1 Accumulation functions

While the definite integral computes a signed area, which is a fixed number, there is a way to turn it into a function.

One thing that you might notice is that an accumulation function seems to have two variables: \(x\) and \(t\). Let’s see if we can explain this. Consider the following graph:

An accumulation function \(F\) measures the signed area in the region \([a,x]\) between \(f\) and the \(t\)-axis. Hence \(t\) is playing the role of a “place-holder” that allows us to evaluate \(f\). On the other hand, \(x\) is the specific number that we are using to bound the region that will determine the area between \(f\) and the \(t\)-axis, and hence the value of \(F\).

Consider the following accumulation function for \(f(x) = x^3\).

\[ F(x) = \int _{-1}^x t^3 dt. \]

Considering the interval \([-1,1]\), where is \(F\) increasing? Where is \(F\) decreasing? When does \(F\) have local extrema?

Below we can see a graph of \(f\) along with the signed area measured by the accumulation function.

The accumulation function starts off at zero, and then as \(x\) grows, \(F\) is increasingdecreasing as the function accumulates negatively signed area.

However when \(x>0\), \(F\) starts to accumulate positively signed area, and hence is increasingdecreasing. Thus \(F\) is increasing on \((0,1)\), decreasing on \((-1,0)\) and hence has a local minimum at \((0,0)\).

Working with the accumulation function leads us to a question, what is

when \(x< a\)? The general convention is that

With this in mind, let’s consider one more example.

Consider the following accumulation function for \(f(x) = x^3\).

\[ F(x) = \int _{-1}^x t^3 dt. \]

Where is \(F\) increasing? Where is \(F\) decreasing? When does \(F\) have local extrema?

From our previous example, we know that \(F\) is increasing on \((0,1)\). Since \(f\) continues to be positive at \(t=1\) and beyond, \(F\) is increasingdecreasing on \((0,\infty )\). On the other hand, we know from our previous example that \(F\) is increasingdecreasing on \((-1,0)\).

For values to the left of \(t=-1\), \(F\) is still decreasing, as less and less positively signed area is accumulated. Hence \(F\) is increasing on \((0,\infty )\), decreasing on \((-\infty ,0)\) and hence has an absolute minimum at \((0,0)\).

The key point to take from these examples is that an accumulation function

is increasing precisely when \(f\) is positive and is decreasing precisely when \(f\) is negative. In short, it seems that \(f\) is behaving in a similar fashion to \(F'\).

2 The First Fundamental Theorem of Calculus

Let \(f\) be a continuous function on the real numbers and consider

From our previous work we know that \(F\) is increasing when \(f\) is positive and \(F\) is decreasing when \(f\) is negative. Moreover, with careful observation, we can even see that \(F\) is concave up when \(f'\) is positive and that \(F\) is concave down when \(f'\) is negative. Thinking about what we have learned about the relationship of a function to its first and second derivatives, it is not too hard to guess that there must be a connection between \(F'\) and the function \(f\). This is a good guess, check out our next theorem:

The First Fundamental Theorem of Calculus says that an accumulation function of \(f\) is an antiderivative of \(f\). Another way of saying this is:

This could be read as:

The rate that accumulated area under a curve grows is described identically by that curve.

Now that we are working with accumulation functions, let’s see what happens when we compose them with other functions.

Let’s practice this once more.

Given the graph of a function \(f\) on the interval \([-1,5]\), sketch the graph of the accumulation function

\[ F(x) = \int _{-1}^x f(t) dt, \hspace {0.2in} -1\le x\le 5. \]

First, we evaluate \(F\) at some significant points.

\[ F(-1) = \int _{-1}^{-1} f(t) dt=\answer [given]{0}. \]

\[ F(0) = \int _{-1}^{0} f(t) dt=\answer [given]{-1.5}. \]

\[ F(1) = \int _{-1}^{1} f(t) dt=\answer [given]{-2}. \]

\[ F(3) = \int _{-1}^{3} f(t) dt=\answer [given]{0}. \]

\[ F(4) = \int _{-1}^{4} f(t) dt=\answer [given]{2}. \]

\[ F(5) = \int _{-1}^{5} f(t) dt=\answer [given]{4}. \]

Since \(F'(x)=f(x)\), it follows that the function \(F\) is increasing on the interval

\[ \left (\answer [given]{1},\answer [given]{5}\right ) \]

and decreasing on the interval

\[ \left (\answer [given]{-1},\answer [given]{1}\right ). \]

Since the function \(f\), the derivative of \(F\), is increasing on \((-1,3)\), \(F\) is concave up on the interval

\[ \left (\answer [given]{-1},\answer [given]{3}\right ). \]

Since \(f\) is constant on the interval \([3,5]\), \(F\) is linear on \([3,5]\).

The function \(F\) has a local and global minimum at

\[ x=\answer [given]{1}, \]

and the global maximum at

\[ x=\answer [given]{5}. \]

Now we are ready to sketch the graph of \(F\), on the same set of axes as the graph of \(f\).

What if the function \(f\) is the velocity function for an object moving along a straight line, i.e. \(f(t)=v(t)\), \(a\le t\le b\)?

What is the meaning of an accumulation function in that case?

We use different variables in that case, since we want \(F\) to be a function of time, \(t\). With this adjustment, we define the accumulation function \(F\) as follows.

Since the integral

gives the displacement of the object on the time interval \([a,t]\), it follows that

where \(s(t)\) gives the position of the object at the time \(t\). If we differentiate this equation with respect to \(t\), we get that

Since \(s'(t)=v(t)\), we have that

This is the the First Fundamental Theorem of Calculus!