6.5 Behavior of accumulation functions

Achievable AP Calculus AB

6. Integration

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Behavior of accumulation functions

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What you’ll learn

Interpreting accumulation functions: Apply the FTC to determine the behavior of an accumulation function from the graph or expression of its derivative.

Every single AP Calculus exam has featured a free-response question centered on a graph of $f (t)$ and a corresponding accumulation function. Suppose we define it as:

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

The absolute first thing you must write on your paper is the Fundamental theorem of calculus link:

$F^{'} (x) = f (x) .$

Treat these problems like the ones from section 5.4: given the graph of the derivative $f$ , what can you conclude about the original function $F$ ?

AP tip:

When given $F (x) = \int_{a}^{x} f (t) d t$ in a problem, apply the FTC and immediately write the following:

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

as a reminder that capital $F$ is the accumulation function and lowercase $f$ is its derivative.

Review of derivative relationships

Property of $F (x)$	where $f (x) ...$
Increasing	$> 0$
Decreasing	$< 0$
Relative min	Changes from $(-)$ to $(+)$
Relative max	Changes from $(+)$ to $(-)$
Concave up	Increasing
Concave down	Decreasing
Inflection point	Has a local min/max

Example 1: Standard problem

Shown below is the graph of $f$ :

Figure 6.5.1 Graph of f

Let $F$ be the function defined by

$F (x) = 2 + \int_{1}^{x} f (t) d t$

Use the graph of $f$ to complete the table describing the behavior of $F$ . Give interval(s) or ordered pair(s) as appropriate.

$F (x)$ Interval or ordered pair

Increasing & decreasing

Relative extrema

Absolute extrema

Concavity

Inflection point(s)

$F (x)$	Interval or ordered pair
Increasing & decreasing
Relative extrema
Absolute extrema
Concavity
Inflection point(s)

$F$ increasing/decreasing

(spoiler)

Increasing: $(4, 6.5)$
Decreasing: $(0, 4) \cup (6.5, 7)$

Why? Because $F^{'} (x) = f (x)$ , $F$ increases where $f > 0$ (above the $x$ -axis) and decreases where $f < 0$ (below the $x$ -axis).

Relative extrema of $F$

(spoiler)

Relative minimum: $(4, - 0.071)$
- Justification: $f$ changes from $(-)$ to $(+)$ at $x = 4$ .

For the $y$ -value, calculate signed area for definite integral:

$F (4) = 2 + \int_{1}^{4} f (t) d t = 2 + triangle (- \frac{1}{2}) + semicircle (- \frac{π}{2}) \approx - 0.071$

Relative maximum: $(6.5, 1.679)$
- Justification: $f$ changes from $(+)$ to $(-)$ at $x = 6.5$ .

Calculation:

$F (6.5) = F (4) + \int_{4}^{6.5} f (t) d t = F (4) + trapezoid (\frac{1 + 2.5}{2}) \approx 1.679$

Absolute extrema of $F$

(spoiler)

Absolute max: $(0, 3.5)$
Absolute min: $(4, - 0.071)$

Because of the closed interval, use the Extreme value theorem and compare endpoints ( $x = 0, 7$ ) and relative extrema values:

Left endpoint ( $x = 0$ ):

$F (0) = 2 + \int_{1}^{0} f (t) d t = 2 - flipped bounds \int_{0}^{1} f (t) d t = 2 - (- 1.5) = 3.5$

Critical point ( $x = 4$ ):

$F (4) \approx - 0.071$

Critical point ( $x = 6.5$ ):

$F (6.5) \approx 1.679$

Right endpoint ( $x = 7$ ):

$F (7) = F (6.5) + \int_{6.5}^{7} f (t) d t \approx F (6.5) + (- 0.25) = 1.429$

Comparing values,

Absolute max: $(0, 3.5)$
Absolute min: $(4, - 0.071)$

Concavity of $F$

(spoiler)

Since $F^{''} (x) = f^{'} (x)$ ,

Concave up: $(0, 2) \cup (3, 5)$ (where $f^{'} > 0$ )
Concave down: $(2, 3) \cup (5, 8)$ (where $f^{'} < 0$ )

Inflection point(s) of $F$

(spoiler)

Answers: $(2, 1.5)$ and $(3, 0.715)$

Inflection point of $F (x) ⟹ F^{''} (x)$ changes sign.

Since $F^{''} (x) = f^{'} (x)$ , look for where:

$f^{'} (x) = 0$ (occurs at $x = 3$ ).
$f^{'} (x)$ is undefined (occurs at sharp corners: $x = 2, 4, 5, 6$ ).

Check signs of $f^{'} (x)$ using slopes of $f$ .

$x$	Sign change in $f^{'}$	Inflection point?
$2$	$(+) \to (-)$	$✓$
$3$	$(-) \to (+)$	$✓$
$4$	None	$\times$
$5$	None	$\times$
$6$	None	$\times$

Note: The last 3 are "none* because going from $(+)$ to $0$ and vice versa is not a sign change.

Lastly, calculate the $y$ -values:

For $x = 2$ :

$F (2) = 2 + \int_{1}^{2} f (t) d t = 2 + (- \frac{1}{2}) = 1.5$

For $x = 3$ :

$F (3) = 2 + \int_{1}^{3} f (t) d t = 2 + (- \frac{1}{2} - \frac{1}{4} π) \approx 0.715$

Inflection points: $(2, 1.5) and (3, 0.715)$

Example 2: FTC with chain rule

The graph of $f$ is shown below:

Figure 6.5.2 Graph of f

Let

$h (x) = \int_{0}^{2 x - 1} f (t) d t$

Find:

a) The $x$ -value at which $h (x)$ has a relative max.

b) $h (3)$

c) $h^{'} (3)$

d) $h^{''} (3)$

Solutions

a) Where $h (x)$ has a relative max

(spoiler)

Relative max of $h (x) ⟹ h^{'} (x)$ changes from $(+)$ to $(-)$ .

Find $h^{'} (x)$ by using FTC with chain rule:

$h^{'} (x) = f (2 x - 1) \cdot \frac{d}{d x} (2 x - 1) = 2 f (2 x - 1)$

Set $h^{'} (x) = 0$ :

$f (2 x - 1) = 0$

From the graph of $f$ , the zeros occur at:

input $t = 3$ : $f$ changes from $(-)$ to $(+) ⟹$ relative min.
input $t = 5$ : $f$ changes from $(+)$ to $(-) ⟹$ relative max.

Now convert $t = 5$ back to $x$ :

$2 x - 1 = 5 ⟹ x = 3$

So $h$ has a relative maximum at $x = 3$ .

b) $h (3)$

(spoiler)

Evaluate the accumulation function:

$h (3) = \int_{0}^{2 (3) - 1} f (t) d t = \int_{0}^{5} f (t) d t$

by adding up the signed areas on $[0, 5]$ .

$\int_{0}^{5} f (t) d t = - \frac{1}{2} (1 + 3) trapezoid on [0, 3] + \frac{1}{2} (1) (2) triangle on [3, 5] = - 1$

c) $h^{'} (3)$

(spoiler)

From part a),

$h^{'} (x) = 2 f (2 x - 1) .$

$h^{'} (3) = 2 f (5) = 0 .$

This matches the fact that $x = 3$ is a relative extremum of $h$ .

d) $h^{''} (3)$

(spoiler)

Differentiate $h^{'} (x)$ again. Since

$h^{'} (x) = 2 f (2 x - 1),$

we get

$h^{''} (x) = 2 f^{'} (2 x - 1) \cdot \frac{d}{d x} (2 x - 1) = 4 f^{'} (2 x - 1) .$

Then

$h^{''} (3) = 4 f^{'} (2 (3) - 1) = 4 f^{'} (5) .$

$f^{'} (5)$ is the slope of the graph of $f$ at input $5$ . At $x = 5$ , the graph lies on the line segment between $(4, 1)$ and $(7, - 2)$ , so the slope is

$\frac{1 - ( - 2 )}{4 - 7} = - 1.$

Therefore $f^{'} (5) = - 1$ , and

$h^{''} (3) = - 4$

Accumulation functions and their properties

Accumulation function: $F (x) = \int_{a}^{x} f (t) d t$
$F^{'} (x) = f (x)$ (Fundamental Theorem of Calculus)
$F^{''} (x) = f^{'} (x)$ (concavity from derivative of $f$ )

Increasing/decreasing behavior

$F (x)$ increasing where $f (x) > 0$
$F (x)$ decreasing where $f (x) < 0$

Relative and absolute extrema

Relative min: where $f (x)$ changes from negative to positive
Relative max: where $f (x)$ changes from positive to negative
Absolute extrema: check endpoints and critical points (where $f (x) = 0$ )

Concavity and inflection points

Concave up: where $f^{'} (x) > 0$ (i.e., $f (x)$ increasing)
Concave down: where $f^{'} (x) < 0$ (i.e., $f (x)$ decreasing)
Inflection point: where $f^{'} (x)$ changes sign

Chain rule with accumulation functions

For $h (x) = \int_{a}^{g (x)} f (t) d t$ :
- $h^{'} (x) = f (g (x)) \cdot g^{'} (x)$
- $h^{''} (x) = f^{'} (g (x)) \cdot (g^{'} (x))^{2} + f (g (x)) \cdot g^{''} (x)$

Graphical interpretation

Signed area under $f$ gives value of accumulation function
Use graph features (zero crossings, increasing/decreasing, slopes) to analyze $F$ or $h$

AP Exam Tip

Write $F^{'} (x) = f (x)$ when given $F (x) = \int_{a}^{x} f (t) d t$ to clarify roles of $F$ and $f$

Behavior of accumulation functions

What you’ll learn

Interpreting accumulation functions: Apply the FTC to determine the behavior of an accumulation function from the graph or expression of its derivative.

Every single AP Calculus exam has featured a free-response question centered on a graph of $f (t)$ and a corresponding accumulation function. Suppose we define it as:

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

The absolute first thing you must write on your paper is the Fundamental theorem of calculus link:

$F^{'} (x) = f (x) .$

Treat these problems like the ones from section 5.4: given the graph of the derivative $f$ , what can you conclude about the original function $F$ ?

AP tip:

When given $F (x) = \int_{a}^{x} f (t) d t$ in a problem, apply the FTC and immediately write the following:

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

as a reminder that capital $F$ is the accumulation function and lowercase $f$ is its derivative.

Review of derivative relationships

Property of $F (x)$	where $f (x) ...$
Increasing	$> 0$
Decreasing	$< 0$
Relative min	Changes from $(-)$ to $(+)$
Relative max	Changes from $(+)$ to $(-)$
Concave up	Increasing
Concave down	Decreasing
Inflection point	Has a local min/max

Example 1: Standard problem

Shown below is the graph of $f$ :

Let $F$ be the function defined by

$F (x) = 2 + \int_{1}^{x} f (t) d t$

Use the graph of $f$ to complete the table describing the behavior of $F$ . Give interval(s) or ordered pair(s) as appropriate.

$F (x)$ Interval or ordered pair

Increasing & decreasing

Relative extrema

Absolute extrema

Concavity

Inflection point(s)

$F (x)$	Interval or ordered pair
Increasing & decreasing
Relative extrema
Absolute extrema
Concavity
Inflection point(s)

$F$ increasing/decreasing

(spoiler)

Increasing: $(4, 6.5)$
Decreasing: $(0, 4) \cup (6.5, 7)$

Why? Because $F^{'} (x) = f (x)$ , $F$ increases where $f > 0$ (above the $x$ -axis) and decreases where $f < 0$ (below the $x$ -axis).

Relative extrema of $F$

(spoiler)

Relative minimum: $(4, - 0.071)$
- Justification: $f$ changes from $(-)$ to $(+)$ at $x = 4$ .

For the $y$ -value, calculate signed area for definite integral:

$F (4) = 2 + \int_{1}^{4} f (t) d t = 2 + triangle (- \frac{1}{2}) + semicircle (- \frac{π}{2}) \approx - 0.071$

Relative maximum: $(6.5, 1.679)$
- Justification: $f$ changes from $(+)$ to $(-)$ at $x = 6.5$ .

Calculation:

$F (6.5) = F (4) + \int_{4}^{6.5} f (t) d t = F (4) + trapezoid (\frac{1 + 2.5}{2}) \approx 1.679$

Absolute extrema of $F$

(spoiler)

Absolute max: $(0, 3.5)$
Absolute min: $(4, - 0.071)$

Because of the closed interval, use the Extreme value theorem and compare endpoints ( $x = 0, 7$ ) and relative extrema values:

Left endpoint ( $x = 0$ ):

$F (0) = 2 + \int_{1}^{0} f (t) d t = 2 - flipped bounds \int_{0}^{1} f (t) d t = 2 - (- 1.5) = 3.5$

Critical point ( $x = 4$ ):

$F (4) \approx - 0.071$

Critical point ( $x = 6.5$ ):

$F (6.5) \approx 1.679$

Right endpoint ( $x = 7$ ):

$F (7) = F (6.5) + \int_{6.5}^{7} f (t) d t \approx F (6.5) + (- 0.25) = 1.429$

Comparing values,

Absolute max: $(0, 3.5)$
Absolute min: $(4, - 0.071)$

Concavity of $F$

(spoiler)

Since $F^{''} (x) = f^{'} (x)$ ,

Concave up: $(0, 2) \cup (3, 5)$ (where $f^{'} > 0$ )
Concave down: $(2, 3) \cup (5, 8)$ (where $f^{'} < 0$ )

Inflection point(s) of $F$

(spoiler)

Answers: $(2, 1.5)$ and $(3, 0.715)$

Inflection point of $F (x) ⟹ F^{''} (x)$ changes sign.

Since $F^{''} (x) = f^{'} (x)$ , look for where:

$f^{'} (x) = 0$ (occurs at $x = 3$ ).
$f^{'} (x)$ is undefined (occurs at sharp corners: $x = 2, 4, 5, 6$ ).

Check signs of $f^{'} (x)$ using slopes of $f$ .

$x$	Sign change in $f^{'}$	Inflection point?
$2$	$(+) \to (-)$	$✓$
$3$	$(-) \to (+)$	$✓$
$4$	None	$\times$
$5$	None	$\times$
$6$	None	$\times$

Note: The last 3 are "none* because going from $(+)$ to $0$ and vice versa is not a sign change.

Lastly, calculate the $y$ -values:

For $x = 2$ :

$F (2) = 2 + \int_{1}^{2} f (t) d t = 2 + (- \frac{1}{2}) = 1.5$

For $x = 3$ :

$F (3) = 2 + \int_{1}^{3} f (t) d t = 2 + (- \frac{1}{2} - \frac{1}{4} π) \approx 0.715$

Inflection points: $(2, 1.5) and (3, 0.715)$

Example 2: FTC with chain rule

The graph of $f$ is shown below:

Let

$h (x) = \int_{0}^{2 x - 1} f (t) d t$

Find:

a) The $x$ -value at which $h (x)$ has a relative max.

b) $h (3)$

c) $h^{'} (3)$

d) $h^{''} (3)$

Solutions

a) Where $h (x)$ has a relative max

(spoiler)

Relative max of $h (x) ⟹ h^{'} (x)$ changes from $(+)$ to $(-)$ .

Find $h^{'} (x)$ by using FTC with chain rule:

$h^{'} (x) = f (2 x - 1) \cdot \frac{d}{d x} (2 x - 1) = 2 f (2 x - 1)$

Set $h^{'} (x) = 0$ :

$f (2 x - 1) = 0$

From the graph of $f$ , the zeros occur at:

input $t = 3$ : $f$ changes from $(-)$ to $(+) ⟹$ relative min.
input $t = 5$ : $f$ changes from $(+)$ to $(-) ⟹$ relative max.

Now convert $t = 5$ back to $x$ :

$2 x - 1 = 5 ⟹ x = 3$

So $h$ has a relative maximum at $x = 3$ .

b) $h (3)$

(spoiler)

Evaluate the accumulation function:

$h (3) = \int_{0}^{2 (3) - 1} f (t) d t = \int_{0}^{5} f (t) d t$

by adding up the signed areas on $[0, 5]$ .

$\int_{0}^{5} f (t) d t = - \frac{1}{2} (1 + 3) trapezoid on [0, 3] + \frac{1}{2} (1) (2) triangle on [3, 5] = - 1$

c) $h^{'} (3)$

(spoiler)

From part a),

$h^{'} (x) = 2 f (2 x - 1) .$

$h^{'} (3) = 2 f (5) = 0 .$

This matches the fact that $x = 3$ is a relative extremum of $h$ .

d) $h^{''} (3)$

(spoiler)

Differentiate $h^{'} (x)$ again. Since

$h^{'} (x) = 2 f (2 x - 1),$

we get

$h^{''} (x) = 2 f^{'} (2 x - 1) \cdot \frac{d}{d x} (2 x - 1) = 4 f^{'} (2 x - 1) .$

Then

$h^{''} (3) = 4 f^{'} (2 (3) - 1) = 4 f^{'} (5) .$

$f^{'} (5)$ is the slope of the graph of $f$ at input $5$ . At $x = 5$ , the graph lies on the line segment between $(4, 1)$ and $(7, - 2)$ , so the slope is

$\frac{1 - ( - 2 )}{4 - 7} = - 1.$

Therefore $f^{'} (5) = - 1$ , and

$h^{''} (3) = - 4$

Achievable AP Calculus AB

6. Integration

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