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Lesson 3.3

THE FUNDAMENTAL THEOREM

For centuries, calculating slopes and calculating areas were thought to be completely unrelated problems. This theorem proves they are two sides of the same coin.

Part 1: The Accumulator

Imagine a function F(x)F(x) that represents "Area Accumulation." As you plug in a larger xx, it sweeps across the graph of f(t)f(t) and calculates the area up to that point.

The first part of the Fundamental Theorem states that the rate at which area accumulates is exactly equal to the height of the curve.

The Accumulator
Integration & Differentiation
Rate: f(t) = 2t
Area = 4.00
Height = 4.00
Accumulation: F(x) = x²
Value = 4.00
Slope = 4.00
Drag Boundary (x)2.00
Area = 4.00Value = 4.00
Notice how the Height on the left is always exactly equal to the Slope on the right.
FTC: Part 1
ddx[∫axf(t) dt]=f(x)\frac{d}{dx} \left[ \int_{a}^{x} f(t) \,dt \right] = f(x)

In plain English: "The derivative of the integral of a function is just the original function." They cancel each other out, like multiplication and division.

Part 2: The Shortcut

This realization completely breaks Calculus wide open. If integration is just "reverse differentiation," we don't need to draw infinite Riemann sums anymore!

To find the exact area under a curve between aa and bb, all you have to do is find the Antiderivative (the function whose derivative is the curve), and subtract the start from the end.

FTC: Part 2
∫abf(x) dx=F(b)−F(a)\int_{a}^{b} f(x) \,dx = F(b) - F(a)

Step 1: Antiderivative

Find the function F(x)F(x) that would result in your current curve if differentiated.

∫2x dx  ⟹  F(x)=x2\int 2x \,dx \implies F(x) = x^2

Step 2: Evaluate

Plug the bounds into the antiderivative and subtract.

F(3)−F(1)=32−12=8F(3) - F(1) = 3^2 - 1^2 = 8
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