MISSION · Calculus
The Fundamental Theorem of Calculus
Why differentiating and integrating undo each other
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The idea in one sentence
Differentiation and integration are inverse operations, so an area that would take infinitely many rectangles to add up can be found by evaluating an antiderivative at two points.
Why it’s called fundamental
Areas and slopes look like completely unrelated problems. One is about totals, the other about rates. They were studied separately for centuries. The fundamental theorem says they are the same problem seen from opposite ends — and that is what turned calculus from a collection of clever tricks into a subject.
The second part first, because it’s the one you use
\[\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F' = f\]To find the area under \(y = 2x\) from 0 to 3: an antiderivative of \(2x\) is \(x^2\), so the answer is \(3^2 - 0^2 = 9\).
Check it geometrically. The region is a triangle with base 3 and height 6, so its area is \(\frac{1}{2} \times 3 \times 6 = 9\). The limit of infinitely many rectangles, replaced by two substitutions.
The constant of integration cancels: using \(x^2 + C\) gives \((9 + C) - (0 + C) = 9\). That’s why definite integrals don’t carry a \(+C\).
The first part, which explains why
Define a function by accumulating area as you go:
\[F(x) = \int_a^x f(t)\,dt\]Then \(F'(x) = f(x)\).
Here’s the reason, in one picture. Push \(x\) along by a tiny amount \(h\). The extra area you sweep up is a sliver of width \(h\) and height about \(f(x)\), so \(F(x+h) - F(x) \approx f(x)h\). Divide by \(h\) and let it shrink:
\[F'(x) = \lim_{h\to 0}\frac{F(x+h)-F(x)}{h} = f(x)\]The rate at which accumulated area grows is simply the height of the function at the moving edge. That’s the whole theorem. Everything else is bookkeeping.
The everyday version
If \(f\) is your speed, then the area under the speed graph is distance travelled. The rate at which distance accumulates is speed. Those two sentences are the two parts of the fundamental theorem, and you already believed both of them before meeting any calculus.
Functions defined by integrals
Part one lets you differentiate things you cannot integrate. \(\int_0^x e^{-t^2}dt\) has no elementary formula, but its derivative is \(e^{-x^2}\) immediately.
With a variable in the upper limit that isn’t just \(x\), the chain rule joins in:
\[\frac{d}{dx}\int_0^{x^2} f(t)\,dt = f(x^2) \cdot 2x\]Signed area
The theorem returns signed area: regions below the axis count as negative. \(\int_0^{2\pi}\sin x\,dx = 0\), because the hump above cancels the hump below. If you want total area rather than net, split the integral at the crossings and take absolute values.
Common mix-ups
- Subtracting the wrong way round. It’s \(F(b) - F(a)\), top minus bottom.
- Adding \(+C\) to a definite integral. It cancels.
- Forgetting the chain rule when the limit is a function of \(x\).
- Reporting a negative “area” when the question wanted total area.
- Assuming every function has an antiderivative you can write down. Plenty don’t, and those get evaluated numerically.
Check yourself
Evaluate \(\int_0^3 2x\,dx\), and state \(F'(x)\) when \(F(x) = \int_0^x f(t)\,dt\).
An antiderivative of 2x is x², so the integral is 9 − 0 = 9 — the same as the triangle’s area. And F′(x) = f(x): the rate of accumulation is the height at the edge.
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