The Fundamental Theorem of Calculus

Two worlds that shouldn't have met

So far you have met two very different machines. The derivative takes a function and tells you its instantaneous slope — how fast it is changing right now. The definite integral takes a function and adds up the area beneath it, built as the limit of a Riemann sum of skinnier and skinnier rectangles. One is about steepness; the other is about accumulation. There is no obvious reason these two should have anything to do with each other.

And yet the Fundamental Theorem of Calculus declares that they are two sides of one coin: differentiation and integration undo each other, the way addition undoes subtraction or squaring undoes square-rooting. This is the result that ties the whole subject together, and it is genuinely astonishing — Newton and Leibniz built modern science on it.

Part 1: the accumulation function and its slope

Start with a function f and define a brand-new function A by accumulating area as you go: A(x) = integral from a to x of f(t) dt. Read that as "how much area has piled up under f, from a fixed left edge a out to a moving right edge x." As you push x to the right, A grows. The question is: how fast does A grow?

Here is the intuition. Nudge x a tiny bit further right, by a width h. The extra area you sweep up is a thin sliver almost exactly a rectangle: height f(x), width h, so roughly f(x)*h. That means the change in A divided by the change in x is about f(x)*h / h = f(x). Take the limit as h -> 0 and the approximation becomes exact. So A's rate of change — its derivative — is just f itself.

                d   ( integral from a to x of f(t) dt )  =  f(x)
               dx

  differentiating an area-so-far function gives back the original function

Part 2: the shortcut that ends the suffering

Part 1 was the deep truth; Part 2 is the everyday tool. It says: to compute a definite integral from a to b, you do not have to grind through a limit of Riemann sums. Instead, find any antiderivative F of f (any function with F'(x) = f(x)), and simply subtract its values at the two ends: integral from a to b of f(x) dx = F(b) - F(a).

Pause on how strange this is. The left side is an infinite, delicate summing process. The right side is two numbers subtracted. The theorem promises they are equal — every time. Why "any" antiderivative? Because two antiderivatives of the same f differ only by a constant, and that constant cancels in the subtraction F(b) - F(a). So you are free to pick the simplest one.

integral from a to b of f(x) dx  =  F(b) - F(a),   where F'(x) = f(x)

shorthand:   integral from a to b of f(x) dx  =  [ F(x) ]  from a to b

A worked example you can trust

Let's find the area under f(x) = x^2 from x = 0 to x = 3 — exactly the kind of thing that would be brutal to do by adding up rectangles, but is a one-liner with Part 2.

1. Find an antiderivative of f(x) = x^2. Since the derivative of x^3 is 3x^2, the function F(x) = x^3 / 3 has F'(x) = x^2. Good — that's our F.
2. Evaluate F at the top end: F(3) = 3^3 / 3 = 27 / 3 = 9.
3. Evaluate F at the bottom end: F(0) = 0^3 / 3 = 0.
4. Subtract: integral from 0 to 3 of x^2 dx = F(3) - F(0) = 9 - 0 = 9. Done — no infinite sum required.

# sanity check: a crude Riemann sum should creep toward 9
def riemann(f, a, b, n):
    dx = (b - a) / n
    return sum(f(a + (i + 0.5) * dx) * dx for i in range(n))

f = lambda x: x**2
print(riemann(f, 0, 3, 1000))   # ~ 8.99999..., closing in on the exact 9