So far area (the
definite integral)
and slope (the
derivative)
have lived in separate rooms. The Fundamental Theorem of Calculus throws open the connecting
door. Part 1 says: accumulating area, then differentiating, returns the integrand
you started with.
Fix a continuous f and define the accumulation function
— the area so far, swept from a base point a up to a moving right
edge x:
F(x) = \int_{a}^{x} f(t)\,dt.
(We write the dummy variable as t so it isn't confused with the
limit x.) The claim is that this area function has a derivative, and
that derivative is exactly f:
F'(x) = f(x).
The proof — line by line
We compute F'(x) straight from the
limit definition
of the derivative. The whole argument is a squeeze.
Step 1 — form the difference quotient. By definition,
F'(x) = \lim_{h\to 0}\frac{F(x+h) - F(x)}{h}.
Step 2 — turn the numerator into one integral. Using additivity of the
definite integral, the area up to x+h minus the area up to
x is just the area of the thin sliver
[x, x+h]:
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.
Step 3 — write the difference quotient as an average. Divide that sliver by its
width:
\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_{x}^{x+h} f(t)\,dt.
The right side is the average value of f over the little
interval [x, x+h].
Step 4 — bound that sliver. Let m_h and
M_h be the smallest and largest values of f
on [x, x+h] (they exist because f is
continuous on a closed interval). The
bounding inequality
m_h\, h \le \int_x^{x+h} f \le M_h\, h, divided by
h > 0, gives
m_h \;\le\; \frac{1}{h}\int_{x}^{x+h} f(t)\,dt \;\le\; M_h.
Step 5 — squeeze as h \to 0. As the sliver shrinks
to the single point x, continuity of f
forces both the minimum and the maximum on it to converge to f(x):
\lim_{h\to 0} m_h = f(x) \qquad\text{and}\qquad \lim_{h\to 0} M_h = f(x).
Step 6 — apply the squeeze theorem. The difference quotient is trapped between
two quantities that both tend to f(x), so it has no choice but to do
the same:
\lim_{h\to 0}\frac{F(x+h) - F(x)}{h} = f(x).
Step 7 — read off the conclusion. The left side is
F'(x). Hence
\boxed{\;F'(x) = f(x).\;}
Differentiation undoes accumulation. (For h < 0 the same argument
runs with the inequalities and the average reversed, giving the identical limit, so the
two-sided derivative exists.)
Think of F(x) as a tank filling with water: at time
x, F is the total volume collected so
far. The instantaneous rate the tank fills is the height of the inflow at this
instant — which is exactly f(x). So F'(x),
the rate of change of accumulated area, equals the current height of the curve. Where
f is tall, area piles up fast and F rises
steeply; where f = 0, the area momentarily stalls and
F is flat; where f < 0, area is being
removed and F falls.
Step 5 is the only place we used continuity, and the proof leans its whole weight there. If
f had a jump at x, the slivers
approaching from the left and the right would average to different values, so
m_h and M_h would not both close on a
single number — the squeeze collapses, and F'(x) need not equal
f(x) there. (The accumulation function F
is still continuous — area can't jump — but it develops a corner, so it isn't differentiable
at the jump.) Continuity of the integrand is exactly what guarantees a clean derivative.
Let f be continuous on [a, b], and define
the accumulation function
F(x) = \int_{a}^{x} f(t)\,dt \qquad (a \le x \le b).
Then F is differentiable on (a, b) and
F'(x) = f(x) \qquad\text{for every } x \in (a, b).
In words: the accumulation function is an antiderivative of its integrand — every continuous
function has an antiderivative, namely its own area-so-far function.