6.5 - Interpreting the Behavior of Accumulation Functions Involving Area

Review

Accumulation function:

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

where

a

is constant and

x

is variable.

Fundamental Theorem of Calculus (Part 2):

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

And if the upper limit is a function

g(x)

F'(x) = f(g(x)) \cdot g'(x)

Behavior of Accumulation Functions

Since

F'(x) = f(x)

, we can analyze

F

just by looking at

f

Property of $F(x)$	Condition on $F'(x)$	Equivalent Condition on $f(x)$
Increasing	$F'(x) > 0$	$f(x) > 0$
Decreasing	$F'(x) < 0$	$f(x) < 0$
Relative maximum	$F'$ changes $(+)$ to $(-)$	$f$ changes $(+)$ to $(-)$
Relative minimum	$F'$ changes $(-)$ to $(+)$	$f$ changes $(-)$ to $(+)$
Concave up	$F''(x) > 0$	$f'(x) > 0$ (slope of $f$ is positive)
Concave down	$F''(x) < 0$	$f'(x) < 0$ (slope of $f$ is negative)
Point of inflection	$F''$ changes sign	$f'$ changes sign

The two relationships that drive everything:

F'(x) = f(x) \quad \text{and} \quad F''(x) = f'(x)

Example 1: Analyzing from the Graph of $f(x)$

Given

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

and the graph of

f(x)

below.

6.5 - Interpreting the Behavior of Accumulation Functions Involving Area

Find the relative minimum of $g(x)$

A relative minimum happens where

g'

changes from

(-)

(+)

. Since

g'(x) = f(x)

, we need

f

to cross from negative to positive.

From the graph,

f

crosses from negative to positive at

x \approx 5.4

Answer: relative minimum at

x \approx 5.4

Find the relative maximum of $g(x)$

A relative maximum happens where

g'

changes from

(+)

(-)

, so we need

f

to cross from positive to negative.

f

crosses from positive to negative at

x = 3

Answer: relative maximum at

x = 3

Where is $g$ concave up?

Concave up means

g''(x) > 0

. Since

g''(x) = f'(x)

, we need the slope of

f

to be positive.

The slope of

f

is positive (the graph is rising) on

(4, 6)

Answer: concave up on

(4, 6)

Where is $g$ concave down?

Concave down means

g''(x) < 0

, so we need the slope of

f

to be negative.

The slope of

f

is negative (the graph is falling) on

(0, 4) \cup (6, 7)

Answer: concave down on

(0, 4) \cup (6, 7)

Points of inflection of $g(x)$

Points of inflection happen where

g''

changes sign, so we need the slope of

f

to change sign.

At $x = 2$ : the slope goes from negative to zero, then back to negative. No sign change.

At $x = 4$ : the slope goes from negative to positive. Sign change ✓

At $x = 6$ : the slope goes from positive to zero, then to negative. Sign change ✓

Both

x = 4 \text{ and } x = 6

have a genuine sign change in slope.

Answer: points of inflection at

x = 4 \text{ and } x = 6

Maximum value of $g$ on $[1, 5]$ , given $g(1) = 5$

Strategy: 1. Find where the maximum occurs (already found:

x = 3

) 2. Calculate

g(3)

using

g(3) = g(1) + \int_1^3 f(t) \, dt

[1, 3]

f

is positive (it doesn’t cross the axis until

x = 3

), and the region breaks into two pieces of area 1 and 2, giving a total area of 3.

g(3) = g(1) + \int_1^3 f(t) \, dt = 5 + 3 = 8

Answer: the maximum value of

g

is 8

Example 2: Finding Critical Points Algebraically

Given

g(x) = \int_2^{x/2 + 5} f(t) \, dt

, where

f

is continuous,

f(t) > 0

for

t < 3

, and

f(t) < 0

for

t > 3

. So

f

crosses from positive to negative at

t = 3

Find the

x

-value where

g

has a relative maximum.

Step 1: find $g'(x)$ using FTC Part 2 with the chain rule:

g'(x) = f\left(\frac{x}{2} + 5\right) \cdot \frac{1}{2}

Step 2: set $g'(x) = 0$ . For a relative maximum,

g'

has to change from

(+)

(-)

, which happens when:

f\left(\frac{x}{2} + 5\right) = 0

Step 3: solve for $x$ . We know

f

changes from positive to negative at

t = 3

, so set the argument equal to 3:

\frac{x}{2} + 5 = 3 \quad \Rightarrow \quad \frac{x}{2} = -2 \quad \Rightarrow \quad x = -4

Answer:

g

has a relative maximum at

x = -4

Verification: - For

x < -4

\frac{x}{2} + 5 < 3

, so

f

is positive (before its sign change), meaning

g' > 0

✓ - For

x > -4

\frac{x}{2} + 5 > 3

, so

f

is negative, meaning

g' < 0

✓ - The sign change from

(+)

(-)

confirms a relative maximum ✓

Example 3: Working from a Table of Values

Let

g(x) = 5 + \int_1^x f(t) \, dt

, where

f

is continuous. Selected values of

f

are in the table.

$x$	1	3	4	7
$f(x)$	6	2	$-1$	$-4$

Find

g(1)

g(1) = 5 + \int_1^1 f(t) \, dt = 5 + 0 = 5

, since an integral from a number to itself is 0.

g

increasing or decreasing at

x = 3

g'(3) = f(3) = 2 > 0

, so

g

is increasing at

x = 3

Must

g

have a critical point between

x = 3

and

x = 4

Yes.

f

is continuous,

f(3) = 2 > 0

, and

f(4) = -1 < 0

, so by the Intermediate Value Theorem,

f(c) = 0

for some

c

(3, 4)

. Then

g'(c) = f(c) = 0

, so

g

has a critical point at

x = c

Approximate

g(7)

with a trapezoidal sum.

The sub-intervals have widths 2, 1, and 3:

\int_1^7 f(t) \, dt \approx 2 \cdot \frac{6 + 2}{2} + 1 \cdot \frac{2 + (-1)}{2} + 3 \cdot \frac{-1 + (-4)}{2} = 8 + 0.5 - 7.5 = 1

g(7) \approx 5 + 1 = 6

Strategy Summary

To analyze

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

Increasing/decreasing -

G

is increasing where

f(x) > 0

(above the x-axis) -

G

is decreasing where

f(x) < 0

(below the x-axis)

Extrema - Relative max where

f

crosses from

(+)

(-)

- Relative min where

f

crosses from

(-)

(+)

Concavity -

G

is concave up where the slope of

f

is positive -

G

is concave down where the slope of

f

is negative - Point of inflection where the slope of

f

changes sign

Finding values - Use

G(b) = G(a) + \int_a^b f(t) \, dt

- Find the area under

f

from

a

b

, then add it to the known value at

a

When the upper limit is a function $h(x)$ 1. Find

G'(x) = f(h(x)) \cdot h'(x)

2. Set it equal to zero, so

f(h(x)) = 0

3. Set

h(x)

equal to the known zeros of

f

and solve 4. Verify the sign change to determine max vs. min

Key Takeaways

Fundamental relationship: if $F(x) = \int_a^x f(t) \, dt$ , then $F'(x) = f(x)$ and $F''(x) = f'(x)$

Increasing/decreasing: analyze the sign of $f(x)$ directly

Extrema: look for where $f$ crosses the x-axis

Concavity: look at the slope of $f$

Points of inflection: find where the slope of $f$ changes sign

Finding values: $F(b) = F(a) + (\text{area under } f \text{ from } a \text{ to } b)$

With a function upper limit: set the argument equal to known features of $f$ and solve

This is Unit 5 material applied. Same concepts, just with an integral defining the function

The graph of $f$ tells the story of $F$ : positive $f$ means $F$ is increasing, negative $f$ means $F$ is decreasing

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6.5 - Interpreting the Behavior of Accumulation Functions Involving Area

Review

Behavior of Accumulation Functions

Example 1: Analyzing from the Graph of f(x)f(x)f(x)

Example 2: Finding Critical Points Algebraically

Example 3: Working from a Table of Values

Strategy Summary

Key Takeaways

Example 1: Analyzing from the Graph of $f(x)$