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Section 9.1

Rate of Change Functions, Accumulation Functions, and Antiderivatives

In Section 8.5 we modeled a situation where we were given acceleration functions for two moving objects.

We defined velocity functions in terms of acceleration functions and we defined distance functions in terms of velocity functions. The functions we defined were:$$\begin{align}a_1(t)&=0.7t+\sin\left(\frac{3\pi t}{5}\right)\\[1ex]
a_2(t)&=3-3\cos\left(\frac{2\pi t}{15}\right)\\[1ex]
v_1(t)&=\int_0^t a_1(u)du\\[1ex]
v_2(t)&=\int_0^t a_2(u)du\\[1ex]
s_1(t)&=\int_0^t v_1(u)du\\[1ex] s_2(t)&=\int_0^t v_2(u)du\end{align}$$

Figure 9.1.1 repeats what you saw when graphing $y=s_1(x)$ and $y=s_2(x)$ in GC. It shows that GC takes a long time to display these graphs. Why?

Figure 9.1.1. GC takes a LONG time to construct the graphs of $s_1$ and $s_2$.

Think about that question ("Why?") before reading on.

For GC to calculate a value of, say, $s_1(4.1)$ it must calculate bits of distance $v_1(t)dt$ for many values of $t$ from 0 to 4.1. But to calculate a value of $v_1(t)dt$ for any value of $t$, GC must also calculate many values of $a_1(u)du$ for values of $u$ ranging from 0 to $t$.

GC must repeat this double-variation process to plot the points associated with all (actually, a large sample) of the values of $x$ from $x=0$ to $x=8$. So for GC to calculate values of $v_1(t)$, GC repeatedly summed an ever increasing number of calculations of $a_1(u)du$ as $u$ varied from 0 to the value of $t$. To calculate values of $s_1(x)$, GC repeatedly summed an ever increasing number of calculations of $v_1(t)$ as the value of $t$ varies from 0 to $x$. All this must happen as $x$ varies from 0 to 8.

Figure 9.1.2 illustrates the above explanation graphically. Suppose that GC computes 2000 incremental values to compute the value of $s_1(4.1)$. Suppose further that GC computes 1500 incremental values to compute each incremental value in the computation of $s_1(4.1)$, and that GC computes 3000 incremental values to compute each of those incremental values. This would explode into $2000\cdot 1500\cdot 3000=9\cdot 10^9$, or 9 billion incremental values to compute the value of $s_1(4.1)$. And GC does this for each value of $x$ that it samples as it varies values of $x$ from 0 to 8.


Figure 9.1.2. GC computes an enormous number of incremental values to compute the value of $s_1(4.1)$.

The fact that GC does so many computations to compute values of $s_1$ and $s_2$, and hence takes so long to generate the graphs in Figure 9.1.1, demonstrates why closed form definitions of accumulation functions can be important. Closed form definitions of $s_1$ and $s_2$ would dramatically reduce the number of computations GC must do.

The closed form definition of $s_1$, renamed $S$, is given below.
$$S(t) =\frac{0.7t^3}{6}+ \frac{-25\sin\left(\dfrac{3\pi t}{5}\right)}{9\pi^2}+\frac{5t}{3\pi}$$
We encourage you to enter the open form definitions of $s_1$ and the closed form definition of $S$ into GC and graph them. This should convince you that the closed form definition of $s_1$ generates the same graph as does the open-form definition of $s_1$, and that GC generates the graph much faster.

The Fundamental Theorem of Calculus

The following discussion of the Fundamental Theorem of Calculus (FTC) relies heavily on you having internalized the terms and meanings reviewed in Section 8.0.

The FTC is fundamental to calculus because it does two things:

It does this by drawing on the fact that rate of change and accumulation are two sides of a coin. When a quantity changes at some rate, the changes accumulate. When a quantity accumulates in relation to changes in another quantity, it changes at some rate with respect to the other quantity.

The FTC's statement is this: Given that $r_f$ is a rate of change function for an accumulation function $f$:

Example

Suppose that values of $r_g(x)=3e^{-2x}$ are the exact rate of change of a function $g$ (which we do not know) in relation to all values of $x$. We know that

$$\int_2^x 3e^{-2t}dt$$

gives the exact net accumulation of g over the interval from $t=2$ to $t=x$ for any value of $x$.

We also know that any value $g(x)$ is the accumulation of $g$ up to $x=2$ plus the accumulation of $g$ from 2 to $x$. We say this symbolically as

$$g(x)=g(2)+\int_2^x 3e^{-2t}dt.$$

Although we do not know the value of $g(2)$ (because we do not know the function $g$), we can say that
$$(*)\qquad\int_2^x 3e^{-2t}dt=g(x)-g(2).$$
However, without a closed-form definition of $g$ we cannot compute values $g(x)$ directly! Without a closed form definition of $g$, the best we can do is to approximate values of $g(x)$ numerically for any value of $x$. GC does this (creates a numerical approximation) when it evaluates $\int_2^x 3e^{-2t}dt$ for any value of $x$.

However, we can make progress by making a connection between the integral and the notion of antiderivative.

By definition of $g$, $r_g(x)=3e^{-2x}$ is the exact rate of change function for $g$, so $g$ is an antiderivative of $r_g(x)=3e^{-2x}$.

But $H(x)=\frac{3}{-2}e^{-2x}$ is also an antiderivative of $r_g(x)=3e^{-2x}$, because $$\frac{d}{dx}\frac{3}{-2}e^{-2x}=3e^{-2x}.$$
Since all antiderivatives of $r_g$ differ by at most a constant, $g(x)=H(x)+C$ for some constant $C$. We therefore have $g(2)=H(2)+C$.

By (*), we now have
$$\begin{align}\int_2^x 3e^{-2t}dt
&=g(x)-g(2)\\[1ex]
&=\left(H(x)+C)\right)-\left(H(2)+C\right)\\[1ex]
&=H(x)-H(2)\\[1ex]
&=\left(\frac{3}{-2}e^{-2x}\right)-\left(\frac{3}{-2}e^{-2\cdot 2}\right)\end{align},$$
We now have a way to compute values of $\int_2^x 3e^{-2t}dt$ directly for any value of $x$ because we have represented values of $\int_2^x 3e^{-2t}dt$ in closed form!

This is what the FTC does for us. It allows us to represent integrals in closed form whenever we know a closed form antiderivative of the integral's rate of change function. Where do antiderivatives come from? They come from us noting the accumulation functions that produce rate of change functions of various forms.

(Gabrielle's Horn)

Exercise Set 9.1 (in progress)

  1. Complete the following statements regarding the relationship between two functions (include the stem in your completion):
    1. The statement, "The function f is an antiderivative of the function g" means that ...

    2. The statement, "The function h is the exact rate of change function for the function j" means that ...

  2. Complete the following table. (Note: "Principal antiderivative" means an antiderivative without a constant.) Use GC to check yourself or when your memory fails you.

    Also, don't forget that the rate of change of a constant function is 0. For example, GC reports that
    $$\frac{d}{dx}\left(a(\cos(bx))^n+c\right)=a\left(\frac{dn}{dx}\ln\cos bx-\frac{n\left(b+x\dfrac{db}{dx}\right)\sin bx}{\cos bx}\right)\cos^nbx+\frac{da}{dx}\cos^n bx+\frac{dc}{dx}.$$
    However, $a, b, c,\text{ and }n$ are constants. So $\dfrac{da}{dx}, \dfrac{db}{dx}, \dfrac{dc}{dx},\text{ and } \dfrac{dn}{dx}$ are all 0. Therefore, GC's complicated expression reduces to
    $$\frac{d}{dx}\left(a(\cos(bx))^n+c\right)=-abn\sin(bx)(\cos(bx))^{n-1}.$$

    The accumulation function f defined as
    has the exact rate of change function $r_f$ defined as
    which means that the function g defined as (below) is the principal antiderivative of $r_f$

    $f(x)=ax+b$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=ax^2+bx+c$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=ax^n+bx^5+cx+d$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=a(\sin(bx))^n+c$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=a\cos^n(bx)$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=a\tan^n(bx)$
     
    $r_f(x)=$ $g(x)=$

    $f(x)=a\sec^n(bx)$
     

    $r_f(x)=$
     

    $g(x)=$
     

    $f(x)=x^n\sin(x)$
     
    $r_f(x)=$ $g(x)=$
  3. Example

    1. Example