The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) says that differentiation and integration are inverse operations. It has two parts --- one proves this from each direction.

Prerequisites

• Derivatives, Differentials, and the Chain Rule --- what $f^{\prime }(x)$ means
• Riemann sums --- approximating the area under a curve by partitioning $[a,b]$ into $n$ strips and summing rectangle areas $\sum f(x_i)\Delta x$; the integral is the limit as $n\to \infty$

Notation

• $f(x)$ --- the function whose area we want to compute (the integrand)
• $F(x)$ --- an antiderivative of $f$, meaning $F^{\prime }(x)=f(x)$. Antiderivatives are not unique: if $F$ is one, then $F+C$ is another for any constant $C$. This doesn’t matter for the FTC because the constant cancels in $F(b)-F(a)$.
• $G(x)={\int }_a^{x}f(t)dt$ --- the “area so far” function (area under $f$ from $a$ to $x$)
• $\Delta x$ --- a small increment in $x$; becomes $dx$ in the limit

Intuition

Two insights, one per direction:

1. Area grows at rate = height. If you sweep a vertical line from left to right under a curve, the rate at which area accumulates at position $x$ is exactly $f(x)$ --- the height of the curve there. (FTC Part 1)

2. Total change = sum of tiny changes. If you chop $F(b)-F(a)$ into many tiny differences, each difference is approximately a rectangle of area $f(x_i)\Delta x$, and the whole thing telescopes so only the endpoints survive. (FTC Part 2)

FTC Part 2: the telescoping direction

Hypothesis: $f$ is continuous on $[a,b]$, and $F$ is an antiderivative of $f$ (i.e., $F^{\prime }(x)=f(x)$ for all $x\in [a,b]$).

Thesis: ${\int }_a^{b}f(x)dx=F(b)-F(a)$

You start with a function $F$ that you already know satisfies $F^{\prime }=f$. The theorem says: then you can evaluate the integral by just plugging $a$ and $b$ into $F$. Without this, computing ${\int }_1^{3}2xdx$ means taking the limit of Riemann sums. With it, you recognize $F(x)=x^2$ satisfies $F^{\prime }(x)=2x$, so the answer is $3^2-1^2=8$.

Proof

$F(b)-F(a)$ splits into tiny differences $\approx f(x_i)\Delta x$. Intermediate values cancel (telescope), leaving only endpoints.

Step 1. Partition $[a,b]$ into $n$ subintervals of width $\Delta x=(b-a)/n$.

$x_0=a,x_1,x_2,\dots ,x_n=b$

Step 2. Write $F(b)-F(a)$ as a telescoping sum (pure arithmetic --- each intermediate $F(x_i)$ cancels).

$F(b)-F(a)={\sum }_{i=0}^{n-1}[F(x_{i+1})-F(x_i)]$

Step 3. Approximate each term by linear approximation.

$F(x_{i+1})-F(x_i)\approx F^{\prime }(x_i)\cdot \Delta x$

Step 4. Substitute the hypothesis $F^{\prime }(x)=f(x)$.

$F^{\prime }(x_i)\cdot \Delta x=f(x_i)\cdot \Delta x$

Step 5. Combine Steps 2—4.

$F(b)-F(a)\approx {\sum }_{i=0}^{n-1}f(x_i)\Delta x$

Step 6. Take $n\to \infty$. The left side is exact (telescope holds for any $n$). The right side $\to {\int }_a^{b}f(x)dx$ by definition of Riemann integral. The approximation error per term is $O(\Delta x^2)$; summing $n$ terms gives $O(n\cdot \Delta x^2)=O(\Delta x)\to 0$.

${\int }_a^{b}f(x)dx=F(b)-F(a)\square$

FTC Part 1: the area-accumulation direction

Hypothesis: $f$ is continuous on $[a,b]$.

Construction: Define $G(x)={\int }_a^{x}f(t)dt$ (area under $f$ from $a$ to $x$).

Thesis: $G^{\prime }(x)=f(x)$ for all $x\in (a,b)$. That is, $G$ is an antiderivative of $f$.

Why this is non-trivial

$G^{\prime }(x)=f(x)$ looks like “the derivative undoes the integral” --- a tautology. It isn’t, because $G$ and $G^{\prime }$ are defined by different limiting processes:

• $G(x)$ is a limit of Riemann sums (partition $[a,x]$, sum rectangles, take $n\to \infty$).
• $G^{\prime }(x)$ is a limit of difference quotients (${\lim }_{h\to 0}\frac{G(x+h)-G(x)}{h}$).

The theorem says these two nested limits compose cleanly: build $G$ via one infinite process, differentiate via another, get back $f(x)$. This requires $f$ to be continuous --- if $f$ has a jump at $c$, then $G$ has a corner at $c$ and $G^{\prime }(c)$ does not exist.

Physical version: Water flows into a tank at rate $f(t)$ litres/second. Total water is $G(t)={\int }_0^{t}f(s)ds$. The rate of change of total water is $f(t)$. Physically obvious --- but the theorem proves that the Riemann-sum definition of $G$ and the difference-quotient definition of $G^{\prime }$ actually produce this result.

Proof

Nudging $x$ by $h$ adds a thin strip of width $h$ and height $\approx f(x)$.

Step 1. Write $G(x+h)-G(x)$ using the definition of $G$.

$G(x+h)-G(x)={\int }_x^{x+h}f(t)dt$

Step 2. Since $f$ is continuous (hypothesis), $f(t)\approx f(x)$ on $[x,x+h]$. Approximate the integral.

${\int }_x^{x+h}f(t)dt\approx f(x)\cdot h$

Step 3. Divide by $h$.

$\frac{G(x+h)-G(x)}{h}\approx f(x)$

Step 4. Bound the error. Continuity at $x$ gives: for any $\epsilon >0$, there exists $\delta >0$ such that $|f(t)-f(x)|<\epsilon$ whenever $|t-x|<\delta$.

$\left|\frac{G(x+h)-G(x)}{h}-f(x)\right|=\left|\frac{1}{h}{\int }_x^{x+h}[f(t)-f(x)]dt\right|\le \epsilon \text{for }|h|<\delta$

Step 5. Take $h\to 0$. Since $\epsilon$ was arbitrary, the limit exists.

$G^{\prime }(x)={\lim }_{h\to 0}\frac{G(x+h)-G(x)}{h}=f(x)\square$

How the two parts connect

	FTC Part 1	FTC Part 2
Says	$G^{\prime }(x)=f(x)$	${\int }_a^{b}f(x)dx=F(b)-F(a)$
Direction	Area $\to$ antiderivative	Antiderivative $\to$ area
Answers	”Why are they related?"	"How do I compute it?”
Key idea	Area grows at rate = height	Total change = sum of tiny changes

The logical dependency: Part 2 assumes an antiderivative $F$ exists and shows how to use it. But who says $f$ even has an antiderivative? Part 1 answers this: it constructs one explicitly --- namely $G(x)={\int }_a^{x}f(t)dt$ --- and proves $G^{\prime }=f$. So Part 1 guarantees existence, Part 2 gives the computation.

This also explains why “any antiderivative works” in Part 2: if $F$ and $G$ are both antiderivatives of $f$, they differ by a constant ($F=G+C$), and the constant cancels in $F(b)-F(a)$.

Worked example

Let $f(x)=2x$. We want ${\int }_1^{3}2xdx$.

Using FTC Part 2: An antiderivative is $F(x)=x^2$ (since $F^{\prime }(x)=2x=f(x)$). Therefore:

${\int }_1^{3}2xdx=F(3)-F(1)=9-1=8$

Verifying geometrically: The region under $f(x)=2x$ from $x=1$ to $x=3$ is a trapezoid with parallel sides $f(1)=2$ and $f(3)=6$, and width $3-1=2$:

$\text{Area}=\frac{(2+6)}{2}\times 2=8✓$

Verifying FTC Part 1: Define $G(x)={\int }_1^{x}2tdt=t^2{|}_1^x=x^2-1$. Then $G^{\prime }(x)=2x=f(x)$. The area function’s derivative is indeed the original function.

Connection to the Mean Value Theorem

Left: green = actual area under $f$. Blue rectangle at height $f(c)$ has the same area, but you don’t know $c$. Right: sum many thin rectangles to compute the same area without needing any special point.

Both theorems express $F(b)-F(a)$ in terms of $f=F^{\prime }$:

	Statement	What you get
MVT	$F(b)-F(a)=f(c)\cdot (b-a)$ for some $c\in (a,b)$	One rectangle, unknown $c$
FTC	$F(b)-F(a)={\int }_a^{b}f(x)dx$	All rectangles, computable

MVT is existential: it guarantees $c$ exists but doesn’t tell you where. FTC is constructive: it gives the answer as a computation. FTC implies MVT (apply the intermediate value theorem to the integral), but not vice versa.

Questions to sit with

1. What happens at discontinuities? If $f$ has a jump at $x=c$, then $G(x)={\int }_a^{x}f(t)dt$ is still continuous (area doesn’t jump) but has a corner at $c$ --- continuous but not differentiable. Does the integral of $f$ still exist? (Yes --- integrability and differentiability of the area function are separate questions. FTC Part 1 fails at $c$, but the integral over $[a,b]$ is still well-defined.)

2. Why does Part 2 feel more useful than Part 1? Part 2 gives you a computation shortcut (evaluate $F$ at endpoints). Part 1 gives you an existence guarantee (continuous $\Rightarrow$ antiderivative exists). In practice you use Part 2 constantly. When would you actually need Part 1? (Answer: when you don’t have a formula for $F$ but need to reason about $G(x)={\int }_a^{x}f(t)dt$ as a function --- e.g., proving that solutions to differential equations exist.)

3. The telescoping argument works for any differentiable $F$. We never used anything special about $F$ being an antiderivative of a nice function. So why can’t we apply FTC to any random differentiable function? (We can --- the only requirement is that $F^{\prime }$ must be integrable. Most “well-behaved” functions satisfy this, but pathological counterexamples exist.)

The Italian variant

At Politecnico, the FTC is often taught as a single theorem combining both statements: if $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$, then ${\int }_a^{b}f(x)dx=F(b)-F(a)$. The Anglo-American tradition (MIT 18.01, Stewart, Spivak) splits this into two numbered parts to emphasize the two different conceptual directions. The mathematical content is the same.

See also

• Derivatives, Differentials, and the Chain Rule
• Linear Approximation --- the key tool in the Part 2 proof (Step 3)
• The Mean Value Theorem