What Is an Accumulation Function?
An accumulation function describes how much of some quantity has accumulated from a starting point up to a variable point. In calculus, we often represent that accumulated amount with a definite integral.
For example, suppose a particle moves with velocity . The amount of displacement accumulated from time to time is represented by an integral:
Notice something important: the upper limit is now a variable, \(x\). That means the integral itself defines a function of \(x\).
We can write that function as:
This is called an accumulation function. As \(x\) changes, the amount accumulated under the graph of \(f\) changes.
The Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus gives us a remarkably simple way to differentiate an accumulation function.
In words: if a function is defined as the integral of another function from a constant to \(x\), its derivative is simply the integrand evaluated at \(x\).
This is one of the most important connections in calculus. Integration accumulates change, while differentiation gives the instantaneous rate of that accumulation.
Why Does This Work?
Suppose
Imagine increasing \(x\) by a very small amount, \(\Delta x\). The additional area accumulated is approximately a rectangle whose:
- width is , and
- height is approximately .
So the change in the accumulated area is approximately:
Dividing by gives:
As , this approximation becomes exact:
So the Second Fundamental Theorem says something intuitive: the rate at which accumulated area is increasing is the height of the function currently being accumulated.
The Variable Is Usually Not the Same Letter
You will often see the accumulation function written as:
Why is the variable inside the integral written as \(t\) instead of \(x\)? Because \(t\) is only a dummy variable of integration.
The value of the integral depends on the upper limit \(x\), not on the name of the variable used inside the integral.
For example, all three expressions below represent the same function:
The variable \(t\), \(u\), or \(s\) is just a placeholder.
A Step-by-Step Method
- Look for an integral that has a variable upper limit. The basic pattern is .
- Identify the integrand. This is the function inside the integral, .
- Replace the integral with its derivative. By the Second Fundamental Theorem, .
- Evaluate the integrand at the variable upper limit. The result is the derivative of the accumulation function.
Worked Examples
Let
Find .
Let
Find .
Suppose
Find .
This is especially important on the AP exam because you do not need to find an antiderivative of .
Now consider an accumulation function whose upper limit is not simply \(x\):
Find .
This is an important extension of the theorem: when the upper limit is a function of \(x\), the chain rule must also be applied.
Suppose
Find .
Accumulation Functions and Motion
Accumulation functions are especially useful in AP Calculus because they connect directly to particle motion.
If a particle has velocity , then its displacement from time to time is:
Define
Then the Second Fundamental Theorem immediately tells us:
That makes sense physically: the derivative of accumulated displacement is velocity.
Likewise, if an accumulation function represents the total amount of water entering a tank, then its derivative represents the instantaneous rate at which water is entering the tank.
Reading an Accumulation Function from a Graph
AP Calculus questions frequently give you a graph of and define an accumulation function such as:
You can immediately conclude:
This gives several useful conclusions without ever evaluating the integral.
- If , then , so \(F\) is increasing.
- If , then , so \(F\) is decreasing.
- If , then \(F'(x)=0\), so \(F\) has a possible local maximum or minimum.
In other words, the graph of \(f\) tells you the slope of the graph of \(F\).
Using the Second Derivative
Once you know that
you can differentiate again:
This lets you determine the concavity of the accumulation function by examining whether \(f\) is increasing or decreasing.
- If \(f\) is increasing, then \(f'(x)>0\), so \(F''(x)>0\) and \(F\) is concave up.
- If \(f\) is decreasing, then \(f'(x)<0\), so \(F''(x)<0\) and \(F\) is concave down.
Common Mistakes
If the question asks for the derivative of an accumulation function, you usually do not need to integrate first. For example, the derivative of \(\int_0^x e^{-t^2}\,dt\) is simply \(e^{-x^2}\), even though \(e^{-x^2}\) does not have an elementary antiderivative.
If the upper limit is something like \(x^2\), \(\sin x\), or \(3x-1\), you must multiply by the derivative of that upper limit. For example, \(\frac{d}{dx}\int_0^{x^2}f(t)\,dt=f(x^2)\cdot2x\).
When the variable appears in the lower limit, the derivative has a negative sign. For example, \(\frac{d}{dx}\int_x^a f(t)\,dt=-f(x)\).
The theorem says that the derivative of the accumulation function is the integrand evaluated at the variable limit. You are not differentiating the entire integral symbol-by-symbol.
Practice Problems
Let \[ F(x)=\int_2^x (t^3-4t)\,dt. \] Find \(F'(x)\).
Show answer
Let \[ G(x)=\int_0^x \cos(t^2)\,dt. \] Find \(G'(x)\).
Show answer
Let \[ H(x)=\int_1^{3x+2}\frac{1}{1+t^2}\,dt. \] Find \(H'(x)\).
Show answer
Let \[ P(x)=\int_x^5 \sqrt{t+1}\,dt. \] Find \(P'(x)\).
Show answer
Let \[ A(x)=\int_0^{x^2}\left(t^4+1\right)\,dt. \] Find \(A'(x)\).
Show answer
Suppose \[ F(x)=\int_1^x f(t)\,dt \] and the graph of \(f\) is below the \(x\)-axis on an interval. What can you conclude about \(F\) on that interval?
Show answer
Test yourself: AP Calculus practice exams
Ready to check what you know? Work an AP-style practice exam, then open the full solutions.
Want all 1,114 practice problems?
This tutorial covers one topic. Calc Pass covers all of AP Calculus AB & BC: full review, every practice question, and step-by-step solutions to each one.
Get the Calc Pass App
Scan to download and practice on the go, with instant scoring.
Vol. 1: 1,114 Questions
The complete practice question bank, organized by AB & BC topic.
View on Amazon