Lesson 0.4 left every integral one step short of a final answer — a family of functions, with an unresolved "+C." A definite integral closes that gap by evaluating an antiderivative at two specific bounds and subtracting. This result is called the Fundamental Theorem of Calculus:
where F is any antiderivative of f. It doesn't matter which one you pick — watch what happens to the "+C":
The C's cancel completely. That's why a definite integral produces a single, specific number instead of a family of functions — the constant that mattered so much in Lesson 0.4 simply doesn't survive the subtraction.
That number has a geometric meaning worth sitting with: for a function f(x) ≥ 0, ∫ₐᵇ f(x) dx is exactly the area of the region trapped between the curve and the x-axis, from x = a to x = b. One way to build that area from scratch is to slice the region into thin rectangles, add up their areas, and then let the rectangles get thinner and thinner — a process called a Riemann sum.
This is a velocity-time graph: v(t) = 3 + 2t (m/s). The shaded rectangles approximate displacement — the area under the curve from t = 0 to t = 4 s. Increase n and watch the rectangle-sum close in on the exact definite integral.
Because v(t) is increasing, using the left edge of each rectangle always underestimates the true area — but that gap shrinks toward zero as the rectangles get thinner.
One more detail: area doesn't always mean a positive number. Where f(x) dips below the x-axis, that region counts as negative area. A definite integral reports the signed area — area above the axis minus area below it.
For a simple enough function, you can check a definite integral's value against a geometric area calculation you already trust — the same "verify a new method against an old one" move from Lesson 0.3.
Evaluate ∫₀⁴ 2x dx using the Fundamental Theorem of Calculus, then verify the result geometrically.
Now try a case where the graph crosses the x-axis inside the interval — this is where signed area actually matters.
Evaluate ∫₁³ (x − 2) dx. Notice the graph of x − 2 crosses zero at x = 2, right in the middle of the interval.
The Riemann sum tool above wasn't really an abstract math example — it was already physics. v(t) was a velocity function, the horizontal axis was time, and the area under that curve was a displacement. That single relationship is worth stating directly:
Displacement is the area under a velocity-time graph. This isn't a coincidence limited to kinematics — the exact same logic reappears twice more this year, just with different quantities on the axes:
A car's velocity is v(t) = 4 + 3t (m/s). Find its displacement from t = 0 to t = 2 s.