Mathwords logoMathwords

accumulation function — Definition, Formula & Examples

An accumulation function gives the total accumulated area under a curve from a fixed starting point to a variable endpoint. It is written as F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt, where xx changes and the output is the net area collected so far.

Given an integrable function ff on an interval [a,b][a, b], the accumulation function is defined as F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt for x∈[a,b]x \in [a, b]. By the First Fundamental Theorem of Calculus, if ff is continuous on [a,b][a, b], then FF is differentiable on (a,b)(a, b) and F′(x)=f(x)F'(x) = f(x).

Key Formula

F(x)=∫axf(t) dtF(x) = \int_a^x f(t)\,dt
Where:
  • F(x)F(x) = The accumulated net area from $a$ to $x$
  • aa = The fixed lower limit of integration
  • xx = The variable upper limit of integration
  • f(t)f(t) = The integrand, evaluated using dummy variable $t$

How It Works

Think of the accumulation function as a running total. As you slide the upper limit xx to the right, you include more area under f(t)f(t), so F(x)F(x) grows (or shrinks, where ff is negative). At the starting point, F(a)=0F(a) = 0 because you have accumulated nothing yet. The key insight from the Fundamental Theorem of Calculus is that the rate of change of FF at any point equals the value of the original function: F′(x)=f(x)F'(x) = f(x). This means you can recover ff by differentiating the accumulation function.

Worked Example

Problem: Let f(t)=2tf(t) = 2t. Find the accumulation function F(x)=∫0x2t dtF(x) = \int_0^x 2t\,dt and evaluate F(3)F(3).
Step 1: Integrate 2t2t with respect to tt from 00 to xx.
F(x)=∫0x2t dt=[t2]0x=x2−0=x2F(x) = \int_0^x 2t\,dt = \left[t^2\right]_0^x = x^2 - 0 = x^2
Step 2: Substitute x=3x = 3 to find the accumulated area.
F(3)=32=9F(3) = 3^2 = 9
Step 3: Verify the derivative relationship: F′(x)=2x=f(x)F'(x) = 2x = f(x), confirming the Fundamental Theorem of Calculus.
F′(x)=ddx(x2)=2xF'(x) = \frac{d}{dx}\left(x^2\right) = 2x
Answer: F(x)=x2F(x) = x^2, and F(3)=9F(3) = 9. The net area under f(t)=2tf(t) = 2t from 00 to 33 is 99 square units.

Why It Matters

Accumulation functions are central to AP Calculus AB/BC Unit 6, where free-response questions frequently ask you to interpret ∫axf(t) dt\int_a^x f(t)\,dt as a quantity that changes over time—such as total distance traveled or total water in a tank. Understanding this concept also underpins the Fundamental Theorem of Calculus, which connects differentiation and integration.

Common Mistakes

Mistake: Using xx as both the upper limit and the variable inside the integral, writing ∫axf(x) dx\int_a^x f(x)\,dx.
Correction: The variable of integration must differ from the upper limit. Use a dummy variable like tt: ∫axf(t) dt\int_a^x f(t)\,dt. The upper limit xx is the input to FF, while tt is just a placeholder that disappears after integration.

Related Terms