6.4 - The Fundamental Theorem of Calculus and Accumulation Functions

Review: The Definite Integral

The definite integral is the area under the curve on a specific interval:

\int_a^b f(x) \, dx

$a$ is the lower limit (a constant)

$b$ is the upper limit (a constant)

The result is a number, the total area

Accumulation Functions

An accumulation function has a variable as its upper limit:

\boxed{F(x) = \int_a^x f(t) \, dt}

Key points: - The lower limit

a

is a constant - The upper limit

x

is a variable - We use a different variable (

t

) inside to avoid confusion -

F(x)

is a function that depends on where we stop - As

x

changes, the area accumulates

Why different variables? Because

x

is already being used as the upper limit. The

t

is a “dummy variable,” and it could be any letter except

x

Example: Building an Accumulation Function

Given a piecewise linear function

f(t)

[0, 5]

6.4 - The Fundamental Theorem of Calculus and Accumulation Functions

Define

F(x) = \int_0^x f(t) \, dt

and accumulate the area:

$x$	$F(x) = \int_0^x f(t) \, dt$	Calculation
0	0	No area from 0 to 0
1	2.5	Area from 0 to 1
2	4	Previous (2.5) + new area (1.5)
3	5	Previous (4) + new area (1)
4	5.5	Previous (5) + new area (0.5)
5	5	Previous (5.5) − area below the axis (0.5)

F(x)

shows the total accumulated area from

t = 0

t = x

Integrals can define new functions. Some functions can’t be written with powers, roots, trig, exponentials, or logs at all. For example:

S(x) = \int_0^x \sin(t^2) \, dt

There’s no algebra formula for

S(x)

, but it’s still a perfectly good function. Every

x

gives one value (the net area from 0 to

x

), so you can graph it, evaluate it with a calculator, and find its derivative with the theorem below:

S'(x) = \sin(x^2)

The Fundamental Theorem of Calculus (Part 2)

\boxed{\frac{d}{dx}\left[\int_a^x f(t) \, dt\right] = f(x)}

In words: the derivative of an integral gives you back the original function.

Condition: this works when

f

is continuous on an interval that contains

a

, for any

x

in that interval.

The big idea here is that derivatives and integrals are inverse operations. They undo each other, the same way division undoes multiplication.

Terminology

F(x) = \int_a^x f(t) \, dt

, then: -

F(x)

is an antiderivative of

f(x)

f(x)

is the integrand (what’s inside the integral) - Integral of

f

= antiderivative =

F

- Derivative of

F

= original function =

f

\text{If } F(x) = \int f(x) \, dx \text{, then } F'(x) = f(x)

Applying the Theorem

Basic case: constant lower limit

Given

F(x) = \int_2^x (3t^2 + 4t) \, dt

, find

F'(x)

Just plug in

x

for

t

F'(x) = 3x^2 + 4x

Steps: 1. Recognize the constant lower limit (2) 2. Variable upper limit (

x

) 3. Replace

t

with

x

in the integrand 4. Done

With the chain rule: upper limit is a function

Given

F(x) = \int_a^{g(x)} f(t) \, dt

\boxed{F'(x) = f(g(x)) \cdot g'(x)}

Steps: 1. Plug

g(x)

in for

t

in the integrand 2. Multiply by the derivative of

g(x)

Example 1

Given

F(x) = \int_1^{x^3} \sin(t) \, dt

, find

F'(x)

F'(x) = \sin(x^3) \cdot \frac{d}{dx}[x^3] = \sin(x^3) \cdot 3x^2 = 3x^2 \sin(x^3)

Example 2

Given

F(x) = \int_0^{4x} h(t) \, dt

, find

F'(x)

F'(x) = h(4x) \cdot \frac{d}{dx}[4x] = 4h(4x)

Notice we never needed to know what

h(t)

actually is. Just plug in and apply the chain rule.

Both Limits Are Functions

Given

F(x) = \int_{h(x)}^{g(x)} f(t) \, dt

\boxed{F'(x) = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)}

Steps: 1. Upper limit: plug in

g(x)

, multiply by

g'(x)

2. Subtract 3. Lower limit: plug in

h(x)

, multiply by

h'(x)

Mnemonic: “top minus bottom.”

Example 3

Given

F(x) = \int_{-x}^{x} 5t \, dt

, find

F'(x)

Upper limit contribution:

5(x) \cdot \frac{d}{dx}[x] = 5x \cdot 1 = 5x

Lower limit contribution:

5(-x) \cdot \frac{d}{dx}[-x] = (-5x)(-1) = 5x

Combine (top minus bottom):

F'(x) = 5x - 5x = 0

Why zero? The integral is symmetric about the origin, so the accumulated value never changes.

Example 4

Given

F(x) = \int_{2x}^{3x} (t^2 - t) \, dt

, find

F'(x)

Upper limit (

3x

): - Plug in:

(3x)^2 - (3x) = 9x^2 - 3x

- Chain rule: multiply by 3 - Result:

27x^2 - 9x

Lower limit (

2x

): - Plug in:

(2x)^2 - (2x) = 4x^2 - 2x

- Chain rule: multiply by 2 - Result:

8x^2 - 4x

Subtract lower from upper:

F'(x) = (27x^2 - 9x) - (8x^2 - 4x) = 19x^2 - 5x

Summary of Cases

Type	Formula	Notes
Constant lower, variable upper	$\frac{d}{dx}\left[\int_a^x f(t) \, dt\right] = f(x)$	Just plug in $x$
Constant lower, function upper	$\frac{d}{dx}\left[\int_a^{g(x)} f(t) \, dt\right] = f(g(x)) \cdot g'(x)$	Use the chain rule
Both variable	$\frac{d}{dx}\left[\int_{h(x)}^{g(x)} f(t) \, dt\right] = f(g(x)) \cdot g'(x) - f(h(x)) \cdot h'(x)$	Top minus bottom

Common Mistakes to Avoid

Forgetting the chain rule when the upper or lower limit isn’t just $x$

Using the same variable inside and outside. Don’t write $\int_a^x f(x) \, dx$ ; use a different variable inside

Sign error with the lower limit. You SUBTRACT the lower limit contribution, you don’t add it

Not applying the derivative to the limits. You have to multiply by $g'(x)$ or $h'(x)$

Key Takeaways

Accumulation function: $F(x) = \int_a^x f(t) \, dt$ accumulates area as $x$ varies

FTC Part 2: $\frac{d}{dx}\left[\int_a^x f(t) \, dt\right] = f(x)$ , so the derivative undoes the integral

Antiderivative: the integral of $f$ gives an antiderivative $F$ where $F' = f$

Basic case: constant lower limit means you just plug in the upper limit

Chain rule: when the upper limit is $g(x)$ , multiply by $g'(x)$

Both limits variable: top contribution MINUS bottom contribution

Different variables: use $t$ (or another letter) inside the integral, save $x$ for the limit

Derivatives and integrals are inverses: they cancel each other out

This theorem is genuinely fundamental. It’s the bridge between differentiation and integration

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