6.4 Accumulation functions

Achievable AP Calculus AB

6. Integration

Our AP Calculus AB course is currently in development and is a work-in-progress.

Accumulation functions

9 min read

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What you’ll learn:

Accumulation functions as the net area accumulated under a curve
How to use the Fundamental theorem of calculus part 1
Apply the chain rule when the upper or lower limit is a function
Use properties of definite integrals 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 $t = x$ .

Examples

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

So the accumulated net area is

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

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

The region under $f (t)$ from $t = 1$ to $t = 5$ consists of:

A $1$ by $3$ unit rectangle
- $A_{1} = 3$ units $^{2}$
A semi-circle of radius $1$
- $A_{2} = \frac{1}{2} π$ units $^{2}$
2 triangles: one above the $x$ -axis and one below the $x$ -axis (so it contributes negative signed area)
- These 2 triangles are congruent, so their signed areas cancel.
- $A_{3} = 0$

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

Suppose $A (x)$ represents the area accumulated up to $x$ . If you increase $x$ by a small amount $Δ x$ , the additional area from $x$ to $x + Δ x$ is approximately the area of a thin rectangle:

$A (x + Δ x) - A (x) \approx Δ x f (x)$

Divide both sides by $Δ x$ and let $Δ x \to 0$ :

$Δ x \to 0 lim \frac{A ( x + Δ x ) - A ( x )}{Δ x} = f (x)$

The left side is the definition of $A^{'} (x)$ , so

$A^{'} (x) = f (x)$

This relationship is the key idea behind the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus (FTC), Part 1

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

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

A helpful way to remember this: differentiation “undoes” accumulation. 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 $x$ .

Examples

A typical problem will be stated as such:

Given:

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

Find $F^{'} (x)$ .

Solution

(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 $t = x$ :

$sin (x^{2})$

When the upper limit is a function

If the upper limit is a function (not just $x$ ), you still use the FTC, but you also multiply by the derivative of the upper limit. This is the chain rule.

FTC with chain rule

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

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

Examples

Given:

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

Find $G^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = x^{3}$ .

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

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

$= 3 x^{2} 1 + x^{12}$

Given:

$H (x) = \int_{2}^{s i n (x)} t^{2} - 3 t + 1 d t$

Find $H^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = sin (x)$ .

Replace $t$ with $sin (x)$ .
Multiply by $g^{'} (x) = cos (x)$ .

$H^{'} (x) = [(sin (x))^{2} - 3 sin (x) + 1] \cdot \frac{d}{d x} (sin (x))$

$= cos (x) (sin^{2} (x) - 3 sin (x) + 1)$

Properties of definite integrals

To apply the FTC in the form $\frac{d}{d x} (\int_{a}^{x} \dots)$ , it helps if one limit is a constant and the other is the variable expression. If your integral isn’t in that form, you can often rewrite it using these properties:

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: This property doesn’t hold when two or more functions are multiplied or divided.

Examples

Given:

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

Find $F^{'} (x)$ .

Solution

(spoiler)

The lower limit is $x^{2}$ , not a constant. Use property 2 (reversing limits) to rewrite the integral with a constant lower limit:

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

Now apply the FTC with the chain rule:

Replace $t$ with $x^{2}$ in the integrand.
Multiply by $\frac{d}{d x} (x^{2}) = 2 x$ .
Keep the negative sign from reversing the limits.

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

$= - 2 x (x^{6} - x^{2})$

Given:

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

Find $g^{'} (x)$ .

Solution

(spoiler)

Both limits depend on $x$ . One approach is to rewrite the integral so each piece has a constant limit.

Use property 1 (splitting up intervals) to insert a constant limit (the $0$ can be replaced by any value):

$g (x) = \int_{3 x}^{0} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Now use property 2 (reversing limits) on the first integral so the lower limit is a constant:

$g (x) = - \int_{0}^{3 x} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Apply the FTC with the chain rule to each term:

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

$= - 3 e^{6 x} - sin (x) e^{2 c o s (x)}$

From the previous example, we split the integral into two parts so we could apply the FTC to each part. You can also use a single generalized rule that handles variable limits on both ends:

FTC when both limits are variable

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

Next are additional examples that require the properties of definite integrals.

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:

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

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

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 as the net area accumulated under a curve
How to use the Fundamental theorem of calculus part 1
Apply the chain rule when the upper or lower limit is a function
Use properties of definite integrals 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 $t = x$ .

Examples

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

So the accumulated net area is

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

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

The region under $f (t)$ from $t = 1$ to $t = 5$ consists of:

A $1$ by $3$ unit rectangle
- $A_{1} = 3$ units $^{2}$
A semi-circle of radius $1$
- $A_{2} = \frac{1}{2} π$ units $^{2}$
2 triangles: one above the $x$ -axis and one below the $x$ -axis (so it contributes negative signed area)
- These 2 triangles are congruent, so their signed areas cancel.
- $A_{3} = 0$

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

Suppose $A (x)$ represents the area accumulated up to $x$ . If you increase $x$ by a small amount $Δ x$ , the additional area from $x$ to $x + Δ x$ is approximately the area of a thin rectangle:

$A (x + Δ x) - A (x) \approx Δ x f (x)$

Divide both sides by $Δ x$ and let $Δ x \to 0$ :

$Δ x \to 0 lim \frac{A ( x + Δ x ) - A ( x )}{Δ x} = f (x)$

The left side is the definition of $A^{'} (x)$ , so

$A^{'} (x) = f (x)$

This relationship is the key idea behind the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus (FTC), Part 1

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

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

A helpful way to remember this: differentiation “undoes” accumulation. 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 $x$ .

Examples

A typical problem will be stated as such:

Given:

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

Find $F^{'} (x)$ .

Solution

(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 $t = x$ :

$sin (x^{2})$

When the upper limit is a function

If the upper limit is a function (not just $x$ ), you still use the FTC, but you also multiply by the derivative of the upper limit. This is the chain rule.

FTC with chain rule

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

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

Examples

Given:

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

Find $G^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = x^{3}$ .

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

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

$= 3 x^{2} 1 + x^{12}$

Given:

$H (x) = \int_{2}^{s i n (x)} t^{2} - 3 t + 1 d t$

Find $H^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = sin (x)$ .

Replace $t$ with $sin (x)$ .
Multiply by $g^{'} (x) = cos (x)$ .

$H^{'} (x) = [(sin (x))^{2} - 3 sin (x) + 1] \cdot \frac{d}{d x} (sin (x))$

$= cos (x) (sin^{2} (x) - 3 sin (x) + 1)$

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: This property doesn’t hold when two or more functions are multiplied or divided.

Examples

Given:

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

Find $F^{'} (x)$ .

Solution

(spoiler)

The lower limit is $x^{2}$ , not a constant. Use property 2 (reversing limits) to rewrite the integral with a constant lower limit:

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

Now apply the FTC with the chain rule:

Replace $t$ with $x^{2}$ in the integrand.
Multiply by $\frac{d}{d x} (x^{2}) = 2 x$ .
Keep the negative sign from reversing the limits.

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

$= - 2 x (x^{6} - x^{2})$

Given:

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

Find $g^{'} (x)$ .

Solution

(spoiler)

Both limits depend on $x$ . One approach is to rewrite the integral so each piece has a constant limit.

Use property 1 (splitting up intervals) to insert a constant limit (the $0$ can be replaced by any value):

$g (x) = \int_{3 x}^{0} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Now use property 2 (reversing limits) on the first integral so the lower limit is a constant:

$g (x) = - \int_{0}^{3 x} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Apply the FTC with the chain rule to each term:

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

$= - 3 e^{6 x} - sin (x) e^{2 c o s (x)}$

From the previous example, we split the integral into two parts so we could apply the FTC to each part. You can also use a single generalized rule that handles variable limits on both ends:

FTC when both limits are variable

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

Next are additional examples that require the properties of definite integrals.

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:

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

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

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 currently in development and is a work-in-progress.

Accumulation functions

9 min read

Font

Discuss

Feedback

What you’ll learn:

Accumulation functions as the net area accumulated under a curve
How to use the Fundamental theorem of calculus part 1
Apply the chain rule when the upper or lower limit is a function
Use properties of definite integrals 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 $t = x$ .

Examples

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

So the accumulated net area is

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

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

The region under $f (t)$ from $t = 1$ to $t = 5$ consists of:

A $1$ by $3$ unit rectangle
- $A_{1} = 3$ units $^{2}$
A semi-circle of radius $1$
- $A_{2} = \frac{1}{2} π$ units $^{2}$
2 triangles: one above the $x$ -axis and one below the $x$ -axis (so it contributes negative signed area)
- These 2 triangles are congruent, so their signed areas cancel.
- $A_{3} = 0$

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

Suppose $A (x)$ represents the area accumulated up to $x$ . If you increase $x$ by a small amount $Δ x$ , the additional area from $x$ to $x + Δ x$ is approximately the area of a thin rectangle:

$A (x + Δ x) - A (x) \approx Δ x f (x)$

Divide both sides by $Δ x$ and let $Δ x \to 0$ :

$Δ x \to 0 lim \frac{A ( x + Δ x ) - A ( x )}{Δ x} = f (x)$

The left side is the definition of $A^{'} (x)$ , so

$A^{'} (x) = f (x)$

This relationship is the key idea behind the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus (FTC), Part 1

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

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

A helpful way to remember this: differentiation “undoes” accumulation. 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 $x$ .

Examples

A typical problem will be stated as such:

Given:

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

Find $F^{'} (x)$ .

Solution

(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 $t = x$ :

$sin (x^{2})$

When the upper limit is a function

If the upper limit is a function (not just $x$ ), you still use the FTC, but you also multiply by the derivative of the upper limit. This is the chain rule.

FTC with chain rule

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

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

Examples

Given:

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

Find $G^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = x^{3}$ .

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

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

$= 3 x^{2} 1 + x^{12}$

Given:

$H (x) = \int_{2}^{s i n (x)} t^{2} - 3 t + 1 d t$

Find $H^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = sin (x)$ .

Replace $t$ with $sin (x)$ .
Multiply by $g^{'} (x) = cos (x)$ .

$H^{'} (x) = [(sin (x))^{2} - 3 sin (x) + 1] \cdot \frac{d}{d x} (sin (x))$

$= cos (x) (sin^{2} (x) - 3 sin (x) + 1)$

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: This property doesn’t hold when two or more functions are multiplied or divided.

Examples

Given:

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

Find $F^{'} (x)$ .

Solution

(spoiler)

The lower limit is $x^{2}$ , not a constant. Use property 2 (reversing limits) to rewrite the integral with a constant lower limit:

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

Now apply the FTC with the chain rule:

Replace $t$ with $x^{2}$ in the integrand.
Multiply by $\frac{d}{d x} (x^{2}) = 2 x$ .
Keep the negative sign from reversing the limits.

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

$= - 2 x (x^{6} - x^{2})$

Given:

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

Find $g^{'} (x)$ .

Solution

(spoiler)

Both limits depend on $x$ . One approach is to rewrite the integral so each piece has a constant limit.

Use property 1 (splitting up intervals) to insert a constant limit (the $0$ can be replaced by any value):

$g (x) = \int_{3 x}^{0} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Now use property 2 (reversing limits) on the first integral so the lower limit is a constant:

$g (x) = - \int_{0}^{3 x} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Apply the FTC with the chain rule to each term:

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

$= - 3 e^{6 x} - sin (x) e^{2 c o s (x)}$

From the previous example, we split the integral into two parts so we could apply the FTC to each part. You can also use a single generalized rule that handles variable limits on both ends:

FTC when both limits are variable

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

Next are additional examples that require the properties of definite integrals.

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:

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

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

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 as the net area accumulated under a curve
How to use the Fundamental theorem of calculus part 1
Apply the chain rule when the upper or lower limit is a function
Use properties of definite integrals 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 $t = x$ .

Examples

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

So the accumulated net area is

$A (3) = 2 + \frac{π}{4}$

c) $A (5)$

(spoiler)

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

The region under $f (t)$ from $t = 1$ to $t = 5$ consists of:

A $1$ by $3$ unit rectangle
- $A_{1} = 3$ units $^{2}$
A semi-circle of radius $1$
- $A_{2} = \frac{1}{2} π$ units $^{2}$
2 triangles: one above the $x$ -axis and one below the $x$ -axis (so it contributes negative signed area)
- These 2 triangles are congruent, so their signed areas cancel.
- $A_{3} = 0$

Adding the signed areas gives

$A (5) = 3 + \frac{π}{2}$

Fundamental theorem of calculus: part 1

Suppose $A (x)$ represents the area accumulated up to $x$ . If you increase $x$ by a small amount $Δ x$ , the additional area from $x$ to $x + Δ x$ is approximately the area of a thin rectangle:

$A (x + Δ x) - A (x) \approx Δ x f (x)$

Divide both sides by $Δ x$ and let $Δ x \to 0$ :

$Δ x \to 0 lim \frac{A ( x + Δ x ) - A ( x )}{Δ x} = f (x)$

The left side is the definition of $A^{'} (x)$ , so

$A^{'} (x) = f (x)$

This relationship is the key idea behind the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus (FTC), Part 1

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

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

A helpful way to remember this: differentiation “undoes” accumulation. 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 $x$ .

Examples

A typical problem will be stated as such:

Given:

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

Find $F^{'} (x)$ .

Solution

(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 $t = x$ :

$sin (x^{2})$

When the upper limit is a function

If the upper limit is a function (not just $x$ ), you still use the FTC, but you also multiply by the derivative of the upper limit. This is the chain rule.

FTC with chain rule

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

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

Examples

Given:

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

Find $G^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = x^{3}$ .

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

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

$= 3 x^{2} 1 + x^{12}$

Given:

$H (x) = \int_{2}^{s i n (x)} t^{2} - 3 t + 1 d t$

Find $H^{'} (x)$ .

Solution

(spoiler)

Here the upper limit is $g (x) = sin (x)$ .

Replace $t$ with $sin (x)$ .
Multiply by $g^{'} (x) = cos (x)$ .

$H^{'} (x) = [(sin (x))^{2} - 3 sin (x) + 1] \cdot \frac{d}{d x} (sin (x))$

$= cos (x) (sin^{2} (x) - 3 sin (x) + 1)$

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: This property doesn’t hold when two or more functions are multiplied or divided.

Examples

Given:

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

Find $F^{'} (x)$ .

Solution

(spoiler)

The lower limit is $x^{2}$ , not a constant. Use property 2 (reversing limits) to rewrite the integral with a constant lower limit:

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

Now apply the FTC with the chain rule:

Replace $t$ with $x^{2}$ in the integrand.
Multiply by $\frac{d}{d x} (x^{2}) = 2 x$ .
Keep the negative sign from reversing the limits.

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

$= - 2 x (x^{6} - x^{2})$

Given:

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

Find $g^{'} (x)$ .

Solution

(spoiler)

Both limits depend on $x$ . One approach is to rewrite the integral so each piece has a constant limit.

Use property 1 (splitting up intervals) to insert a constant limit (the $0$ can be replaced by any value):

$g (x) = \int_{3 x}^{0} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Now use property 2 (reversing limits) on the first integral so the lower limit is a constant:

$g (x) = - \int_{0}^{3 x} e^{2 t} d t + \int_{0}^{c o s (x)} e^{2 t} d t$

Apply the FTC with the chain rule to each term:

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

$= - 3 e^{6 x} - sin (x) e^{2 c o s (x)}$

From the previous example, we split the integral into two parts so we could apply the FTC to each part. You can also use a single generalized rule that handles variable limits on both ends:

FTC when both limits are variable

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

Next are additional examples that require the properties of definite integrals.

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:

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

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

Then use property 3 (constant multiples):

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

$= - 14$