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# Section 9.1The 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 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 varies accumulate. When a quantity accumulates in relation to variations in another quantity, it varies 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 $f$:

• The net accumulation function $\int_a^x r_f(t)dt$ is an antiderivative of $r_f$.
• All antiderivatives of $r_f$ differ by at most a constant. Put another way, if $H$ is also an antiderivative of $r_f$, then $$\int_a^x r_f(t)dt=H(x)+C\text{, for some constant }C.$$It turns out, as we will see below, that $C=-H(a)$, so that$$\int_a^x r_f(t)dt=H(x)-H(a).$$

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