Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Question bank

Reminders of the two statements you are reasoning about (from the parent note):

Part 1: $f$ continuous, $F (x) = \int_{a}^{x} f (t) d t$ $\Rightarrow$ $F^{'} (x) = f (x)$ .
Part 2: $G^{'} = f$ (with $f$ continuous) $\Rightarrow$ $\int_{a}^{b} f = G (b) - G (a)$ .

True or false — justify

Every item is a claim; the answer says which of the two theorems / hypotheses is really doing the work.

FTC Part 1 needs

f

to be differentiable, since we take a derivative of

F

False — we take the derivative of $F$ , not of

f

. We only need

f

continuous; then EVT and Squeeze force

F^{'} = f

. In fact

F

ends up smoother than

f

f

is merely integrable but has a jump, then

F (x) = \int_{a}^{x} f

is still continuous.

True — accumulating area over a shrinking interval

\to 0

, so

F

is continuous everywhere. But

F

need not be differentiable at the jump, so Part 1's conclusion

F^{'} = f

can fail there.

Part 2 requires you to use the specific antiderivative

F (x) = \int_{a}^{x} f

; any other antiderivative gives a different answer.

False — any antiderivative

G

works. Two antiderivatives differ by a constant (proved via the Mean Value Theorem), and that constant cancels in

G (b) - G (a)

The equation

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

says differentiation and integration are inverse operations.

True in the "integrate-then-differentiate" direction: Part 1 shows differentiating the accumulation function returns the original height

f

. The reverse direction (

\int_{a}^{x} F^{'} d t = F (x) - F (a)

) is Part 2.

Since

\int_{a}^{a} f (t) d t = 0

, the lower limit

a

is irrelevant to the value of

F^{'} (x)

True for the derivative: changing

a

shifts

F

by a constant (the area between the two lower limits), and constants vanish under differentiation. So

F^{'} (x) = f (x)

regardless of

a

If two functions have the same derivative on an interval, they are equal.

False — they differ by a constant, not necessarily zero. This "

+ C

" is exactly what makes indefinite antiderivatives a family, and it is what cancels in Part 2.

FTC lets us compute every definite integral by finding an antiderivative.

False in practice — Part 2 needs an antiderivative you can write down. Functions like

\int_{0}^{x} e^{- t^{2}} d t

have no elementary antiderivative, so Part 2 is useless; only Part 1 (or Riemann sums) applies.

Spot the error

Each statement contains one flawed step. Name it.

\frac{d}{d x} \int_{0}^{x^{2}} f (t) d t = f (x^{2})

Missing the Chain Rule factor. The upper limit is

x^{2}

, not

x

, so the answer is

f (x^{2}) \cdot 2 x

. Part 1 gives the rate per unit of the limit; the limit itself moves at speed

2 x

(F - G)^{'} = 0

, therefore

F - G = 0

, so

F = G

Zero derivative gives constant, not zero:

F - G = C

. Concluding

C = 0

needs a specific data point. This is why Part 2 subtracts endpoints instead of claiming

F = G

"Since

f

attains a min

m_{h}

and max

M_{h}

[x, x + h]

, we get

m_{h} h \leq \int_{x}^{x + h} f \leq M_{h} h

— this holds for all

h

The inequality direction assumes

h > 0

. For

h < 0

the interval is

[x + h, x]

and the width factor flips sign, so the inequalities reverse. The conclusion still follows, but you must handle

h < 0

separately.

\frac{d}{d x} \int_{x}^{b} f (t) d t = f (x)

by Part 1."

Wrong sign. Here

x

is the lower limit. Write it as

- \int_{b}^{x} f

, then Part 1 gives

- f (x)

. Increasing the lower limit removes area, so the derivative is negative of the height.

f

is continuous, so by EVT it has a max on

(x, x + h)

EVT needs a closed, bounded interval

[x, x + h]

; on the open interval the max may not be attained. The proof deliberately uses the closed sliver.

"To prove Part 2 we can just cite that antiderivatives differ by a constant — no theorem needed, it's obvious."

"Derivative zero

\Rightarrow

constant" is a theorem, not an axiom, and its honest proof on an interval uses the Mean Value Theorem. Skipping it hides the real logical hinge.

Why questions

Why must

f

be continuous, not merely integrable, for Part 1's conclusion

F^{'} = f

The proof twice uses continuity: EVT needs it to guarantee a max/min on the sliver, and the final Squeeze needs

f (u), f (v) \to f (x)

u, v \to x

, which is exactly Continuity at

x

Why does the

+ C

from the antiderivative never appear in a definite integral's answer?

Part 2 evaluates

G (b) - G (a)

; if

G

is replaced by

G + C

, both endpoints gain

C

and it subtracts away. The definite integral is a difference, blind to vertical shifts.

Why does Part 1 guarantee that a continuous

f

has an antiderivative, even one with no formula?

It explicitly constructs one:

F (x) = \int_{a}^{x} f (t) d t

is a genuine function whose derivative is

f

. Existence comes from the integral, not from being able to write a closed form.

Why do we need Part 1 before Part 2 in the proof?

Part 2's Step 1 uses the integral-defined

F

and needs to know

F^{'} = f

— that fact is precisely Part 1. Part 1 supplies the existence that Part 2 then compares an arbitrary

G

against.

Why is the average height of

f

[x, x + h]

trapped between

m_{h}

and

M_{h}

The integral divided by width

h

is an average value; a continuous function's average over an interval can never dip below its minimum or exceed its maximum on that interval, so it lies between

m_{h}

and

M_{h}

Why does the "water tank" picture (level rises at the rate of the curve's height) capture Part 1 exactly?

The height

f (x)

is the incoming flow rate; the water level is accumulated volume

F (x)

. Rate-of-level-change equals inflow means

F^{'} (x) = f (x)

— Part 1 in physical clothing.

Why can Part 1 differentiate

\int_{0}^{x} sin (t^{2}) d t

even though the integral has no elementary formula?

Part 1 reads off the derivative directly from the integrand at the upper limit (

sin (x^{2})

) without ever evaluating the integral. That is its superpower: it bypasses antidifferentiation.

Edge cases

What is

\frac{d}{d x} \int_{a}^{x} f

x = a

? Does Part 1 still apply?

Yes — at the left endpoint you use the one-sided derivative (only

h > 0

makes sense), and the same Squeeze argument gives the right-hand derivative

f (a)

. The proof's

h < 0

branch is simply omitted here.

f (x) = c

(a constant), what does Part 1 predict, and does it match direct computation?

F (x) = \int_{a}^{x} c d t = c (x - a)

, so

F^{'} (x) = c = f (x)

. It matches — the accumulation of a flat height grows linearly, and the growth rate is that height.

What happens to Part 2 on an unbounded interval, e.g.

\int_{1}^{\infty} x^{- 2} d x

Ordinary Part 2 assumes a finite

[a, b]

with

f

continuous there. Infinite limits need Improper Integrals: compute

[G (x)]_{1}^{b}

then take

lim_{b \to \infty}

. Here

lim_{b \to \infty} (1 - 1/ b) = 1

f

has a single jump discontinuity at

x_{0} \in (a, b)

, can you still find

\int_{a}^{b} f

Yes by splitting:

\int_{a}^{x_{0}} + \int_{x_{0}}^{b}

, applying Part 2 on each piece where

f

is continuous. But

F

is not differentiable at

x_{0}

, so you cannot use one antiderivative across the jump.

What does

\int_{a}^{a} f (t) d t

equal and why is it structurally important?

It equals

0

— zero width, zero area. This makes

F (a) = 0

, giving

F

the correct base value so that

F (b) - F (a) = \int_{a}^{b} f

needs no correction term.

For

b < a

, is

\int_{a}^{b} f = G (b) - G (a)

still valid?

Yes, by the orientation convention

\int_{a}^{b} = - \int_{b}^{a}

. Part 2's algebra

G (b) - G (a)

automatically produces the sign flip, so the formula holds regardless of order.

G^{'} = f

but

G

is discontinuous somewhere in

[a, b]

, can Part 2 fail?

An antiderivative satisfying

G^{'} = f

everywhere on

[a, b]

is automatically differentiable, hence continuous, there — so this cannot happen. The danger is a "piecewise antiderivative" glued with a jump, which does not have

G^{'} = f

at the gluing point.