An accumulation function measures the total amount built up from a changing rate. In calculus, this is usually written as an integral whose upper limit is a variable, such as A(x) = ∫_a^x f(t) dt. As x moves to the right, the function adds signed area under the graph of f(t).
This idea matters because many real quantities, like distance, total charge, and accumulated profit, come from adding up rates over time.
The graph of an accumulation function is connected to the original function in a precise way. If A(x) = ∫_a^x f(t) dt, then the Fundamental Theorem of Calculus says A'(x) = f(x). This means the original function gives the slope of the accumulation function at each x-value.
Positive f(x) makes A(x) increase, negative f(x) makes A(x) decrease, and zeros of f(x) can mark local maxima or minima of A(x).
Key Facts
- An accumulation function has the form A(x) = ∫_a^x f(t) dt.
- The lower limit a sets the starting value, so A(a) = 0.
- The Fundamental Theorem of Calculus gives d/dx [∫_a^x f(t) dt] = f(x).
- If f(x) > 0, then A(x) is increasing; if f(x) < 0, then A(x) is decreasing.
- The value A(b) = ∫_a^b f(t) dt is the signed area from t = a to t = b.
- For G(x) = ∫_a^x f(t) dt, concavity depends on f'(x), so G''(x) = f'(x).
Vocabulary
- Accumulation function
- A function defined by an integral with a variable limit that gives the total signed area accumulated from a starting point.
- Variable upper limit
- The input value x in an integral such as ∫_a^x f(t) dt that controls how much area is included.
- Signed area
- Area counted as positive when the graph is above the axis and negative when the graph is below the axis.
- Fundamental Theorem of Calculus
- The theorem that connects derivatives and integrals by showing that differentiating an accumulation function recovers the original rate function.
- Rate of change
- A quantity that describes how fast another quantity is increasing or decreasing, often represented by the derivative.
Common Mistakes to Avoid
- Treating ∫_a^x f(t) dt as f(x), which is wrong because the integral gives accumulated area while f(x) gives the instantaneous height or rate.
- Ignoring negative area, which is wrong because regions below the x-axis subtract from the accumulation instead of adding positive area.
- Forgetting that A(a) = 0, which is wrong because the integral from a to a covers no interval and therefore accumulates no area.
- Assuming the accumulation graph has the same shape as f, which is wrong because f gives the slope of the accumulation graph, not its height.
Practice Questions
- 1 Let A(x) = ∫_0^x 3t^2 dt. Find A(2) and A'(2).
- 2 Let B(x) = ∫_1^x (4t - 2) dt. Find B(3), then state whether B is increasing or decreasing at x = 3.
- 3 Suppose C(x) = ∫_0^x f(t) dt and the graph of f is positive on 0 < x < 2, negative on 2 < x < 5, and positive again on 5 < x < 7. Describe where C is increasing and decreasing, and explain what could happen at x = 2 and x = 5.