Module 5 · Lesson 5.3

The Fundamental Theorem of Calculus

Discover why the "area-so-far" function's slope is exactly the original curve — the fact that ties antiderivatives and Riemann sums together.

Intuition

Imagine sliding a wall rightward across the region under a curve, keeping track of how much area you've swept up so far. Call that running total $A(x)$. As the wall moves one tiny step further right, it sweeps up one thin new sliver of area — and that sliver's height is just the curve's height at that point.

Core concept

A tiny step forward adds area roughly equal to $f(x)\cdot\Delta x$, so the rate at which $A(x)$ grows is $f(x)$ itself. In other words, $A'(x) = f(x)$ — the area-so-far function is an antiderivative of the curve. That single fact means you never have to grind through a Riemann sum by hand again: find any antiderivative $F$, and the exact definite integral is just $F(b) - F(a)$.

1 3 f(x)=2x
Integral to evaluate∫₁³ 2x dx
Find an antiderivativeF(x) = x²
Evaluate at upper limitF(3) = 9
Evaluate at lower limitF(1) = 1
Subtract9 − 1
Answer8

Interactive Lab

Drag the slider to move the right edge of the shaded region. Watch the top graph's shaded area grow, and watch the bottom graph trace out $A(x)$ — the running area total. The dashed tangent line on the bottom graph always matches the curve's height on the top graph at the same $x$.

f(t) A(x) = area so far
x = 2.0 f(x) = 2.0 A(x) = 2.0 slope of A at x = 2.0

Key Idea

$$ \frac{d}{dx}\int_a^x f(t)\,dt = f(x) \qquad\Longrightarrow\qquad \int_a^b f(x)\,dx = F(b) - F(a) $$

The left equation says the area-so-far function's derivative is the original curve. The right equation is the practical payoff: pick any antiderivative $F$ of $f$, plug in the two endpoints, and subtract. No rectangles, no limits, no infinite sums required at evaluation time.

$a$The fixed starting point where the area-so-far count begins
$x$The moving right edge of the accumulated region
$\int_a^x f(t)\,dt$The area-so-far function — total area from $a$ up to $x$
$F$Any antiderivative of $f$, used as a shortcut evaluation tool

Think Further

  1. If two different antiderivatives $F_1$ and $F_2$ of the same $f$ differ by a constant, would $F_1(b) - F_1(a)$ and $F_2(b) - F_2(a)$ give the same answer? Why does the constant not matter here?
  2. A physics engine tracks an object's kinetic energy as the area under a force-versus-distance curve. Why would the Fundamental Theorem let it compute total energy without simulating every tiny step?
  3. Predict what happens to $A(x)$ on an interval where $f(x)$ is negative. Does the area-so-far total go up, go down, or stay flat there?
Show suggested answers
  1. Yes, they give the same answer. If $F_2(x) = F_1(x) + C$, then $F_2(b) - F_2(a) = (F_1(b)+C) - (F_1(a)+C) = F_1(b) - F_1(a)$ — the constant cancels out in the subtraction.
  2. Because the theorem replaces a step-by-step accumulation with a single evaluation of an antiderivative at two points, turning what would be a costly simulation loop into one fast, exact calculation.
  3. It goes down. Since $A'(x) = f(x)$, a negative $f(x)$ means $A(x)$ is decreasing there — each new sliver subtracts from the running total instead of adding to it.