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One interesting way of defining functions is when using a variable as the integration limit of an integral. This is particularly useful because it helps us build the bridge between derivatives and integrals, which is a central part of the Fundamental Theorem of Calculus. For this reason, it is important to study what is known as the accumulation function.
Definition of the Accumulation Function
You will usually face two types of integrals when studying Calculus: definite integrals and indefinite integrals. Usually, when finding a definite integral you get a number as an answer.
The value of the definite integral
\[ \int_0^2 x\,\mathrm{d}x\]
can be found by first finding its antiderivative, or indefinite integral (with no need of adding an integration constant), that is
\[ \int x\,\mathrm{d}x = \frac{1}{2}x^2,\]
and then using the evaluation part of the Fundamental Theorem of Calculus, so
\[ \int_0^2 x\,\mathrm{d}x = \left( \frac{1}{2}(2)^2\right) - \left( \frac{1}{2}(0)^2\right),\]
which will yield you a number as a result, that is
\[ \int_0^2 x\,\mathrm{d}x = 2.\]
However, you can leave one of the integration limits as a variable, turning the definite integral into a function. This function is known as an accumulation function.
Let \( f\) be a function that is integrable in the interval \( [a,b]\). An accumulation function is a function \( F(x) \), for
\(a<x<b\), such that
\[ F(x) = \int_a^x f(t)\,\mathrm{d}t,\]
or
\[ F(x) = \int_x^b f(t)\,\mathrm{d}t.\]
Basically, an accumulation function is a function obtained by solving a definite integral while leaving one of the integration limits as a variable.
Find the accumulation function
\[ F(x)=\int_0^x t\,\mathrm{d}t.\]
Solution:
To find an accumulation function you should treat the involved integral the same way you treat a definite integral, so begin by finding the antiderivative, that is
\[ \int t\,\mathrm{d}t = \frac{1}{2}t^2,\]
and then use the evaluation part of the Fundamental Theorem of Calculus, giving you
\[ \begin{align} \int_0^x t\,\mathrm{d}t &= \left( \frac{1}{2}(x)^2 \right) - \left( \frac{1}{2}(0)^2\right) \\ &= \frac{1}{2}x^2. \end{align}\]
This means that
\[ F(x)= \frac{1}{2}x^2.\]
In general, the accumulation function depends on the limits of integration, so having different integration limits should modify the accumulation function.
Find the accumulation function
\[ G(x)=\int_1^x t\,\mathrm{d}t.\]
Solution:
You already found the indefinite integral in the previous example, so
\[ \int t\,\mathrm{d}t = \frac{1}{2}t^2.\]
This time you have to use the evaluation part of the Fundamental Theorem of Calculus with different integration limits, doing so will give you
\[ \begin{align} \int_1^x t\,\mathrm{d}t &= \left( \frac{1}{2}(x)^2 \right) - \left( \frac{1}{2}(1)^2\right) \\ &= \frac{1}{2}x^2-\frac{1}{2}. \end{align}\]
This means that, this time, the accumulation function for these integration limits is
\[ G(x)= \frac{1}{2}x^2-\frac{1}{2}.\]
In the last examples, the accumulation functions coincide with antiderivatives of \(f(x)\), that is
\[ F'(x) = x,\]
and
\[ G'(x) = x.\]
This will not always be the case. Consider the next example.
Find the accumulation function
\[ H(x)=\int_x^1 t\,\mathrm{d}t.\]
Solution:
You can go straight to the evaluation part of the Fundamental Theorem of Calculus, this time with different integration limits, so
\[ \begin{align} \int_x^1 t\,\mathrm{d}t &= \left( \frac{1}{2}(1)^2 \right) - \left( \frac{1}{2}(x)^2\right) \\ &= \frac{1}{2}-\frac{1}{2}x^2. \end{align}\]
This means that this time, the accumulation function for these integration limits is
\[ H(x)= \frac{1}{2}-\frac{1}{2}x^2.\]
This time you can find out that
\[ H'(x) = -x,\]
so if the \(x\) variable is in the lower integration limit, the accumulation function will differ from an antiderivative by a sign!
The Accumulation Function Integral
You might have noticed that an accumulation function is defined by using two variables, yet, the answer is a function of just one variable. How is this possible?
To illustrate this process, consider the definite integral
\[ \int_1^2 x^2 \,\mathrm{d}x.\]
By solving the above definite integral, you will get a number as a result, that is
\[ \int_1^2 x^2 \,\mathrm{d}x = \frac{7}{3}.\]
Where did the \(x \) go? Remember that in definite integrals, the integration variable goes off, which means that you can use any variable you want inside a definite integral, so
\[ \int_1^2 t^2\,\mathrm{d}t\]
and
\[ \int_1^2 y^2\,\mathrm{d}y\]
will yield the same result as well. Since the most usual variable used in functions is \(x\), to avoid any confusion you need to use another variable in the definite integral, and usually, the letter \(t\) is chosen. This way the value of the definite integral
\[ \int_a^x f(t) \, \mathrm{d}t \]
depends on the value of \(x,\) so it is written as a function of \(x\), that is
\[ F(x) = \int_a^x f(t) \, \mathrm{d}t.\]
Graph of the Accumulation Function
One of the basic interpretations of definite integrals is that they give you a measure of the area below a curve in a given interval. In other words, the definite integral
\[ \int_a^b f(x)\,\mathrm{d}x\]
gives you the net signed area bound between \(x=a\), \( f(x)\), \( x=b\), and the \(x-\)axis.
Figure 1. Area bound between \(x=a,\) \( f(x),\) \( x=b,\) and the \(x-\)axis
Remember that a signed area is positive if it lies above the \(x-\)axis, and negative if it lies below the \(x-\)axis. The net signed area is obtained by subtracting any areas below the \(x-\)axis from the areas above the \(x-\)axis.
This means that an accumulation function is a function that gives you the net signed area below the curve \( f \) in terms of one of its integration limits.
If the variable is in the upper integration limit, then the accumulation function \( \int_a^x f(t)\,\mathrm{d}t \) relates to the signed area to the right of \(a\).
Figure 2. Signed area below \( f(t) \) to the right of \(a\)
If the variable is in the lower integration limit instead, then the accumulation function \( \int_x^b f(t)\,\mathrm{d}t \) relates to the signed area to the left of \(b\).
Figure 3. Signed area below \( f(t) \) to the left of \(b\)
Finding an Accumulation Function from a Graph
Sometimes you will be asked to find some values of an accumulation function given a graph rather than a mathematical expression. Here you can look at the most common requested tasks.
Finding the Value of the Accumulation Function
Suppose you are given the accumulation function
\[ g(x) = \int_{-2}^x f(t)\,\mathrm{d}t\]
and you are also given the graph of \( f \).
Figure 4. Graph of the function \( f\)
Even though you do not know the mathematical expression for \(f\), you can still find the accumulation function by looking at the graph and relating it to geometrical shapes! For example, you can find \(g(0)\) by noting that this will become
\[ g(0) = \int_{-2}^0 f(t)\,\mathrm{d}t,\]
which is the area between \( x=-2 \) and \( x=0\). This area is highlighted in the following figure.
Figure 5. Area below \( f(t) \) in the interval from \(-2 \) to \(0\)
Note that this area is the same as the area of a triangle with breadth \( 2 \) and height \(2,\) so its area is given by
\[ \begin{align} A &= \frac{1}{2}(2)(2) \\ &= 2, \end{align}\]
which gives you the value of the accumulation function as well, that is
\[ \begin{align} g(0) &= \int_{-2}^0 f(t) \mathrm{d}t \\ &= 2. \end{align} \]
Finding the Derivative of the Accumulation Function at a Point
You might also be asked to find the value of the derivative of an accumulation function under the same conditions. This task might seem difficult at first since you are not given a mathematical expression to differentiate.
Suppose that, for the same accumulation function, you are asked to find \( g'(1)\). You should begin by noting that the graphs of \( f\) and \( g' \) are the same! This is because the variable is in the upper integration limit of the accumulation function.
Since the goal is to find \( g'(1)\), you just need to find \( f(1) \), which you can do by looking at the \(f\) value when \( t=1\).
Figure 6. The function passes through the point (1,2)
This means that \(g'(1)=2\).
Finding the Second Derivative of the Accumulation Function at a Point
Following the same reasoning as the previous example, you can find the second derivative of \(g\) by looking at the first derivative of \(f,\) that is \( g''=f'\). You can use this fact to find the second derivative of \(g\) at a point.
Suppose that you need to find \( g''(1)\). This is the same as finding \( f'(1)\), so look at the slope of \( f \) at the point when \( t=1\). This might seem hard at first, but when you look at the graph you will find out that the function is constant between \(0\) and \(2\), so its slope is this interval is equal to \(0\).
Figure 7. The function is horizontal at \(t=1\), so its slope is equal to \(0\)
From here, you can conclude that \( f'(1) = 0,\) which means that \( g''(1)=0\).
Please note that you have to be careful when finding derivatives this way. If the function is not smooth at a point, then its derivative does not exist! This is the case of the function at \(t=0\). Note the edge of the graph as it switches from being linear to being constant.
Figure 8. The derivative of the function at \(t=0 \) does not exist
Formula of the Accumulation Function
To find an accumulation function you need to find the definite integral
\[ \int_a^x f(t)\,\mathrm{d}t,\]
or
\[ \int_x^b f(t)\,\mathrm{d}t,\]
which means that rather than a formula for the accumulation function, you need a method to solve the definite integral. Check out more in our Definite Integrals article!
Examples of the Accumulation Function
Calculus is all about practice. Finding accumulation functions will also help you practice solving definite integrals!
Find the accumulation function
\[ F(x) = \int_\pi^x \cos{t} \, \mathrm{d}t.\]
Solution:
In order to find an accumulation function, you should start by solving the indefinite integral. Since the integrand is the cosine function, you can solve the indefinite integral by using the fact that the derivative of the sine function is the cosine function. As usual, you do not require adding an integration constant, so
\[ \int \cos{t} \,\mathrm{d}t = \sin{t}.\]
Next, you need to evaluate the definite integral, which will give you
\[ \int_\pi^x \cos{t}\,\mathrm{d}t = \sin{x}-\sin{\pi}.\]
Since \( \sin{\pi}=0\), you can simplify the above expression and obtain the accumulation function, that is
\[ \begin{align} \int_\pi^x\cos{t}\,\mathrm{d}t &= \sin{x}-0 \\ &= \sin{x}. \end{align} \]
This means that the accumulation function is
\[ F(x) = \sin{x}.\]
The integration limits can also be special numbers, like \(e\).
Find the accumulation function
\[ G(x) = \int_e^x \frac{5}{t} \, \mathrm{d}t.\]
Solution:
This time you will need to solve the indefinite integral
\[ \int \frac{5}{t} \, \mathrm{d}t,\]
which is one of the Integrals Involving Logarithmic Functions. Doing so will give you
\[ \begin{align} \int \frac{5}{t} \, \mathrm{d}t &= 5 \int \frac{1}{t} \, \mathrm{d}t \\ &= 5\ln{t}. \end{align}\]
Next, evaluate the definite integral as usual, so
\[ \int_e^x \frac{5}{t} \, \mathrm{d}t = 5\ln{x}-5\ln{e}.\]
The natural logarithm of the number \( e \) is equal to \(1\). Knowing this you can simplify the above expression and get
\[ \int_e^x \frac{5}{t} \, \mathrm{d}t =5\ln{x}-5, \]
so the accumulation function is
\[ G(x) = 5\ln{x}-5.\]
What if the variable is in a lower integration limit? No problem! You just got to pay attention to its sign.
Find the accumulation function
\[ H(x) = \int_x^3 \left(\frac{1}{2}t-3\right) \, \mathrm{d}t.\]
Solution:
You can use the Power Rule to find the indefinite integral
\[ \int \left(\frac{1}{2}t-3\right) \, \mathrm{d}t, \]
so
\[ \int \left(\frac{1}{2}t-3\right) \, \mathrm{d}t = t^2-3t.\]
Knowing this, you can evaluate the definite integral, that is
\[ \begin{align} \int_x^3 \left( \frac{1}{2}t-3t\right) \, \mathrm{d}t &= \left((3)^2-3(3) \right) - \left((x)^2-3(x) \right) \\ &= 0-(x^2-3x) \\ &= -x^2+3x. \end{align}\]
This means that the accumulation function is
\[ H(x) = -x^2+3x.\]
It's time to tackle an accumulation function defined through a graph!
Let \( f \) be a continuous function defined in the interval \( [-3,4] \) by the following graph.
Figure 9. Graph of f
Let \( g\) be an accumulation function defined as
\[ g = \int_{-3}^x f(t)\,\mathrm{d}t.\]
- Find \( g(4) \).
- Find \( g'(0) \).
- Find \( g''(-1) \).
Solution:
a. Find \( g(4) \).
To find \( g(4) \) you will need to find the area of the whole curve. Begin by inspecting the geometrical shapes present in the graph of \(f\).
Figure 10. The area below \( f\) consists of two triangles, a square, an a quarter of a circle
From here, you can find the area of each geometric figure.
Figure 11. Dimensions of the figures present in the graph of \(f \)
Finally, to find the area you need to add all the above values while noting that the triangle in the interval \( [-3,2] \) goes below the \(t-\)axis, so this area has to be subtracted from the rest.
\[ \begin{align} A &= -\frac{1}{2}+2+4+\frac{1}{4}\pi(2)^2 \\ &= \frac{11}{2}+\pi. \end{align} \]
This means that the value of the accumulation function evaluated at \( 4 \) is
\[ g(4) = \frac{11}{2}+\pi\].
b. Find \( g'(0)\).
This point is straightforward, since the graphs of \( g'\) and \( f \) are the same, you just need to find the value of \(f \) at \(0\). This means that
\[ \begin{align} g'(0) &= f(0) \\ &= 2. \end{align}\]
c. Find \( g''(-1)\).
Since the second derivative of \(g\) equals the derivative of \(f\), you need to find the slope of a line tangent to \(f \) at the requested point, that is
\[ g''(-1)=f'(-1).\]
Note that \(f\) is a straight line in the interval from \(-3\) to \(0,\) so its slope is constant in this interval. You can find this slope by noting that as \(t\) increases by \(1\) in this portion of the graph, \( f(t)\) increases by \(1\) as well. This means that the slope is \(1\). From here you can conclude that
\[ \begin{align} g''(-1) &= f'(-1) \\ &= 1. \end{align}\]
The Accumulation Function - Key takeaways
- An accumulation function is a function \( F(x) \) such that\[ F(x) = \int_a^x f(t) \, \mathrm{d}t,\]or\[ F(x) = \int_x^b f(t) \, \mathrm{d}t,\]where \( f \) is a function that is integrable on the interval \( [a,b]\), and \( a<x<b\).
- An accumulation function is a function obtained by solving a definite integral while leaving one of the integration limits as a variable.
- Accumulation functions give you information about the net signed area of a function \( f \) in terms of one of its integration limits.
- Since the letter \( x\) is usually reserved for a variable, the integrand of an accumulation function usually does not use it as the integration variable, so \( t \) is commonly used instead.
- Accumulation functions are defined by means of solving a definite integral, so the function will depend on which function you are integrating, as well as the integration limits.
- You can also evaluate accumulation functions by looking at a graph. For this you need to keep in mind that if\[ F(x)= \int_a^x f(t)\,\mathrm{d}t,\]then\[ F'(x)=f(x),\]so you can relate both graphs.
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Frequently Asked Questions about Accumulation Function
What is the accumulation function formula?
To obtain an accumulation function you need to solve a definite integral while leaving one of its integration limits as a varaible.
How do you graph an accumulation function?
To graph an accumulation function you need to solve the definite integral involved. The resulting operation should be a function of x, which you can graph accordingly.
Is an integral an accumulation function?
Accumulation functions are defined by using definite integrals, so you can see a definite integral as an accumulation function that has already been evaluated. You can also see an indefinite integral as an accumulation function that has not been evaluated yet.
What are accumulation functions used for?
There are many applications for accumulation functions. These are used to model net changes, net signed areas, the work done by a force, population growths, and many more applications.
Is the accumulation function the Antiderivative?
Usually, the accumulation function can be seen as an antiderivative of the function in the integrand. However, in order for this to be true, the variable must be in the upper integration limit because if it is in the lower integration limit, the function will differ from the antiderivative by a sign.
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