Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

WHAT are we proving?

WHY two parts? Part 1 guarantees an antiderivative exists (it constructs one as an integral). Part 2 uses an antiderivative to compute a definite integral. Part 1 supplies the existence; Part 2 supplies the shortcut.

Figure — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

HOW to prove Part 1 — from first principles

We use the definition of the derivative, nothing else.

Step 1 — Write the difference quotient. $F'(x) = \lim_{h\to 0}\frac{F(x+h)-F(x)}{h}.$ Why this step? The derivative IS this limit by definition — we are deriving, not assuming.

Step 2 — Simplify the numerator using the additivity of integrals. $F(x+h)-F(x) = \int_a^{x+h} f(t)\,dt - \int_a^{x} f(t)\,dt = \int_x^{x+h} f(t)\,dt.$ Why this step? $\int_a^{x+h}=\int_a^x+\int_x^{x+h}$ (splitting an interval). The "area up to $x$ " cancels, leaving the thin sliver from $x$ to $x+h$ .

Step 3 — Squeeze the sliver using the Extreme Value Theorem. Since $f$ is continuous on the closed interval $[x,x+h]$ , it attains a minimum $m_h=f(u)$ and maximum $M_h=f(v)$ there. The area of the sliver is between the smallest and largest rectangle: $m_h\, h \le \int_x^{x+h} f(t)\,dt \le M_h\, h \quad (h>0).$ Divide by $h$ : $m_h \le \frac{F(x+h)-F(x)}{h} \le M_h.$ Why this step? A continuous function can't dip below its min or exceed its max, so the average height of the sliver is trapped between them.

Step 4 — Take the limit (Squeeze Theorem). As $h\to 0$ , the points $u,v\in[x,x+h]\to x$ , and by continuity $f(u)\to f(x)$ , $f(v)\to f(x)$ . So both bounds collapse: $\lim_{h\to 0} m_h = f(x) = \lim_{h\to 0} M_h.$ By the Squeeze Theorem the middle term is forced to $f(x)$ : $\boxed{F'(x) = f(x).}$ (The $h<0$ case is identical with inequalities flipped — same conclusion.) $\blacksquare$

HOW to prove Part 2 — building on Part 1 + MVT

Step 1 — Use Part 1 to get one antiderivative. Define $F(x)=\int_a^x f(t)\,dt$ . By Part 1, $F'=f$ .

Step 2 — Compare to an arbitrary antiderivative $G$ . Suppose $G'=f$ too. Then $(F-G)'=f-f=0$ on $[a,b]$ . Why this step? We must connect the specific integral-defined $F$ to any $G$ the student might find.

Step 3 — Zero derivative ⇒ constant (this needs the Mean Value Theorem). If $(F-G)'=0$ everywhere, then for any $x$ , MVT gives $(F-G)(x)-(F-G)(a)=(F-G)'(c)(x-a)=0$ . Hence $F-G$ is constant: $G(x)=F(x)+C$ . Why this step? "Derivative zero ⇒ constant" is not free — it is a consequence of the MVT. This is the honest hinge of the proof.

Step 4 — Evaluate at the endpoints. $G(b)-G(a) = \big(F(b)+C\big)-\big(F(a)+C\big) = F(b)-F(a).$ Now $F(b)=\int_a^b f\,dt$ and $F(a)=\int_a^a f\,dt = 0$ . Therefore $\boxed{\int_a^b f(x)\,dx = G(b)-G(a).}$ $\blacksquare$

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Feynman: explain to a 12-year-old

Imagine pouring water into a tank while watching the water level. The height of the water curve at this instant is how fast the level rises — that's Part 1 (rate of filling = the curve's height). Now, the total water added from start to end is just (level at the end) minus (level at the start) — you don't need to watch every drop, just check the two readings — that's Part 2. Filling up (integrating) and measuring the rising speed (differentiating) are two sides of the same coin.

What does FTC Part 1 state?

f

is continuous and

F(x)=\int_a^x f(t)dt

, then

F'(x)=f(x)

What does FTC Part 2 state?

G'=f

with

f

continuous, then

\int_a^b f\,dx = G(b)-G(a)

In the Part 1 proof, what simplifies

F(x+h)-F(x)

\int_x^{x+h}f

Additivity of integrals (interval splitting).

Which theorem provides

m_h h \le \int_x^{x+h}f \le M_h h

Extreme Value Theorem (max/min on closed interval), then bounded rectangles.

Which theorem forces the difference quotient to

f(x)

h\to 0

The Squeeze Theorem, using continuity of

f

In Part 2, why is

G(x)=F(x)+C

(F-G)'=f-f=0

, and zero derivative ⇒ constant by the Mean Value Theorem.

Why does the constant

C

not appear in the final answer of Part 2?

It cancels in

G(b)-G(a)

Compute

\frac{d}{dx}\int_0^{x^2}\sin t\,dt

\sin(x^2)\cdot 2x

(Part 1 + chain rule).

What single hypothesis on

f

powers both proofs?

Continuity on

[a,b]

What is

\int_a^a f(t)dt

and why does it matter?

0

; it sets the base value of the area function

F

Connections

Riemann Sums and the Definite Integral — what $\int_a^b$ means before FTC.
Antiderivatives and Indefinite Integrals — Part 2's $G$ .
Mean Value Theorem — the hinge of Part 2, Step 3.
Extreme Value Theorem / Squeeze Theorem — engines of Part 1.
Continuity — the indispensable hypothesis.
Chain Rule — extends Part 1 to variable limits (Leibniz rule).
Improper Integrals — where continuity fails and FTC needs care.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, FTC ka core idea simple hai: integration aur differentiation ek-doosre ko undo karte hain. Imagine karo ek tank mein paani bhar rahe ho. Curve $f(t)$ ki height batati hai ki paani kitni tezi se bhar raha hai. Ab "ab tak kitna paani jama hua" — woh hai area function $F(x)=\int_a^x f(t)\,dt$ . Part 1 kehta hai: is jama hue paani ke badhne ki rate exactly curve ki height ke barabar hai, yani $F'(x)=f(x)$ . Proof mein humne sirf derivative ki definition li, area ko thin sliver $\int_x^{x+h}f$ tak simplify kiya, EVT se usko $m_h h$ aur $M_h h$ ke beech trap kiya, aur Squeeze theorem se limit nikaal di. Continuity yahan zaroori hai — usi se $f(u),f(v)\to f(x)$ hota hai.

Part 2 practical jaadu hai. Jab koi bhi antiderivative $G$ mil jaaye (yani $G'=f$ ), to definite integral bas $G(b)-G(a)$ ho jata hai — har chhoti strip add karne ki zaroorat nahi, sirf do endpoint readings. Iska proof: Part 1 se humein ek antiderivative $F$ mil gaya; phir $(F-G)'=0$ , aur MVT se zero derivative ka matlab constant — to $G=F+C$ . Endpoints pe subtract karo, $C$ cancel ho jata hai, aur answer aa jata hai.

Exam mein do galtiyan sabse common hain: (1) variable upper limit ke saath chain rule bhool jaana — $\frac{d}{dx}\int_0^{x^2}f = f(x^2)\cdot 2x$ , woh $2x$ mat chhodo; (2) Part 2 mein lower limit ka term subtract karna bhool jaana. Yaad rakho: Part 1 antiderivative ka existence prove karta hai, Part 2 usse count karta hai. Bas yeh do lines pakad lo, baaki sab clear ho jayega.

Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

WHAT are we proving?

HOW to prove Part 1 — from first principles

HOW to prove Part 2 — building on Part 1 + MVT

Worked examples

Common mistakes (steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections