6.4 Accumulation functions

Achievable AP Calculus AB

6. Integration

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Accumulation functions

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What you’ll learn

Accumulation functions: Define and evaluate as net signed area.
FTC Part 1: Find derivatives of integrals.
Chain rule: Differentiate integrals with functional bounds.
Integral properties: Apply properties to simplify expressions.

An accumulation function is defined using a definite integral with a variable upper limit:

$A (x) = \int_{a}^{x} f (t) d t$

$f (t)$ is a known function (often interpreted as a rate of change).
$A (x)$ is a new function that accumulates the net area from $a$ to $x$ .

You can think of $A (x)$ as the net signed area added between $t = a$ and $t = x$ .

Example

Graphs consisting of piecewise linear segments and standard geometric shapes often appear in the free-response section of the AP exam.

Shown below is the graph of $f$ , which consists of two line segments and a semicircle. Let $A (x) = \int_{1}^{x} f (t) d t$

Graph of f

Find:

a) $A (1)$

b) $A (3)$

c) $A (5)$

Solutions

a) $A (1)$

(spoiler)

$A (1) = \int_{1}^{1} f (t) d t$

This integral covers an interval of length $0$ , so no area is accumulated.

$A (1) = 0$

b) $A (3)$

(spoiler)

$A (3) = \int_{1}^{3} f (t) d t$

The region under $f (t)$ from $t = 1$ to $t = 3$ consists of:

A $1 \times 2$ unit rectangle ( $A_{1} = 2$ ).
A quarter-circle of radius $1$ ( $A_{2} = \frac{π}{4}$ ).

Therefore,

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

$A (5) = \int_{1}^{5} f (t) d t$

The region from $t = 1$ to $t = 5$ consists of:

A $1 \times 3$ rectangle ( $A_{1} = 3$ ).
A semicircle of radius $1$ ( $A_{2} = \frac{π}{2}$ ),.
Two congruent triangles (one positive, one negative) that cancel each other out ( $A_{3} = 0$ ).

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

The Fundamental theorem of calculus (FTC), Part 1 establishes that differentiation and integration are inverse operations. Essentially, taking the derivative “undoes” the integral.

Fundamental Theorem of Calculus (FTC), Part 1

Differentiation and integration are inverse operations.

$\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$

where $a$ is a constant and $f$ is a continuous function.

When you differentiate $\int_{a}^{x} f (t) d t$ , you get back the original integrand, evaluated at the upper limit.

You won’t have to prove this on the AP exam. As long as the lower bound $a$ is a real number, you can apply the rule directly: remove the integral, and replace $t$ with the upper limit $x$ .

A typical problem will be stated as such:

Find $F^{'} (x)$ given:

$F (x) = \int_{0}^{x} sin (t^{2}) d t$

(spoiler)

Start by writing the derivative:

$F^{'} (x) = \frac{d}{d x} (\int_{0}^{x} sin (t^{2}) d t)$

By the FTC (Part 1), this becomes the integrand evaluated at $t = x$ :

$sin (x^{2})$

FTC variations (with chain rule)

When the limits of integration are functions instead of a single variable $x$ , apply the chain rule. Evaluate the integrand at the upper limit, then multiply by the derivative of that limit.

Variation 1: Functional upper limit

If the upper limit is a function $g (x)$ , evaluate the integrand at $g (x)$ and multiply by $g^{'} (x)$ .

$\frac{d}{d x} (\int_{a}^{g (x)} f (t) d t) = f (g (x)) \cdot g^{'} (x)$

where $a$ is a constant and $g (x)$ is a function.

Example 1. Find $G^{'} (x)$ given

$G (x) = \int_{1}^{x^{3}} 1 + t^{4} d t$

(spoiler)

Here the upper limit is $g (x) = x^{3}$ , so we:

Replace $t$ in the integrand with $x^{3}$ .
Multiply by the entire expression by $g^{'} (x) = 3 x^{2}$ .

$G^{'} (x) = 1 + (x^{3})^{4} \cdot \frac{d}{d x} (x^{3}) = 3 x^{2} 1 + x^{12}$

Variation 2: Functional lower limit

If the lower limit is a function $h (x)$ and the upper limit is a constant, reverse the limits of integration. This introduces a negative sign.

$\frac{d}{d x} (\int_{h (x)}^{a} f (t) d t) = - f (h (x)) \cdot h^{'} (x)$

Example 2. Find $F^{'} (x)$ given

$F (x) = \int_{x^{2}}^{- 1} (t^{3} - t) d t$

(spoiler)

Flip the bounds of the integral and make it negative:

$F (x) = - \int_{- 1}^{x^{2}} (t^{3} - t) d t$

Then, apply the chain rule when differentiating by:

Replacing every $t$ in the integrand with $x^{2}$ .
Multiplying the entire expression by its derivative, $2 x$ .

$F^{'} (x) = - [(x^{2})^{3} - (x^{2})] \cdot \frac{d}{d x} (x^{2}) = - (x^{6} - x^{2}) \cdot 2 x = - 2 x^{7} + 2 x^{3}$

Variation 3: Two variable limits

When both limits are functions of $x$ , apply the chain rule to both boundaries and subtract the lower boundary result from the upper boundary result.

$\frac{d}{d x} (\int_{h (x)}^{g (x)} f (t) d t) = f (g (x)) \cdot g^{'} (x) - f (h (x)) \cdot h^{'} (x)$

Example 3. Find $f^{'} (x)$ given

$f (x) = \int_{3 x}^{c o s (x)} e^{2 t} d t$

Solution

(spoiler)

For the upper bound:

Replace $t$ with $cos (x)$ .
Multiply by its derivative of $- sin (x)$ .

Subtract the result of the lower bound:

Replace $t$ with $3 x$ .
Multiply by its derivative of $3$ .

$f^{'} (x) = [e^{2 c o s (x)} \cdot (- sin (x))] - [e^{2 (3 x)} \cdot 3] = - sin (x) e^{2 c o s (x)} - 3 e^{6 x}$

Properties of definite integrals

Splitting an interval $(a < b < c)$

$\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x$

Use this to break up integrals over intervals or to combine pieces.

Reversing limits

$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$

Switching the upper and lower bounds changes the sign.

Constant multiples

$\int_{a}^{b} k \cdot f (x) d x = k \cdot \int_{a}^{b} f (x) d x$

You can factor out constants.

Linearity

$\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$

Addition and subtraction work inside integrals.

Caution: The final property doesn’t hold when two or more functions are multiplied or divided.

Given $\int_{1}^{3} f (x) d x = 4$ and $\int_{3}^{5} f (x) d x = - 2$ , find:

a) $\int_{1}^{5} f (x) d x$

b) $\int_{5}^{1} f (x) d x$

Solutions

(spoiler)

a) Add over intervals:

$\int_{1}^{5} f (x) d x = \int_{1}^{3} f (x) d x + \int_{3}^{5} f (x) d x = 4 + (- 2) = 2$

b) Reverse the bounds:

$\int_{5}^{1} f (x) d x = - \int_{1}^{5} f (x) d x = - 2$

Given $\int_{3}^{10} f (x) d x = - 5$ and $\int_{6}^{10} f (x) d x = 2$ , find:

$\int_{3}^{6} 2 f (x) d x$

Solution

(spoiler)

Using property 1 (splitting up intervals),

$\int_{3}^{10} f (x) d x = \int_{3}^{6} f (x) d x + \int_{6}^{10} f (x) d x$

Substitute the given values:

$- 5 - 7 = \int_{3}^{6} f (x) d x + 2 = \int_{3}^{6} f (x) d x$

Then use property 3 (constant multiples):

$\int_{3}^{6} 2 f (x) d x = 2 (- 7) = - 14$

Accumulation functions

Defined as $A (x) = \int_{a}^{x} f (t) d t$
Represents net signed area under $f (t)$ from $a$ to $x$
$f (t)$ is the rate of change; $A (x)$ accumulates this rate

Fundamental Theorem of Calculus (FTC), Part 1

$\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$
Differentiation “undoes” accumulation
Applies when lower bound is constant and $f$ is continuous

FTC with chain rule (variable upper limit)

$\frac{d}{d x} (\int_{a}^{g (x)} f (t) d t) = f (g (x)) \cdot g^{'} (x)$
Replace $t$ with $g (x)$ in $f (t)$ , multiply by $g^{'} (x)$

FTC with both limits variable

$\frac{d}{d x} (\int_{h (x)}^{g (x)} f (t) d t) = f (g (x)) g^{'} (x) - f (h (x)) h^{'} (x)$
Subtract lower limit contribution from upper limit

Properties of definite integrals

Splitting: $\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x$
Reversing limits: $\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$
Constant multiples: $\int_{a}^{b} k f (x) d x = k \int_{a}^{b} f (x) d x$
Linearity: $\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$
- Does not apply to products or quotients

Common strategies

Rewrite integrals so one limit is constant before applying FTC
Use properties to combine, split, or reverse integrals as needed
When both limits are variable, apply generalized FTC formula

Accumulation functions

What you’ll learn

Accumulation functions: Define and evaluate as net signed area.
FTC Part 1: Find derivatives of integrals.
Chain rule: Differentiate integrals with functional bounds.
Integral properties: Apply properties to simplify expressions.

An accumulation function is defined using a definite integral with a variable upper limit:

$A (x) = \int_{a}^{x} f (t) d t$

$f (t)$ is a known function (often interpreted as a rate of change).
$A (x)$ is a new function that accumulates the net area from $a$ to $x$ .

You can think of $A (x)$ as the net signed area added between $t = a$ and $t = x$ .

Example

Graphs consisting of piecewise linear segments and standard geometric shapes often appear in the free-response section of the AP exam.

Shown below is the graph of $f$ , which consists of two line segments and a semicircle. Let $A (x) = \int_{1}^{x} f (t) d t$

Find:

a) $A (1)$

b) $A (3)$

c) $A (5)$

Solutions

a) $A (1)$

(spoiler)

$A (1) = \int_{1}^{1} f (t) d t$

This integral covers an interval of length $0$ , so no area is accumulated.

$A (1) = 0$

b) $A (3)$

(spoiler)

$A (3) = \int_{1}^{3} f (t) d t$

The region under $f (t)$ from $t = 1$ to $t = 3$ consists of:

A $1 \times 2$ unit rectangle ( $A_{1} = 2$ ).
A quarter-circle of radius $1$ ( $A_{2} = \frac{π}{4}$ ).

Therefore,

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

$A (5) = \int_{1}^{5} f (t) d t$

The region from $t = 1$ to $t = 5$ consists of:

A $1 \times 3$ rectangle ( $A_{1} = 3$ ).
A semicircle of radius $1$ ( $A_{2} = \frac{π}{2}$ ),.
Two congruent triangles (one positive, one negative) that cancel each other out ( $A_{3} = 0$ ).

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

The Fundamental theorem of calculus (FTC), Part 1 establishes that differentiation and integration are inverse operations. Essentially, taking the derivative “undoes” the integral.

Fundamental Theorem of Calculus (FTC), Part 1

Differentiation and integration are inverse operations.

$\frac{d}{d x} (\int_{a}^{x} f (t) d t) = f (x)$

where $a$ is a constant and $f$ is a continuous function.

When you differentiate $\int_{a}^{x} f (t) d t$ , you get back the original integrand, evaluated at the upper limit.

You won’t have to prove this on the AP exam. As long as the lower bound $a$ is a real number, you can apply the rule directly: remove the integral, and replace $t$ with the upper limit $x$ .

A typical problem will be stated as such:

Find $F^{'} (x)$ given:

$F (x) = \int_{0}^{x} sin (t^{2}) d t$

(spoiler)

Start by writing the derivative:

$F^{'} (x) = \frac{d}{d x} (\int_{0}^{x} sin (t^{2}) d t)$

By the FTC (Part 1), this becomes the integrand evaluated at $t = x$ :

$sin (x^{2})$

FTC variations (with chain rule)

When the limits of integration are functions instead of a single variable $x$ , apply the chain rule. Evaluate the integrand at the upper limit, then multiply by the derivative of that limit.

Variation 1: Functional upper limit

If the upper limit is a function $g (x)$ , evaluate the integrand at $g (x)$ and multiply by $g^{'} (x)$ .

$\frac{d}{d x} (\int_{a}^{g (x)} f (t) d t) = f (g (x)) \cdot g^{'} (x)$

where $a$ is a constant and $g (x)$ is a function.

Example 1. Find $G^{'} (x)$ given

$G (x) = \int_{1}^{x^{3}} 1 + t^{4} d t$

(spoiler)

Here the upper limit is $g (x) = x^{3}$ , so we:

Replace $t$ in the integrand with $x^{3}$ .
Multiply by the entire expression by $g^{'} (x) = 3 x^{2}$ .

$G^{'} (x) = 1 + (x^{3})^{4} \cdot \frac{d}{d x} (x^{3}) = 3 x^{2} 1 + x^{12}$

Variation 2: Functional lower limit

If the lower limit is a function $h (x)$ and the upper limit is a constant, reverse the limits of integration. This introduces a negative sign.

$\frac{d}{d x} (\int_{h (x)}^{a} f (t) d t) = - f (h (x)) \cdot h^{'} (x)$

Example 2. Find $F^{'} (x)$ given

$F (x) = \int_{x^{2}}^{- 1} (t^{3} - t) d t$

(spoiler)

Flip the bounds of the integral and make it negative:

$F (x) = - \int_{- 1}^{x^{2}} (t^{3} - t) d t$

Then, apply the chain rule when differentiating by:

Replacing every $t$ in the integrand with $x^{2}$ .
Multiplying the entire expression by its derivative, $2 x$ .

$F^{'} (x) = - [(x^{2})^{3} - (x^{2})] \cdot \frac{d}{d x} (x^{2}) = - (x^{6} - x^{2}) \cdot 2 x = - 2 x^{7} + 2 x^{3}$

Variation 3: Two variable limits

When both limits are functions of $x$ , apply the chain rule to both boundaries and subtract the lower boundary result from the upper boundary result.

$\frac{d}{d x} (\int_{h (x)}^{g (x)} f (t) d t) = f (g (x)) \cdot g^{'} (x) - f (h (x)) \cdot h^{'} (x)$

Example 3. Find $f^{'} (x)$ given

$f (x) = \int_{3 x}^{c o s (x)} e^{2 t} d t$

Solution

(spoiler)

For the upper bound:

Replace $t$ with $cos (x)$ .
Multiply by its derivative of $- sin (x)$ .

Subtract the result of the lower bound:

Replace $t$ with $3 x$ .
Multiply by its derivative of $3$ .

$f^{'} (x) = [e^{2 c o s (x)} \cdot (- sin (x))] - [e^{2 (3 x)} \cdot 3] = - sin (x) e^{2 c o s (x)} - 3 e^{6 x}$

Properties of definite integrals

Splitting an interval $(a < b < c)$

$\int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x$

Use this to break up integrals over intervals or to combine pieces.

Reversing limits

$\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x$

Switching the upper and lower bounds changes the sign.

Constant multiples

$\int_{a}^{b} k \cdot f (x) d x = k \cdot \int_{a}^{b} f (x) d x$

You can factor out constants.

Linearity

$\int_{a}^{b} [f (x) \pm g (x)] d x = \int_{a}^{b} f (x) d x \pm \int_{a}^{b} g (x) d x$

Addition and subtraction work inside integrals.

Caution: The final property doesn’t hold when two or more functions are multiplied or divided.

Given $\int_{1}^{3} f (x) d x = 4$ and $\int_{3}^{5} f (x) d x = - 2$ , find:

a) $\int_{1}^{5} f (x) d x$

b) $\int_{5}^{1} f (x) d x$

Solutions

(spoiler)

a) Add over intervals:

$\int_{1}^{5} f (x) d x = \int_{1}^{3} f (x) d x + \int_{3}^{5} f (x) d x = 4 + (- 2) = 2$

b) Reverse the bounds:

$\int_{5}^{1} f (x) d x = - \int_{1}^{5} f (x) d x = - 2$

Given $\int_{3}^{10} f (x) d x = - 5$ and $\int_{6}^{10} f (x) d x = 2$ , find:

$\int_{3}^{6} 2 f (x) d x$

Solution

(spoiler)

Using property 1 (splitting up intervals),

$\int_{3}^{10} f (x) d x = \int_{3}^{6} f (x) d x + \int_{6}^{10} f (x) d x$

Substitute the given values:

$- 5 - 7 = \int_{3}^{6} f (x) d x + 2 = \int_{3}^{6} f (x) d x$

Then use property 3 (constant multiples):

$\int_{3}^{6} 2 f (x) d x = 2 (- 7) = - 14$

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.