the slope of F at any point equals the value of f at that point. Therefore, if a question gives you the graph of f and asks you to sketch the graph of F, you only need to “read the slope”: where f is positive, F is increasing; where f is negative, F is decreasing; where f is zero, F has a horizontal tangent (a stationary point).
再进一步：F 的凹凸性由 F” = f’ 决定。f 递增的地方 F 是凹向上的，f 递减的地方 F 是凹向下的，而 f 的极值点对应 F 的拐点。这类”由 f 的图像推 F 的图像”的题型是 IB AA HL Paper 2 的高频题，也是同学们最容易丢分的地方，因为大家习惯于直接算解析式，而忽略了图像对应关系。
Going further: the concavity of F is governed by F” = f’. Where f is increasing, F is concave up; where f is decreasing, F is concave down; and the stationary points of f correspond to inflection points of F. This type of question, “deduce the graph of F from the graph of f”, is a high-frequency item in IB AA HL Paper 2, and a common source of lost marks, because students habitually compute explicit formulas and overlook the graph correspondence.
还有一个重要的等价视角：变上限积分正是”反导数的图像定义”。∫ₐˣ f(t) dt 是 f 的无数个反导数之一，而改变常数 a 只是把整条面积函数上下平移。这解释了为什么所有反导数只相差一个常数：不同的 a 对应不同的平移量。
There is also an important equivalent viewpoint: the variable-limit integral is precisely the graphical definition of an antiderivative. ∫ₐˣ f(t) dt is one of the infinitely many antiderivatives of f, and changing the constant a merely shifts the whole area function vertically. This explains why all antiderivatives differ only by a constant: different choices of a correspond to different vertical shifts.
五、上下限都是函数的推广：莱布尼茨积分法则 | The Generalisation When Both Limits Are Functions: The Leibniz Integral Rule
IB 真题中极少直接给”上限就是 x”这种最朴素的形式，更常见的写法是 g(x) = ∫_{u(x)}^{v(x)} f(t) dt，上下限本身也是 x 的函数。这时需要用到莱布尼茨积分法则（Leibniz integral rule）：
IB exam questions rarely use the simplest form where the upper limit is exactly x. Far more common is g(x) = ∫_{u(x)}^{v(x)} f(t) dt, where both limits are themselves functions of x. This requires the Leibniz integral rule:
d/dx ∫_{u(x)}^{v(x)} f(t) dt = f(v(x)) · v'(x) − f(u(x)) · u'(x)。
d/dx ∫_{u(x)}^{v(x)} f(t) dt = f(v(x)) · v'(x) − f(u(x)) · u'(x).
这个公式的推导只需三步：首先把积分拆成两项，∫_{u(x)}^{v(x)} = ∫_{a}^{v(x)} − ∫_{a}^{u(x)}；其次对每一项应用 FTC-1，例如 d/dx ∫ₐ^{v(x)} f(t) dt = f(v(x)) · v'(x)，这里对上限 v(x) 使用了链式法则；最后把两项相减，得到带负号的下限项。
The derivation takes only three steps. First, split the integral into two parts: ∫_{u(x)}^{v(x)} = ∫_{a}^{v(x)} − ∫_{a}^{u(x)}. Second, apply FTC-1 to each part; for example d/dx ∫ₐ^{v(x)} f(t) dt = f(v(x)) · v'(x), where the chain rule is used on the upper limit v(x). Third, subtract the two results, which produces the negative sign in front of the lower-limit term.
记忆口诀：上导正、下导负、先代入再乘内层导数（”代入上限乘上限导数，减去代入下限乘下限导数”）。特别地，当下限是常数 u(x) = a 时，u'(x) = 0，公式自动退化为 FTC-1 的形式，所以莱布尼茨法则是 FTC-1 的完整版，考试中建议直接使用完整版以避免疏漏。
Memory aid: upper limit positive, lower limit negative; substitute first, then multiply by the inner derivative (“substitute the upper limit and multiply by its derivative, minus substitute the lower limit and multiply by its derivative”). In particular, when the lower limit is a constant u(x) = a, we have u'(x) = 0 and the formula automatically reduces to FTC-1. The Leibniz rule is therefore the complete version of FTC-1, and in exams it is safer to use the full version to avoid omissions.
六、经典例题一：直接应用 FTC-1 | Worked Example 1: Direct Application of FTC-1
例题：设 F(x) = ∫₂ˣ (t³ + sin t) dt，求 F'(x)。
Example: Let F(x) = ∫₂ˣ (t³ + sin t) dt. Find F'(x).
解答：这是 FTC-1 的最直接应用。上限就是 x 本身，内层导数为 1，所以直接把被积函数中的 t 换成 x 即可：F'(x) = x³ + sin x。注意下限 2 完全没有进入答案，因为它是一个常数，常数下限对导数没有贡献。
Solution: This is the most direct application of FTC-1. The upper limit is exactly x, whose derivative is 1, so we simply replace t by x in the integrand: F'(x) = x³ + sin x. Note that the lower limit 2 does not appear in the answer at all, because it is a constant and a constant lower limit contributes nothing to the derivative.
变式思考：如果题目改为 F(x) = ∫₂ˣ (t³ + sin t) dt 求 F'(3)，仍然先算出 F'(x) = x³ + sin x，再代入 x = 3，得到 F'(3) = 27 + sin 3。很多同学会错误地把 3 直接代入被积函数中的 t，即 3³ + sin 3 的算法其实结果相同，但那是运气：正确的理解是”先求导函数，再代值”。
Variant: if the question instead asks for F'(3) where F(x) = ∫₂ˣ (t³ + sin t) dt, you still first compute F'(x) = x³ + sin x, then substitute x = 3 to obtain F'(3) = 27 + sin 3. Many students wrongly substitute 3 directly into the t in the integrand; the answer happens to coincide, but that is luck. The correct reasoning is “differentiate first, then evaluate”.
另一种高频变式是”先求导再求驻点”：例如求 F 的极值点。令 F'(x) = x³ + sin x = 0，解出驻点后，再通过 F”(x) = 3x² + cos x 的符号判断极大还是极小。这类题把 FTC 与函数分析无缝衔接，是 IB 特别喜欢的综合考法。
Another high-frequency variant is “differentiate, then find stationary points”: for example, find the extrema of F. Setting F'(x) = x³ + sin x = 0 gives the stationary points, and the sign of F”(x) = 3x² + cos x then decides whether each is a maximum or a minimum. Such questions connect the FTC seamlessly with function analysis, a combination IB particularly likes.
七、经典例题二：上下限均为函数（莱布尼茨法则） | Worked Example 2: Both Limits Are Functions (Leibniz Rule)
例题：设 g(x) = ∫_{x²}^{x³} e^{t²} dt，求 g'(x)。
Example: Let g(x) = ∫_{x²}^{x³} e^{t²} dt. Find g'(x).
解答：上限 v(x) = x³，下限 u(x) = x²。代入莱布尼茨法则：
Solution: The upper limit is v(x) = x³ and the lower limit is u(x) = x². Substituting into the Leibniz rule:
g'(x) = e^{(x³)²} · 3x² − e^{(x²)²} · 2x = 3x² e^{x⁶} − 2x e^{x⁴}。
g'(x) = e^{(x³)²} · 3x² − e^{(x²)²} · 2x = 3x² e^{x⁶} − 2x e^{x⁴}.
三步走的执行顺序非常关键：第一步，把上限 x³ 代入被积函数的 t，得到 e^{x⁶}（注意是 (x³)² = x⁶，而不是 x⁵，这是最常见的计算错误）；第二步，乘以上限的导数 3x²；第三步，处理下限 x²，代入得 e^{x⁴}，乘以下限导数 2x，并且整体带负号。
The order of execution is critical. Step 1: substitute the upper limit x³ for t in the integrand, obtaining e^{x⁶} (note that (x³)² = x⁶, not x⁵, which is the most common calculation error). Step 2: multiply by the derivative of the upper limit, 3x². Step 3: handle the lower limit x², substitute to get e^{x⁴}, multiply by its derivative 2x, and put a minus sign in front of the whole term.
如果这道题再进一步，例如求 g'(1) 或 g'(0)，直接代入即可：g'(1) = 3e − 2e = e，g'(0) = 0。注意即使被积函数 e^{t²} 没有初等反导数（无法用基本函数写出不定积分），我们依然能够求出 g’ 的解析式，这正是 FTC/莱布尼茨法则强大的地方：我们不需要真的算出那个积分。
If the question goes further, for example asking for g'(1) or g'(0), just substitute: g'(1) = 3e − 2e = e, and g'(0) = 0. Note that even though e^{t²} has no elementary antiderivative (its indefinite integral cannot be written with basic functions), we can still find an explicit formula for g’. This is the power of the FTC and the Leibniz rule: we never need to actually compute the integral.
八、经典例题三：与洛必达法则结合（IB 压轴题模式） | Worked Example 3: Combining with L’Hopital’s Rule (IB Final-Question Pattern)
例题：求极限 lim_{x→0} (∫₀ˣ e^{t²} dt) / x。
Example: Evaluate the limit lim_{x→0} (∫₀ˣ e^{t²} dt) / x.
解答：先检查形式。当 x → 0 时，分子 ∫₀ˣ e^{t²} dt → ∫₀⁰ e^{t²} dt = 0，分母 x → 0，因此这是 0/0 型不定式，可以使用洛必达法则。对分子求导时恰好用上 FTC-1：
Solution: First check the form. As x → 0, the numerator ∫₀ˣ e^{t²} dt → ∫₀⁰ e^{t²} dt = 0 and the denominator x → 0, so this is a 0/0 indeterminate form and L’Hopital’s rule applies. Differentiating the numerator uses FTC-1 exactly:
lim_{x→0} (∫₀ˣ e^{t²} dt) / x = lim_{x→0} e^{x²} / 1 = e⁰ = 1。
lim_{x→0} (∫₀ˣ e^{t²} dt) / x = lim_{x→0} e^{x²} / 1 = e⁰ = 1.
进阶例题：求 lim_{x→0} (∫₀ˣ sin(t²) dt) / x³。同样先判断为 0/0 型，用洛必达法则一次：分子导数 = sin(x²)，分母导数 = 3x²，得到 lim sin(x²)/(3x²)。再使用等价无穷小 sin u ~ u（u → 0），sin(x²) ~ x²，故极限 = x²/(3x²) = 1/3。注意这里若继续用洛必达法则也可以：lim 2x·cos(x²)/(6x) = 1/3，殊途同归。
Advanced example: evaluate lim_{x→0} (∫₀ˣ sin(t²) dt) / x³. This is again 0/0. Applying L’Hopital’s rule once, the derivative of the numerator is sin(x²) and the derivative of the denominator is 3x², giving lim sin(x²)/(3x²). Using the equivalent infinitesimal sin u ~ u as u → 0, we have sin(x²) ~ x², so the limit is x²/(3x²) = 1/3. Alternatively, applying L’Hopital again gives lim 2x·cos(x²)/(6x) = 1/3; both routes agree.
这类”洛必达 + FTC”的组合是 AA HL Paper 2 与 Paper 3 的压轴常客。解题口诀：看到积分符号与极限并存，先检验是否为 0/0 或 ∞/∞ 型，再对分子用 FTC 求导，必要时重复使用洛必达法则，最后代值或使用等价无穷小化简。
This combination of “L’Hopital + FTC” is a frequent final-question guest in AA HL Papers 2 and 3. Memory aid: when you see an integral symbol together with a limit, first check whether the form is 0/0 or ∞/∞, then differentiate the numerator with the FTC, repeat L’Hopital’s rule if necessary, and finally substitute the value or simplify with equivalent infinitesimals.
九、五个高频易错点与对策 | Five High-Frequency Mistakes and How to Avoid Them
易错点一：忘记链式法则。上限是 2x、x² 等函数时，必须乘以上限的导数。对策：写完答案后，用”上限代入 × 上限导数 − 下限代入 × 下限导数”的模板逐项核对。
Mistake 1: forgetting the chain rule. When the upper limit is a function such as 2x or x², you must multiply by the derivative of that limit. Countermeasure: after writing the answer, check term by term against the template “substituted upper limit × derivative of upper limit, minus substituted lower limit × derivative of lower limit”.
易错点二：下限项漏掉负号。很多同学只处理上限，把下限当成常数忽略。对策：把公式完整写出来再代值，尤其注意 u(x) 存在时 −f(u(x))·u'(x) 这一项不能丢。
Mistake 2: missing the minus sign on the lower-limit term. Many students handle only the upper limit and treat the lower limit as a constant. Countermeasure: write the complete formula before substituting values, and remember that the term −f(u(x))·u'(x) must not be dropped when u(x) is present.
易错点三：哑变量混淆。题目写成 ∫₀ˣ x·t dt 时，被积函数中的 x 是”参数”，只有 t 是积分变量；求导前应先正确区分。若对 x 和 t 的角色混淆，后续所有计算都会出错。
Mistake 3: confusing the dummy variable. When a question is written as ∫₀ˣ x·t dt, the x inside the integrand is a parameter and only t is the integration variable; you must distinguish them correctly before differentiating. Confusing the roles of x and t corrupts every later step.
易错点四：认为变上限积分可以直接用”上底 + 下底 × 高”之类的近似公式。F(x) = ∫ₐˣ f(t) dt 是精确的面积，不是 f(x)·(x − a) 的矩形近似。只有在题目明确要求近似（如梯形法则、辛普森法则）时才使用近似公式。
Mistake 4: believing a variable-limit integral can be replaced by crude “rectangle” approximations. F(x) = ∫ₐˣ f(t) dt is the exact area, not the rectangle approximation f(x)·(x − a). Use approximation formulas only when a question explicitly asks for them (for example, the trapezium rule or Simpson’s rule).
易错点五：代入顺序错误。正确的顺序永远是”先代入上下限，再乘导数”，而不是先对上下限求导再代入。例如例题二中，必须先算 e^{(x³)²}，再乘 3x²；若先乘后代入，虽然结果常常相同，但在更复杂的嵌套函数中会出现错误。
Mistake 5: wrong substitution order. The correct order is always “substitute the limits first, then multiply by their derivatives”, never differentiate the limits first and substitute afterwards. In Example 2, you must first compute e^{(x³)²} and then multiply by 3x²; substituting after multiplying often gives the same result here, but it fails with more complicated nested functions.
十、IB AA HL 真题考法分析：这类题在试卷中如何出现 | How This Topic Appears in Real IB AA HL Papers
在近几年的 AA HL 试卷中，”变上限积分”主要出现在三个位置。其一是 Paper 1 的短答题：直接给出 F(x) = ∫ₐˣ f(t) dt，求 F'(x) 或 F'(k)，属于基础送分题，但要求书写规范。其二是 Paper 2 的图像题：给出 f 的图像与若干关键点，要求判断面积函数 F 的增减、极值、凹凸与拐点，或求 F 在特定点的值（此时需要借助梯形或几何面积）。
In recent AA HL papers, “variable-limit integrals” appear in three main positions. First, Paper 1 short questions: given F(x) = ∫ₐˣ f(t) dt, find F'(x) or F'(k); these are basic mark-earners but require neat presentation. Second, Paper 2 graph questions: given the graph of f with key points, decide where the area function F is increasing, where its extrema, concavity and inflection points are, or evaluate F at specific points using geometric areas (trapezia or simple shapes).
其三是 Paper 2/3 的综合题：把变上限积分与洛必达法则、微分方程、级数甚至概率密度函数结合。例如在连续概率分布中，累积分布函数 F(x) = ∫_{−∞}^{x} f(t) dt 本质上就是一个变上限积分，其导数 f(x) 正是概率密度函数；理解 FTC 的同学能立刻看出这一层关系，而只靠背公式的同学往往无法迁移。
Third, Paper 2/3 extended questions: variable-limit integrals combined with L’Hopital’s rule, differential equations, series, or probability density functions. For example, in continuous probability distributions, the cumulative distribution function F(x) = ∫_{−∞}^{x} f(t) dt is essentially a variable-limit integral, whose derivative f(x) is exactly the probability density function. Students who understand the FTC see this connection immediately, while students who only memorise formulas often fail to transfer the idea.
评分标准方面，IB 阅卷对”是否写出 FTC 名称或公式”不作强制要求，但对”中间步骤是否完整”非常严格：直接写答案不给满分，必须展示”代入上限 → 乘导数 → 减去下限项”的完整过程。建议同学们在答卷上至少写出莱布尼茨法则的通用形式，再代入具体函数。
Regarding marking, IB examiners do not require you to name the FTC, but they are strict about complete intermediate steps: writing only the final answer earns no full marks. You must show “substitute upper limit, multiply by its derivative, subtract the lower-limit term” in full. It is recommended to write down the general Leibniz rule first, then substitute the specific functions.
十一、分层练习题与完整解答 | Tiered Practice Problems with Full Solutions
基础题 1：设 F(x) = ∫₁ˣ ln t dt，求 F'(x)。解答：由 FTC-1，F'(x) = ln x。
Foundation question 1: Let F(x) = ∫₁ˣ ln t dt. Find F'(x). Solution: by FTC-1, F'(x) = ln x.
基础题 2：设 h(x) = ∫_{x}^{2x} (t² + 1) dt，求 h'(x)。解答：h'(x) = ((2x)² + 1)·2 − (x² + 1)·1 = (4x² + 1)·2 − x² − 1 = 7x² + 1。
Foundation question 2: Let h(x) = ∫_{x}^{2x} (t² + 1) dt. Find h'(x). Solution: h'(x) = ((2x)² + 1)·2 − (x² + 1)·1 = (4x² + 1)·2 − x² − 1 = 7x² + 1.
进阶题 1：求 lim_{x→0} (∫₀ˣ cos(t²) dt) / x。解答：0/0 型，洛必达 + FTC：lim cos(x²)/1 = 1。
Challenge question 1: Evaluate lim_{x→0} (∫₀ˣ cos(t²) dt) / x. Solution: 0/0 form; L’Hopital plus FTC gives lim cos(x²)/1 = 1.
进阶题 2：已知 f 是连续函数，且 ∫₀^{x²} f(t) dt = x·sin x，求 f(4)。解答：两边对 x 求导：f(x²)·2x = sin x + x·cos x。代入 x = 2：f(4)·4 = sin 2 + 2·cos 2，故 f(4) = (sin 2 + 2·cos 2)/4。这类”等式两端同时求导”的技巧是隐藏的 FTC 应用题，请务必掌握。
Challenge question 2: Given that f is continuous and ∫₀^{x²} f(t) dt = x·sin x, find f(4). Solution: differentiate both sides with respect to x: f(x²)·2x = sin x + x·cos x. Substituting x = 2 gives f(4)·4 = sin 2 + 2·cos 2, hence f(4) = (sin 2 + 2·cos 2)/4. This “differentiate both sides of the equation” technique is a disguised FTC application, and you must master it.
综合题：已知 g(x) = ∫₀ˣ f(t) dt，且 f 的图像在 [0, 4] 上为一条经过 (0, 0)、(2, 3)、(4, 0) 的抛物线（开口向下）。求 g'(2)、g”(2)，并判断 g 在 x = 2 处是否有极值。解答：g'(x) = f(x)，故 g'(2) = f(2) = 3；g”(x) = f'(x)，抛物线 f 在 x = 2 处达到最大值，故 f'(2) = 0，即 g”(2) = 0。又因 f 在 2 附近先增后减，g’ = f 由正变负，故 g 在 x = 2 处取得极大值。此题同时考查 FTC、极值判定与图像理解，是典型的 IB 综合题。
Integrated question: Let g(x) = ∫₀ˣ f(t) dt, where the graph of f on [0, 4] is a downward-opening parabola through (0, 0), (2, 3) and (4, 0). Find g'(2) and g”(2), and decide whether g has an extremum at x = 2. Solution: g'(x) = f(x), so g'(2) = f(2) = 3; g”(x) = f'(x), and since the parabola f reaches its maximum at x = 2, f'(2) = 0, hence g”(2) = 0. Moreover, because f increases then decreases around 2, g’ = f changes from positive to negative, so g has a local maximum at x = 2. This question tests the FTC, extremum classification and graphical understanding together, a typical IB integrated question.
Summary | 总结
本文围绕”积分上限为变量的定积分”展开了完整的知识框架：第一，我们明确了变上限积分 F(x) = ∫ₐˣ f(t) dt 的本质是一个函数，其几何意义是带符号面积；第二，我们给出了微积分基本定理第一部分的严格表述与直观证明，理解了”积分再求导还原被积函数”的互逆本质；第三，我们推广到上下限均为函数的莱布尼茨积分法则，并总结出”代入上限乘上限导数，减去代入下限乘下限导数”的操作模板；第四，通过三道层层递进的例题，我们展示了直接应用、复合函数应用以及与洛必达法则结合三类核心考法；第五，我们归纳了五个高频易错点与对策，并分析了 IB 真题中该知识点的三个典型出现位置；最后，通过分层练习题检验了掌握程度。
This article has built a complete framework around “definite integrals with a variable upper limit”. First, we clarified that the variable-limit integral F(x) = ∫ₐˣ f(t) dt is fundamentally a function whose geometric meaning is signed area. Second, we gave the exact statement and an intuitive proof of FTC Part 1, understanding the inverse relationship that integrating and then differentiating recovers the integrand. Third, we generalised to the Leibniz rule where both limits are functions, and summarised the operating template “substitute the upper limit and multiply by its derivative, minus substitute the lower limit and multiply by its derivative”. Fourth, through three progressively harder worked examples, we demonstrated the three core question types: direct application, composite-function application, and combination with L’Hopital’s rule. Fifth, we listed five high-frequency mistakes with countermeasures and analysed the three typical positions of this topic in real IB papers. Finally, tiered practice problems tested your mastery.
掌握了变上限积分，你就掌握了连接微分与积分的那座桥梁：它不仅是一个考点，更是理解累积分布函数、微分方程初值问题与物理中位移-速度-加速度关系的通用工具。建议同学们在复习时，把本文的例题与练习题独立重做一遍，并把莱布尼茨法则的公式抄写在自己的公式本上，考试前再快速浏览一遍易错点清单。
Once you master variable-limit integrals, you have mastered the bridge connecting differentiation and integration: it is not merely an exam topic but a universal tool for understanding cumulative distribution functions, initial-value problems in differential equations, and the displacement-velocity-acceleration relationships in physics. For revision, we recommend redoing the worked examples and practice problems independently, copying the Leibniz rule into your formula booklet, and skimming the mistake list once more before the exam.
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