Fundamental Theorem of Calculus

Why this theorem exists (motivation)

Before FTC, you might experience calculus as two separate worlds:

• •Derivatives: turn a function into its instantaneous slope / rate function.
• •Definite integrals: turn a function into a single number measuring net accumulation (signed area).

FTC answers a natural question:

If integration is “adding up tiny pieces,” and differentiation measures the “instantaneous change,” are these operations related?

They are related in the strongest possible way: integration and differentiation are inverse processes (up to a constant).

The statement (two parts)

FTC is commonly presented in two parts.

FTC Part 1 (Accumulation → derivative)

Let $f$ be continuous on an interval containing $a$. Define an accumulation function

Then

Interpretation: If $F(x)$ measures “area accumulated up to $x$,” then the rate at which that area grows at position $x$ equals the height of the curve there.

FTC Part 2 (Antiderivative → definite integral)

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ (meaning $F'(x)=f(x)$), then

Interpretation: Net accumulated change equals the antiderivative’s endpoint difference.

A quick “two-way bridge” picture

• •Part 1: Start with an integral that depends on $x$ → differentiate → get back $f(x)$.
• •Part 2: Start with $f(x)$ → find an antiderivative $F$ → evaluate at endpoints → get the definite integral.

Visual intuition (diagram you can map to the interactive canvas)

Imagine the interactive canvas shows the curve $y=f(x)$, a fixed vertical line at $x=a$, and a movable vertical line at $x$.

As you drag $x$ to the right, you shade the region under $f$ from $a$ to $x$. The shaded area is $F(x)$.

Canvas state mapping:

• •State A: curve $f$, left bound $a$, right bound $x$, shaded area.
• •State B: a readout showing $F(x)$ as a number.
• •State C: a second plot showing the graph of $F$ being traced as you move $x$.

The key claim of Part 1 is that the slope of the traced $F$ graph at the current $x$ equals the current height $f(x)$.

That is the heart of FTC: slope-of-area equals height-of-curve.

Why Part 1 is true (intuition before formulas)

Think of $F(x)=\int_a^x f(t)\,dt$ as “net area accumulated.”

If you increase $x$ by a tiny amount $h$, you add a thin vertical slice of area from $x$ to $x+h$.

If $h$ is very small, that slice’s area is approximately a rectangle:

• •width ≈ $h$
• •height ≈ $f(x)$ (because $f$ doesn’t change much over a tiny interval)

So the added area is approximately $f(x)\,h$.

But the added area is also exactly $F(x+h)-F(x)$.

So we expect

which implies

Taking the limit as $h\to 0$ gives $F'(x)=f(x)$.

Inline diagram 1: “thin slice” picture

Use this mental picture even without the canvas.

          y
          ^
          |
          |          /\
          |         /  \        curve y=f(x)
          |        /    \
          |_______/______\____________> x
                 a   x  x+h

        Shaded area from a to x is F(x)
        Thin slice from x to x+h has area ≈ f(x)·h

Canvas tie-in: In an interactive version, the thin slice is the last sliver of shaded region as you nudge $x$ to $x+h$.

Show the work (a careful derivation)

Start with

Compute the difference:

Now form the difference quotient:

If $f$ is continuous, then on the tiny interval $[x, x+h]$, $f(t)$ is close to $f(x)$.

A standard argument (using the Mean Value Theorem for Integrals) says there exists some $c\in[x,x+h]$ such that

So

As $h\to 0$, the point $c\to x$, and by continuity $f(c)\to f(x)$. Therefore

What Part 1 does not say

• •It does not say $\int f(x)\,dx = f(x)$.
• •It does not say “integral cancels derivative” without conditions.

It says something precise:

If you define $F$ as the integral from a constant $a$ up to the variable endpoint $x$, then differentiating returns the original integrand.

A pacing checkpoint: interpret $F$ as a new function

When you see

read it as: “$F$ is a function whose input is the upper limit.”

So $F$ has its own graph, its own slope, and its own values.

In the interactive canvas, you can imagine two linked plots:

• •Plot 1: $f(x)$ with shaded area up to $x$.
• •Plot 2: $F(x)$ being drawn point-by-point as $x$ moves.

FTC Part 1 is the rule that synchronizes these plots: the slope on Plot 2 equals the height on Plot 1.

Why Part 2 matters (motivation)

Definite integrals look like they should require adding infinitely many tiny rectangles.

But in practice, we almost never compute them that way.

FTC Part 2 is the practical miracle:

If you can find any antiderivative $F$ of $f$, then the definite integral is just $F(b)-F(a)$.

This turns an accumulation problem into an endpoint subtraction problem.

Connecting Part 2 to Part 1

Part 1 gives you a way to create an antiderivative.

Define

Then Part 1 says $G'(x)=f(x)$. So $G$ is an antiderivative of $f$.

Now evaluate $G$ at $b$:

Also, $G(a)=\int_a^a f(t)\,dt=0$.

So

But any two antiderivatives differ by a constant. If $F'(x)=f(x)$, then $F(x)=G(x)+C$.

So

That’s Part 2.

Inline diagram 2: “area-to-antiderivative endpoints”

Think of $F$ as a height function whose change equals area under $f$.

F(x)
^
|            • (b, F(b))
|          /
|        /
|      /
|  • (a, F(a))
+----------------------------> x
   a                          b

Vertical change: F(b)-F(a)
Equals signed area under f from a to b.

Canvas tie-in: Imagine toggling a mode where instead of shading area under $f$, you watch a point move on the graph of an antiderivative $F$. The theorem says the vertical difference between the endpoint values equals the shaded area you would have gotten.

Signed area and orientation

Remember the definite integral is signed:

• •Area above the $x$-axis contributes positive.
• •Area below contributes negative.

FTC Part 2 preserves that: if $f$ is negative on much of $[a,b]$, then $F(b)-F(a)$ will reflect a net decrease.

Also note the direction:

That matches the endpoint rule:

A pacing checkpoint: what you’re allowed to do

FTC Part 2 gives a recipe:

1. 1)Find $F$ such that $F'(x)=f(x)$.
2. 2)Compute $F(b)-F(a)$.

This is why learning antiderivative patterns (power rule, trig, substitution later) is so valuable.

Mini table: what changes between Part 1 and Part 2

1) Turning geometry/accumulation into algebra

Suppose $f(x)$ is a rate:

• •water flowing into a tank (liters/min)
• •velocity (m/s)
• •marginal cost ($/unit)

Then the accumulated total from $a$ to $b$ is

FTC Part 2 says you can compute that total if you can find an antiderivative.

2) The derivative of “area so far” is the current rate

This is FTC Part 1 in words:

If $F(x)$ is total accumulated amount up to time $x$, then $F'(x)$ is the instantaneous rate at time $x$.

So FTC explains why “total” and “rate” are inverse ideas:

• •Differentiate total → get rate.
• •Integrate rate → get total change.

3) Interpreting an integral-defined function without computing it fully

Often you’ll see a function defined by an integral that you cannot evaluate in elementary terms, like

Even if you can’t find a closed-form antiderivative, FTC Part 1 still gives

That’s extremely useful for analyzing monotonicity, slopes, and local behavior.

4) A common “interactive canvas” mental model

If the canvas had three layers:

• •Layer 1 (Curve): plot $y=f(x)$
• •Layer 2 (Accumulation): shaded region from $a$ to current $x$
• •Layer 3 (Accumulation graph): a point $(x, F(x))$ on the graph of the area function

Then moving $x$ produces two simultaneous changes:

• •The shaded area increases by a thin slice.
• •The point on $F$ rises by exactly that slice area.

FTC Part 1 says: the instantaneous rise-per-run of that point equals the current function height.

5) The “constant of integration” appears naturally

Indefinite integrals represent families of antiderivatives:

FTC Part 2 avoids worrying about $C$ because subtraction cancels it:

So definite integrals are often cleaner than indefinite integrals: the constant disappears automatically.

6) Where this connects next

FTC is a hub concept:

• •It justifies many integration techniques (substitution later feels less like magic).
• •It underlies the idea that integrals solve differential equations by “undoing” derivatives.

If you keep one sentence in mind, make it this:

Net accumulation over an interval equals antiderivative change across the endpoints.