Behavior of Accumulation Functions

Part 1 · Hook

Why this matters

On a free-response question you are handed a graph of

f

made of lines and half-circles, and asked about a function

g

you never see. Where does

g

peak? Where does it bend? How big does it get? Everything is in the graph of

f

, once you know that

g

's slope is

f

and

g

's value is the area so far.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $g(x)$ $= \int_0^x f(t)\,dt$ and $f$ is continuous, then $g'(x)$ is

$f(x)$
$f'(x)$
$f(x) - f(0)$
$g(x)$

Show the answer

The Fundamental Theorem of Calculus.

Correct: $f(x)$ :
$f'(x)$ :
$f(x) - f(0)$ :
$g(x)$ :

2. $h'$ changes from positive to negative at $x = 2$ . What happens to $h$ at $x = 2$ ?

a relative maximum
a relative minimum
a point of inflection
nothing can be said

Show the answer

The First Derivative Test: h rises, then falls.

Correct: a relative maximum:
a relative minimum:
a point of inflection:
nothing can be said:

3. On a closed interval, where can a continuous function have its absolute maximum?

at a critical point or an endpoint
only where the derivative is 0
only at an endpoint
only at a point of inflection

Show the answer

The candidates test compares the endpoints and the critical points.

Correct: at a critical point or an endpoint:
only where the derivative is 0:
only at an endpoint:
only at a point of inflection:

Part 4 · See it

See it first

Top: the graph of f on [0, 8]: a line from (0, 2) to (2, 0), a semicircle of radius 2 below the axis from x = 2 to 6, and a line from (6, 0) to (8, 2); the triangles have area 2 each and the half-disc area 2 pi. Bottom: g(x), the integral of f from 0 to x, rises to a relative maximum of 2 at x = 2, falls to a relative minimum of 2 - 2 pi at x = 6, and has an inflection point at x = 4, where f changes from decreasing to increasing. — Area above the axis adds to g; area below subtracts. g peaks where f changes from + to -, and bends where f turns around. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$g(x)$ $= \int_a^x f(t)\,dt$ $g'(x) = f(x)$ and $g''(x) = f'(x)$
$f > 0$ (graph of $f$ above the axis) $g$ is increasing; where $f$ changes sign, $g$ has a relative extremum
$f$ is increasing (graph of $f$ rising) $g$ is concave up; where $f$ turns around, $g$ has a point of inflection
Regions below the axis count as negativevalues of $g$ are signed areas: add what is above, subtract what is below

Part 6 · Key ideas

Key ideas

Signed area: $\int_a^b f(x)\,dx$ = (area above the axis) minus (area below).
$g' = f$ : $g$ increases where $f > 0$ , decreases where $f < 0$ , and has a relative extremum where $f$ changes sign.
$g'' = f'$ : $g$ is concave up where $f$ increases and concave down where $f$ decreases; inflection points sit where $f$ changes direction.
For an absolute extremum of g, compare g at the endpoints and at the sign changes of f (the candidates test).

Part 7 · Misconception

A common mistake

The wrong idea: $g$ has its inflection points where $f(x) = 0$ .

What actually happens: Where $f = 0$ and changes sign, $g$ has a relative extremum. Inflection points of $g$ are where $f' = g''$ changes sign: where the graph of $f$ turns from rising to falling or back.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f$ , with $g(x)$ $= \int_0^x f(t)\,dt$

The graph of the continuous function

f

on

[0, 8]

is made of a line segment from

(0, -2)

to

(1, 0)

, a semicircle above the axis from

x = 1

to

x = 5

, a segment from

(5, 0)

to

(7, -2)

, and a segment from

(7, -2)

to

(8, -2)

. Let

g(x)

= \int_0^x f(t)\,dt

.

1. Find $g(5)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

From 0 to 1: a triangle below the axis, $-\frac{1}{2}(1)(2) = -1$ . From 1 to 5: a half-disc above, $\frac{1}{2}\pi(2)^2 = 2\pi$ . So $g(5) = 2\pi - 1$ .

Answer: $2\,\pi - 1$ ≈ 5.283

2. At what value of $x$ in $(0, 8)$ does $g$ have a relative maximum? Why?

$x = 3$ , because $f$ has its largest value on $[0, 8]$ there
$x = 1$ , because $g' = f$ changes sign from $-$ to $+$ there
$x = 7$ , because $f$ stops decreasing there
$x = 5$ : there $g' = f$ changes from positive to negative

Show the answer

$g'(x) = f(x)$ . $f > 0$ on $(1, 5)$ and $f < 0$ on $(5, 8)$ , so $g'$ changes from positive to negative at $x = 5$ : a relative maximum.

$x = 3$ , because $f$ has its largest value on $[0, 8]$ there: At $x = 3$ , $g' = f$ is largest, so $g$ is rising fastest. $g$ keeps rising until $f$ turns negative.
$x = 1$ , because $g' = f$ changes sign from $-$ to $+$ there: At $x = 1$ , $f$ changes from negative to positive, so $g$ stops falling and starts rising: a relative minimum.
$x = 7$ , because $f$ stops decreasing there: Where $f$ changes direction tells about the concavity of $g$ , not its extrema. $f$ is negative on both sides of 7.
Correct: $x = 5$ : there $g' = f$ changes from positive to negative: Correct: area stops adding and starts subtracting at $x = 5$ .

3. At what value of $x$ in $(0, 8)$ does the graph of $g$ have a point of inflection?

$x = 5$ , where $f$ changes from positive to negative
$x = 7$ , where $f$ stops decreasing and becomes constant
$x = 3$ , where $f$ changes from increasing to decreasing
$x = 1$ , where $f$ changes from negative to positive

Show the answer

$g'' = f'$ . $f$ increases on $(0, 3)$ and decreases on $(3, 7)$ , so $g$ changes from concave up to concave down at $x = 3$ .

$x = 5$ , where $f$ changes from positive to negative: That is where $g$ has its relative maximum. $f$ is decreasing on both sides of 5, so $g$ stays concave down there.
$x = 7$ , where $f$ stops decreasing and becomes constant: On $(5, 7)$ , $f$ decreases so $g$ is concave down; on $(7, 8)$ , $f$ is constant so $g'' = 0$ . Concavity does not change from down to up.
Correct: $x = 3$ , where $f$ changes from increasing to decreasing: Correct: $g'' = f'$ changes from positive to negative at 3.
$x = 1$ , where $f$ changes from negative to positive: A sign change of $f = g'$ makes an extremum of $g$ . An inflection point needs $g'' = f'$ to change sign.

4. Where on $[0, 8]$ does $g$ reach its absolute minimum value?

$x = 7$ , where $f$ has its lowest value
$x = 1$ , where $g(1) = -1$
$x = 0$ , where $g(0) = 0$
$x = 8$ , where $g(8) = 2\pi - 5$

Show the answer

Candidates: endpoints and where $g' = f$ changes sign. $g(0) = 0$ , $g(1) = -1$ , $g(5) = 2\pi - 1 \approx 5.28$ , $g(8)$ $= 2\pi - 1 - 2 - 2$ $= 2\pi - 5$ $\approx 1.28$ . The minimum is $-1$ , at $x = 1$ .

$x = 7$ , where $f$ has its lowest value: $f$ is least on $[7, 8]$ , but $f$ is the rate. The value of $g$ keeps dropping until 8, and stays above $-1$ .
Correct: $x = 1$ , where $g(1) = -1$ : Correct: the candidates are $g(0) = 0$ , $g(1) = -1$ , $g(5) = 2\pi - 1$ and $g(8) = 2\pi - 5$ ; the least is $-1$ .
$x = 0$ , where $g(0) = 0$ : $g$ decreases right after 0 (because $f < 0$ there), so values just past 0 are less than $g(0)$ .
$x = 8$ , where $g(8) = 2\pi - 5$ : $g$ does fall after 5, but $g(8) = 2\pi - 5 \approx 1.28$ is still above $g(1) = -1$ .

5. Let $g(x)$ $= \int_0^x (t - 2)^2\,dt$ . Does $g$ have a relative extremum at $x = 2$ ?

No, because $g(2)$ is not 0
No: $g'(x) = (x - 2)^2$ is 0 at 2 but does not change sign
Yes, a relative minimum, because $g'(2) = 0$
Yes, a relative maximum, because $(t - 2)^2$ is least at 2

Show the answer

$g'(x) = (x - 2)^2 \ge 0$ , and it is positive on both sides of $x = 2$ . $g$ is increasing through 2, so there is no relative extremum. A critical point is a candidate, not a guarantee.

No, because $g(2)$ is not 0: The value of $g(2)$ does not decide extrema. Look at whether $g'$ changes sign.
Correct: No: $g'(x) = (x - 2)^2$ is 0 at 2 but does not change sign: Correct: $g' \ge 0$ on both sides, so $g$ keeps increasing through $x = 2$ .
Yes, a relative minimum, because $g'(2) = 0$ : $g'(2) = 0$ is not enough. $g' = (x - 2)^2$ is positive on both sides of 2, so there is no extremum.
Yes, a relative maximum, because $(t - 2)^2$ is least at 2: The integrand is smallest at 2, so $g$ rises most slowly there, but it still rises: no maximum.

6. For the function $g$ in a free-response question, $g(x)$ $= \int_0^x f(t)\,dt$ and $f$ is graphed. Which sentence earns the point for " $g$ has a relative maximum at $x = 5$ "?

$f(5) = 0$ , so $g'(5) = 0$ .
$f$ is decreasing at $x = 5$ , so $g$ is concave down.
The graph changes from going up to going down at $x = 5$ .
$g'(x) = f(x)$ changes from positive to negative at $x = 5$ .

Show the answer

A relative maximum of $g$ needs $g'$ to change from positive to negative, said in words and tied to the given function: " $g'(x) = f(x)$ changes from positive to negative at $x = 5$ ."

$f(5) = 0$ , so $g'(5) = 0$ .: A zero of the derivative only makes x = 5 a candidate; the sign change is the reason.
$f$ is decreasing at $x = 5$ , so $g$ is concave down.: Concave down at a point does not make it a maximum unless g prime is also 0 there, and this sentence does not connect the two.
The graph changes from going up to going down at $x = 5$ .: This does not say which graph. The scoring guidelines need the function named: g prime, which equals f.
Correct: $g'(x) = f(x)$ changes from positive to negative at $x = 5$ .: Correct: it names g prime, says what changes sign and in which direction.

Part 9 · Summary

Summary

A definite integral is a signed area: regions below the axis subtract. For

g(x)

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

, read

g

's values from signed areas, its direction from the sign of

f

, and its concavity from whether

f

rises or falls. Justify with

g' = f

named, and use the candidates test for absolute extrema.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of ff, with g(x)g(x)=∫0xf(t)dt= \int_0^x f(t)\,dt

Summary

What comes next

Connections

The graph of $f$ , with $g(x)$ $= \int_0^x f(t)\,dt$