Unit 6: Integration & Accumulation of Change

Applying the Fundamental Theorem

Using the Fundamental Theorem of Calculus to evaluate definite integrals and define accumulation functions.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Definite Integral

The Bridge Between Area and Antiderivatives

Every definite integral evaluated so far has needed either a geometric shortcut or a full Riemann-sum limit. Before reading on: is there a faster connection between a function f and the area function it generates — one that ties area-so-far directly back to f itself, without ever taking a limit of sums again?

Definition — The Fundamental Theorem of Calculus

Part 1: for A(x) = ∫ₐˣ f(t) dt, A'(x) = f(x) — the derivative of an accumulation function is the original function itself. Part 2: if F is any antiderivative of f (meaning F' = f), then ∫ₐᵇ f(x) dx = F(b) − F(a) — a definite integral can be evaluated directly from an antiderivative, with no limit process needed.

\int_a^b f(x)\,dx = F(b) - F(a), \quad \text{where } F' = f

Part 1 follows from a direct rate-of-change argument: moving x forward by a small amount h adds a thin sliver of area, approximately f(x)·h (a rectangle of height f(x) and width h). So [A(x+h) − A(x)]/h ≈ f(x), and as h → 0, this approximation becomes exact — A'(x) = f(x). The accumulation function's instantaneous rate of change is exactly the height of the function generating it.

Worked Example — Evaluating a Definite Integral with an Antiderivative

Evaluate ∫₁⁴ (3x² − 2x) dx. An antiderivative is F(x) = x³ − x² (check: d/dx[x³ − x²] = 3x² − 2x, matching). By FTC Part 2: ∫₁⁴(3x² − 2x)dx = F(4) − F(1) = (64 − 16) − (1 − 1) = 48 − 0 = 48.

Worked Example — Differentiating an Accumulation Function

Let A(x) = ∫₂ˣ (t² + 1) dt. By FTC Part 1, A'(x) = x² + 1 directly. Now let B(x) = ∫₂^(x²) (t² + 1) dt — the upper bound is x², not x itself. Since B(x) = A(x²), the chain rule applies: B'(x) = A'(x²) · d/dx[x²] = ((x²)² + 1) · 2x = (x⁴ + 1)(2x) = 2x⁵ + 2x.

Integral Visualizer

f(x) =

x^2

Integration bounds

a — lower bound0

b — upper bound2

Animate b (the play button above) to watch the shaded area — and the accumulation curve below — grow.

Approximation method

Method

Subdivisions (n)10

Show approximationShow exact area

Trace mode

f(x), with the region between a and b shaded

Show accumulation function A(x)

Accumulation function A(x) = ∫ₐˣ f(t) dt — linked to the graph above; panning or zooming either moves both

What the Fundamental Theorem of Calculus tells us

A(x) = ∫ from a to x of f(t) dt is the accumulation function. At x = b ≈ 2, A(b) ≈ 2.67 — the signed area under f from a ≈ 0 to b.
The Fundamental Theorem of Calculus says A'(x) = f(x): differentiating the accumulation function gives the original function back.
Confirmed numerically here: differentiating the accumulation curve gives A'(b) ≈ 4, matching f(b) ≈ 4.

Numerical approximation comparison

Every method's estimate of ∫ₐᵇ f(x) dx at n = 10 subdivisions, against a high-resolution reference — all should converge toward the same value.
Method	Estimate
Left Riemann sum	2.28
Right Riemann sum	3.08
Midpoint Riemann sum	2.66
Trapezoidal Rule	2.68
Simpson's Rule	2.66667
Exact (high-resolution reference)	2.66667

Analysis of f(x)

y-intercept: (0, 0)
x-intercepts: (0, 0)
Extrema: local min at (0, 0)
Inflection points: none found in view
Vertical asymptotes: none found in view
Horizontal asymptotes: none found
Domain: all real numbers in view
Range (estimated): approximately [0, 100]

Tip

When an accumulation function's upper bound is something other than x itself, apply FTC Part 1 first (treating that upper bound as a placeholder variable), then multiply by the derivative of the actual upper bound — the same chain rule pattern as any other composite function.

Common Mistakes

Differentiating ∫₂^(x²)(t²+1)dt as (x²)² + 1, forgetting the extra chain rule factor from the non-x upper bound.
FTC Part 1 alone only handles a plain x upper bound — a upper bound like x² needs an additional chain rule factor, d/dx[x²] = 2x, multiplied in afterward.
Using the wrong antiderivative sign or forgetting a term when finding F(x) before applying FTC Part 2.
Always verify F'(x) actually equals the original integrand before using F in FTC Part 2 — a single missing or mis-signed term in F produces a wrong definite integral even though the rest of the evaluation is done correctly.

Key Takeaways

FTC Part 1 says the derivative of an accumulation function A(x) = ∫ₐˣf(t)dt is f(x) itself — area and rate of change are two sides of the same relationship.
FTC Part 2 evaluates a definite integral directly from any antiderivative: ∫ₐᵇf(x)dx = F(b) − F(a), with no Riemann sum needed.
When an accumulation function's upper bound is a function of x rather than x itself, differentiating it needs an extra chain rule factor.

Summary

The Fundamental Theorem connects integration and differentiation directly. The next lesson develops the main technique for finding antiderivatives of composite functions — u-substitution, the integration counterpart of the chain rule.

Track your progress

← The Definite Integral Back to Unit 6: Integration & Accumulation of Change U-Substitution→