Accumulation functions | Integration | Achievable AP Calculus AB

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 by a definite integral with a variable upper limit:

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

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

Think of $A (x)$ as measuring the net area that has been added so far 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)

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

On the graph, this is the area covered from $t = 1$ to $t = 1$ , which means no area covered. So

$A (1) = 0$

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 $1$
- $A_{2} = \frac{1}{4} π (1)^{2}$

Then the area, or $A (3)$ , 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, the other below the $x$ -axis with a negative signed area.
- These 2 triangles happen to be congruent, so the signed areas cancel out.
- $A_{3} = 0$

Then the sum of the areas is

$A (5) = 3 + \frac{π}{2}$

Fundamental Theorem of Calculus: Part 1

Suppose a function has already accumulated some area $A (x)$ . If it accumulates a little more, then the additional area over a small interval $Δ x$ can be approximated by the area of a rectangle:

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

If we divide both sides by $Δ x$ and let $Δ x$ become very small, notice that

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

The left side is the limit definition of the derivative, so we conclude

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

This remarkable relationship - where the derivative of the accumulated area function equals the original function - leads us to a central result: 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.

The core idea is that derivatives and integrals are inverse processes, just as addition and subtraction or multiplication and division are. The FTC connects the geometric idea of area under a curve with the analytic concept of a derivative.

You won’t have to prove this on the AP exam - as long as the lower bound $a$ is a real number, all you have to do is let the $\frac{d}{d x}$ “undo” the $\int_{a}^{x}$ and replace the $t$ with the upper limit $x$ in these types of problems.

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)

Since

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

then by the FTC, this equals

$sin (x^{2})$

When the upper limit is a function

When the upper limit is a function and not just $x$ , use 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)

The upper limit is a function $g (x) = x^{3}$ . So any $t$ that appears in the integrand should be replaced with $x^{3}$ , and the entire expression should be multiplied by $g^{'} (x)$ .

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

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

Given:

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

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

Solution

(spoiler)

The upper limit is a function $g (x) = sin (x)$ . So any $t$ that appears in the integrand should be replaced with $sin (x)$ , and the entire expression should be multiplied by $g^{'} (x)$ .

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

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

Properties of definite integrals

In order for the FTC to apply, the lower bound must be a real number with the upper bound as the variable one. If it isn’t, then a few properties can be applied to achieve that form:

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 reassemble parts.

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)

Flip the integral so that the lower limit is the number using property 2 (reversing limits).

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

Then by the FTC with the chain rule and replacing all instances of $t$ with $x^{2}$ ,

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

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

Given:

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

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

Solution

(spoiler)

Use property 1 (splitting up intervals) to rewrite the integral so that each has a real number as one of the limits (the $0$ can be replaced by any value):

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

Using property 2 (reversing limits), flip the 1st integral so that the lower limit is a number:

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

Then applying the FTC with the chain rule on each integral,

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

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

From the previous example, we broke the integral into two parts to apply the FTC to each individually. But we can skip the rewriting step entirely by with a 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.

1. 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),

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

Substituting,

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

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

Then using property 3 (constant multiples),

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

$= - 14$

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 by a definite integral with a variable upper limit:

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

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

Think of $A (x)$ as measuring the net area that has been added so far 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)

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

On the graph, this is the area covered from $t = 1$ to $t = 1$ , which means no area covered. So

$A (1) = 0$

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 $1$
- $A_{2} = \frac{1}{4} π (1)^{2}$

Then the area, or $A (3)$ , 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, the other below the $x$ -axis with a negative signed area.
- These 2 triangles happen to be congruent, so the signed areas cancel out.
- $A_{3} = 0$

Then the sum of the areas is

$A (5) = 3 + \frac{π}{2}$

Fundamental Theorem of Calculus: Part 1

Suppose a function has already accumulated some area $A (x)$ . If it accumulates a little more, then the additional area over a small interval $Δ x$ can be approximated by the area of a rectangle:

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

If we divide both sides by $Δ x$ and let $Δ x$ become very small, notice that

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

The left side is the limit definition of the derivative, so we conclude

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

This remarkable relationship - where the derivative of the accumulated area function equals the original function - leads us to a central result: 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.

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)

Since

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

then by the FTC, this equals

$sin (x^{2})$

When the upper limit is a function

When the upper limit is a function and not just $x$ , use 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)

The upper limit is a function $g (x) = x^{3}$ . So any $t$ that appears in the integrand should be replaced with $x^{3}$ , and the entire expression should be multiplied by $g^{'} (x)$ .

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

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

Given:

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

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

Solution

(spoiler)

The upper limit is a function $g (x) = sin (x)$ . So any $t$ that appears in the integrand should be replaced with $sin (x)$ , and the entire expression should be multiplied by $g^{'} (x)$ .

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

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

Properties of definite integrals

In order for the FTC to apply, the lower bound must be a real number with the upper bound as the variable one. If it isn’t, then a few properties can be applied to achieve that form:

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 reassemble parts.

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)

Flip the integral so that the lower limit is the number using property 2 (reversing limits).

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

Then by the FTC with the chain rule and replacing all instances of $t$ with $x^{2}$ ,

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

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

Given:

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

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

Solution

(spoiler)

Use property 1 (splitting up intervals) to rewrite the integral so that each has a real number as one of the limits (the $0$ can be replaced by any value):

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

Using property 2 (reversing limits), flip the 1st integral so that the lower limit is a number:

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

Then applying the FTC with the chain rule on each integral,

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

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

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.

1. 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),

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

Substituting,

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

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

Then using property 3 (constant multiples),

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

$= - 14$