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Calculus AB · Unit 6 · 6.5
AP Calculus · Integration and Accumulation of Change
Interpreting the Behavior of Accumulation Functions
Read the sign of $f$ for direction, its slope for concavity, and its area for value.
key idea$g'=f:\ \text{zeros of }f\to\text{extrema of }g;\ \text{extrema of }f\to\text{inflections of }g$
Reading the graph
Given a graph of $f$, every question about $g(x)=\int_a^x f$ is answered by reading areas and signs — never by finding a formula. The accumulation function is one derivative level above the graph you are shown.
the accumulation $g(x)=\int_1^x f$
reading $g$ from the sign of $f$
| $x\lt 1$ | $1\lt x\lt 2$ | $2\lt x\lt 3$ | $x\gt 3$ |
|---|
| $f=g'$ | − | + | + | − |
| $f'=g''$ | + | + | − | − |
| verdict | $g$ falling, cup | $g$ rising, cup | $g$ rising, cap | $g$ falling, cap |
| Feature of $g$ | Look at $f$ for | Condition |
|---|
| $g$ increasing | the sign of $f$ | $f\gt 0$ |
| local max of $g$ | a sign change in $f$ | $f$ goes $+\to-$ |
| local min of $g$ | a sign change in $f$ | $f$ goes $-\to+$ |
| $g$ concave up | whether $f$ is rising | $f$ increasing |
| inflection of $g$ | an extremum of $f$ | $f$ turns around |
| $g(a)=0$ | — | always — zero-width interval |
| value of $g(b)$ | signed area from $a$ to $b$ | count boxes, subtract below-axis |
Everything about $g$ comes from $f$ one level up: sign of $f$ gives direction, slope of $f$ gives concavity, area under $f$ gives value.
Direction of $g$the sign of $f$
Concavity of $g$whether $f$ is rising or falling
Value of $g$the signed area under $f$
Worked examples
EX 1
Find the maximum of $g$
sign changeFor $g(x)=\displaystyle\int_{1}^{x}f(t)\,dt$ with $f$ as shown, where is $g$ greatest on $[0,4]$?
1
Use $g'=f$
Critical points of $g$ are the zeros of $f$: $x=1$ and $x=3$.
2
Test the signs
$f\lt 0$ before $1$, $f\gt 0$ on $(1,3)$, $f\lt 0$ after $3$.
3
Classify
At $x=1$: $-\to+$, a local minimum. At $x=3$: $+\to-$, a local maximum.
4
Check the endpoints
$g(0)$ and $g(4)$ are both smaller, since area is being subtracted on both outer stretches.
5
Answer
$g$ attains its maximum at $x=3$.
EX 2
Evaluate $g$ by counting area
geometry$f$ consists of a triangle of area $6$ above the axis on $[0,4]$ and a rectangle of area $2$ below the axis on $[4,6]$. Find $g(6)$ where $g(x)=\int_0^x f$.
1
Accumulate the first region
Area $6$, above the axis, so it contributes $+6$.
2
Accumulate the second
Area $2$, below the axis, so it contributes $-2$.
3
Add the signed contributions
$g(6)=6-2=4$.
4
Note $g(4)$ as well
$g(4)=6$ — the maximum, since $f$ turns negative after that.
5
The method
No antiderivative was used. Reading areas off a graph is the intended technique here.
EX 3
Concavity of $g$
one level upOn $(1,2)$ the graph of $f$ is increasing; on $(2,3)$ it is decreasing. What does that say about $g$?
1
Relate the derivatives
$g'=f$, so $g''=f'$.
2
Translate the first interval
$f$ increasing means $f'\gt 0$, so $g''\gt 0$ — $g$ is concave up on $(1,2)$.
3
Translate the second
$f$ decreasing gives $g''\lt 0$ — $g$ is concave down on $(2,3)$.
4
Identify the inflection
The concavity of $g$ switches at $x=2$, which is where $f$ has its maximum.
5
State the rule
An extremum of $f$ is an inflection point of $g$. Every feature shifts one derivative level.
EX 4
Negative accumulation
below the axis$h(x)=\displaystyle\int_{2}^{x}f(t)\,dt$ and $f\lt 0$ throughout $[0,2]$. Is $h(0)$ positive or negative?
1
Note the reversed limits
$h(0)=\displaystyle\int_{2}^{0}f(t)\,dt$, with the upper limit below the lower.
2
Reverse them
$=-\displaystyle\int_{0}^{2}f(t)\,dt$.
3
Sign the inner integral
$f\lt 0$ on $[0,2]$, so $\int_{0}^{2}f$ is negative.
4
Apply the outer minus
Negative of a negative, so $h(0)$ is positive.
5
The two sign sources
One from $f$ being below the axis, one from the reversed limits. Both must be tracked.
Watch out
The graph shown is $g'$, not $g$
Heights on the picture are slopes of the accumulation function. The largest value of $f$ is where $g$ is steepest, not where $g$ is largest.
$g(a)=0$ at the lower limit
Whatever $f$ does, the accumulation starts from zero at $x=a$. That single value anchors the whole curve.
Below the axis subtracts
$g$ decreases wherever $f$ is negative. A large area below the axis can drive $g$ negative.
Reversed limits flip the sign
If the upper limit is below the lower, swap them and prepend a minus — then read the area as usual.