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Fundamental Theorem
of Calculus

Discover how accumulation functions connect derivatives and integrals: the most powerful bridge in all of calculus.

AP Calculus AB

What Is an Accumulation Function?

An accumulation function measures the total accumulated area under a curve from a fixed starting point \(a\) up to a variable endpoint \(x\). We write it as:

As \(x\) changes, \(F(x)\) grows or shrinks depending on whether \(f(t)\) is positive (area adds) or negative (area subtracts).

Visualizing Accumulated Area

t f(t) a x F(x) f(t) 0

Why Does This Matter?

Physics
Velocity accumulates to give displacement: \(\int_0^t v(\tau)\,d\tau = s(t)\)
Biology
A rate of growth integrates to total population change over time.
Economics
Marginal cost accumulates to total cost: \(\int_0^q MC(u)\,du = TC(q)\)

FTC Part 1: The Derivative of an Integral

In words: the derivative of the accumulation function equals the original integrand evaluated at the upper limit. The integral "remembers" the function, and differentiation "recovers" it.

Follow the salt through a working mixing rig

A stirred transparent tank has one upper inlet and one lower outlet with a valve. The exact diagram links its 100 L salt balance to a computed approach to equilibrium.For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.

Signed net flow and inventory

For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.For net flow R(t) = 4t − t² gal/min and 10 gal initially, V(t) = 10 + 2t² − t³/3. Volume peaks at t = 4 minutes, when the net rate changes from positive to negative.

Why \(F'(x) = f(x)\)?

t f(t) f(x) h a x x+h F(x) ≈ f(x)·h

FTC Part 2: Evaluating Definite Integrals

Part 2 says: to compute a definite integral, find any antiderivative and subtract its values at the endpoints. No Riemann sums required.

Part 1 vs Part 2: Key Distinction

FTC Part 1 FTC Part 2
Statement \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\) \(\int_a^b f(x)\,dx = F(b)-F(a)\)
Direction Integral → Derivative Derivative → Integral
What it finds Instantaneous rate of accumulation Total net accumulation
Upper limit Variable \(x\) Constant \(b\)
Result type A function of \(x\) A number

When the Upper Limit Is Not Just \(x\)

If the upper limit is a function \(g(x)\) instead of plain \(x\), apply the chain rule:

1
Evaluate \(f\) at the upper limit: replace \(t\) with \(g(x)\).
2
Multiply by \(g'(x)\): the derivative of the upper limit.

How to Apply the FTC

1
Identify the type: Are you differentiating an integral (Part 1) or evaluating a definite integral (Part 2)?
2
Check continuity: Confirm \(f\) is continuous on the relevant interval.
3
Part 1: Write \(F'(x) = f(x)\). If the upper limit is \(g(x)\), apply the chain rule: multiply by \(g'(x)\).
4
Part 2: Find an antiderivative \(F\), then compute \(F(b) - F(a)\).
5
Simplify and interpret the result in context (area, displacement, total change).

Example 1: Basic FTC Part 1

Find \(\displaystyle\frac{d}{dx}\int_2^x (3t^2 + 1)\,dt\).

1
The integrand is \(f(t) = 3t^2 + 1\), which is continuous everywhere.
2
The upper limit is simply \(x\): no chain rule needed.
3
By FTC Part 1: replace \(t\) with \(x\).

Example 2: Chain Rule Extension

Find \(\displaystyle\frac{d}{dx}\int_0^{\sin x} e^{t^2}\,dt\).

1
Identify: upper limit is \(g(x) = \sin x\), so chain rule applies.
2
Evaluate \(f\) at \(g(x)\): \(f(\sin x) = e^{(\sin x)^2} = e^{\sin^2 x}\).
3
Multiply by \(g'(x)\): \(g'(x) = \cos x\).

Example 3: Evaluating a Definite Integral

Evaluate \(\displaystyle\int_1^4 \left(2\sqrt{x}\right)\,dx\).

1
Rewrite: \(2\sqrt{x} = 2x^{1/2}\).
2
Antiderivative: \(F(x) = 2 \cdot \frac{x^{3/2}}{3/2} = \frac{4}{3}x^{3/2}\).
3
Evaluate: \(F(4) - F(1) = \frac{4}{3}(8) - \frac{4}{3}(1) = \frac{32}{3} - \frac{4}{3}\).

Example 4: Reading \(F(x)\) from a Graph

t f(t) 1 2 3 4 5 2 -2 +3 -2

Properties of \(F(x)\)

Increasing
\(F'(x) = f(x) > 0\): the integrand is positive, so area is being added.
Decreasing
\(F'(x) = f(x) < 0\): the integrand is negative, so accumulated area shrinks.
Extrema
\(F'(x) = f(x) = 0\) and \(f\) changes sign: \(F\) has a local max or min.

Common Mistakes to Avoid

✕ Forgetting the Chain Rule
When the upper limit is \(g(x)\), you must multiply by \(g'(x)\). Writing just \(f(g(x))\) is incomplete.
✕ Confusing Variable and Dummy
The \(t\) in \(\int_a^x f(t)\,dt\) is a dummy variable. The answer is a function of \(x\), not \(t\).
✕ Wrong Sign with Flipped Limits
\(\int_x^a f(t)\,dt = -\int_a^x f(t)\,dt\). Differentiating gives \(-f(x)\), not \(f(x)\).

Summary Cheat Sheet

Scenario Formula Key Detail
Basic FTC Part 1 \(\frac{d}{dx}\int_a^x f(t)\,dt = f(x)\) Upper limit is \(x\)
Chain Rule Version \(f(g(x))\cdot g'(x)\) Upper limit is \(g(x)\)
Flipped Limits \(-f(x)\) Lower limit is \(x\)
FTC Part 2 \(F(b) - F(a)\) Both limits are constants

Question 1 of 5

What is \(\displaystyle\frac{d}{dx}\int_3^x \cos(t^2)\,dt\)?

Question 2 of 5

What is \(\displaystyle\frac{d}{dx}\int_1^{x^3} \ln(t)\,dt\)?

Question 3 of 5

If \(F(x) = \displaystyle\int_0^x f(t)\,dt\) and \(f\) changes from positive to negative at \(x = 4\), what happens to \(F\) at \(x = 4\)?

Question 4 of 5

Evaluate \(\displaystyle\int_0^{\pi} \sin(x)\,dx\) using FTC Part 2.

Question 5 of 5

What is \(\displaystyle\frac{d}{dx}\int_x^5 e^{t}\,dt\)?