Interpreting the Behavior of Accumulation Functions Involving Area
Introduced| English | Meaning |
|---|---|
| extrema/ekˈstremə/ | Maximum and minimum values of a function, either locally or over a specified domain. |
Reading $g$ from the graph of $f$
- If $g(x)=\int_a^x f(t)\,dt$, then by FTC Part 1, $g'=f$ — so $f$ is the derivative of $g$.
- That means everything from Unit 5 applies: use $f$ (as $g'$) to analyze $g$.
- The sign of $f$ tells you where $g$ increases and decreases.
- The zeros of $f$ (where it changes sign) locate the extrema of $g$.
When analyzing $g(x)=\int_a^x f\,dt$, the graph of $f$ acts as the graph of $g'$.
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FTC Part 1: $g'=f$.
Increase/decrease from the sign of $f$
- Where $f(x)>0$: $g$ is increasing (area is being added).
- Where $f(x)<0$: $g$ is decreasing (area is being subtracted).
- So look at the graph of $f$: above the axis → $g$ climbs; below → $g$ falls.
- $g$ is the running total of signed area, so its direction follows $f$'s sign.
An accumulation function adds area where f > 0 and subtracts it where f < 0
g rises where f is positive
y = f(t) (the integrand = g′)
While $f$ (the shaded height) is positive, the accumulated area $g$ increases; when $f$ turns negative, $g$ decreases.
For $g(x)=\int_a^x f(t)\,dt$, $g$ is increasing exactly where...
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$g'=f$, so $g$ increases where $f>0$.
Extrema of $g$ where $f$ changes sign
- $g$ has a local max where $f$ (its derivative) changes from positive to negative.
- $g$ has a local min where $f$ changes from negative to positive.
- These are just the First Derivative Test applied to $g$, using $f$ as $g'$.
- The extrema 极值 of the accumulation function sit at the $x$-axis crossings of $f$.
$g$ has a local maximum where $f$ changes from...
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$g'=f$ goes $+\to-$ → local max of $g$.
From the graph of $f$, you can read which features of $g=\int_a^x f\,dt$?
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Increase/decrease, extrema, concavity — all from $f$. Exact values need the actual area.
Concavity of $g$ from the slope of $f$
- Since $g'=f$, we get $g''=f'$ — so the concavity of $g$ follows the slope of $f$.
- Where $f$ is increasing, $g$ is concave up; where $f$ is decreasing, $g$ is concave down.
- An inflection point of $g$ sits where $f$ has a local max or min (where $f'$ changes sign).
- One graph of $f$ tells you increase/decrease and concavity of $g$.
Since $g''=f'$, $g$ is concave up where $f$ is ____.
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$g''=f'>0$ where $f$ rises.
A large positive value of $f$ at $x$ means that $g$ is...
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$f=g'$ is the rate; a large $f$ means $g$ rises fast, not that $g$ is big.
To analyze $g(x)=\int_a^x f\,dt$, treat the graph of $f$ as the graph of $g'$. Don't read $f$'s height as $g$'s height — read $f$'s sign for $g$'s increase/decrease, and $f$'s sign changes for $g$'s extrema. A big positive $f$ means $g$ is rising fast, not that $g$ is large.
$g(x)=\int_0^x f(t)\,dt$, where $f>0$ on $(0,3)$, $f(3)=0$, and $f<0$ on $(3,5)$.
- On $(0,3)$: $f>0$ → $g$ increasing. On $(3,5)$: $f<0$ → $g$ decreasing.
- At $x=3$: $f$ changes $+\to-$ → $g$ has a local maximum.
- So $g$ climbs to a peak at $x=3$, then falls.
For $g(x)=\int_a^x f(t)\,dt$, use $g'=f$: $g$ increases where $f>0$, decreases where $f<0$, and has extrema where $f$ changes sign. Since $g''=f'$, $g$ is concave up where $f$ is increasing. Read the graph of $f$ as the graph of $g'$.
Write from a graph of the integrand
- f is positive on (0, 3) and negative on (3, 5).
- g is the integral of f from 0 to x.
Where does g increase, and where does g have an extremum?
g prime is f, so the sign of f is the direction of g.
f changes from positive to negative at 3. For g, the point 3 is what?
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g prime changing from positive to negative is a maximum of g.