Section 9.1
Rate of Change Functions, Accumulation Functions, and Antiderivatives

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. See Section 6.2 for a full discussion of the Fundamental Theorem of Calculus.

The Fundamental Theorem of Calculus (FTC) is fundamental to calculus because it does two things:

• It relates the ideas of exact rate of change function, exact accumulation function, and antiderivative.
• It provides a way, in principle, to efficiently compute values of exact accumulations functions.

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

The Fundamental Theorem of Calculus is:

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
$$\color{red}{\text{(Eq. 9.1.1)}}\qquad\int_2^x 3e^{-2t}dt=g(x)-g(2).$$

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.

We can make progress toward computing this integral directly 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 Equation 9.1.1, 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 you remembering accumulation functions that produce rate of change functions having specific forms.

For the future: Gabriel's Horn

Exercise Set 9.1

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. Do this for each of a - k below:
• In (i), write the rate of change function $r_f$ in closed form for the given accumulation function in closed form
• In (ii), write the principle antiderivative for the given closed form rate of change function.

1. (i)   $f(x)=-20x$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=7$

2. (i)   $f(x)=3\cos(x)$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=-6\sin(x)$

3. (i)   $f(x)=x^3$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=8x^7$

4. (i)   $f(x)=9^x$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=\ln(4)4^x$

5. (i)   $f(x)=6x^{-10}$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=-15x^{-6}$

6. (i)   $f(x)=-2e^x$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=\dfrac{13}{8}e^x$

7. (i)   $f(x)=\sin^8(x)$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=5\left(\sin^4(x)\right)\cos(x)$

8. (i)   $f(x)=\dfrac{-2}{9}\ln(x)$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=\dfrac{21}{x}$

9. (i)   $f(x)=x^2$	$r_f(x)=?$
   (ii) $g(x)=\, ?$	$r_g(x)=x$

10. (i)   $f(x)=\sqrt{\tan(x)}$	$r_f(x)=?$
    (ii) $g(x)=\, ?$	$r_g(x)=\dfrac{\sec^2\sqrt x}{2\sqrt x}$
    (iii) $h(x)=\, ?$	$r_h(x)=\dfrac{\cos(x)}{2\sqrt {\sin(x)}}$

11. (i)   $f(x)=e^{\cos(x)}$	$r_f(x)=?$
    (ii) $g(x)=\, ?$	$r_g(x)=5x^4 e^{x^5}$
    (iii) $h(x)=\, ?$	$r_h(x)=e^{6x}$

3. Complete the following table.
   
   Don't forget that the rate of change of a constant function is 0. For example, GC reports that
   $$\dfrac{d}{dx}\left(a(\cos(bx))^n+c\right)=a\left(\frac{dn}{dx}\ln\cos bx-\frac{n\left(b+x\dfrac{db}{dx}\right)\sin bx}{\cos bx}\right)\cos^nbx+\frac{da}{dx}\cos^n bx+\frac{dc}{dx}.$$
   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
   $$\frac{d}{dx}\left(a(\cos(bx))^n+c\right)=-a\cdot b\cdot n\cdot\sin(bx)(\cos(bx))^{n-1}.$$
   

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 $r_f$
$f(x)=ax+b$	$r_f(x)=$	$g(x)=$
$f(x)=ax^2+bx+c$	$r_f(x)=$	$g(x)=$
$f(x)=ax^n+bx^5+cx+d$	$r_f(x)=$	$g(x)=$
$f(x)=a(\sin(bx))^n+c$	$r_f(x)=$	$g(x)=$
$f(x)=a\cos^n(bx)$	$r_f(x)=$	$g(x)=$
$f(x)=a\tan^n(bx)$	$r_f(x)=$	$g(x)=$
$f(x)=a\sec^n(bx)$	$r_f(x)=$	$g(x)=$
$f(x)=x^n\sin(x)$	$r_f(x)=$	$g(x)=$