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.

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 $$\int _a^x f(t) dt$$ when $x< a$? The general convention is that $$\int _a^b f(t) dt = -\int _b^a f(t) dt.$$ 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 First Fundamental Theorem of Calculus

Let $f$ be a continuous function on the real numbers and consider $$F(x) = \int _a^x f(t) dt.$$ 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:

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. $$F(t)=\int _a^t v(z) dz.$$ Since the integral $$\int _a^t v(z) dz$$ gives the displacement of the object on the time interval $[a,t]$, it follows that $$s(t)-s(a)= \int _a^t v(z) dz,$$ 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 $$s'(t)=\frac {d}{dt} \int _a^t v(z) dz.$$ Since $s'(t)=v(t)$, we have that $$v(t)=\frac {d}{dt} \int _a^t v(z) dz.$$ This is the the First Fundamental Theorem of Calculus!