< Previous Section | Home | Next Section > |
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 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
The statement, "The function f is an antiderivative of the function g" means that ...
The statement, "The function h is the exact rate of change function for the function j" means that ...
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 $r_f$ |
$f(x)=ax+b$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=ax^2+bx+c$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=ax^n+bx^5+cx+d$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=a(\sin(bx))^n+c$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=a\cos^n(bx)$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=a\tan^n(bx)$ |
$r_f(x)=$ | $g(x)=$ |
$f(x)=a\sec^n(bx)$ |
$r_f(x)=$ |
$g(x)=$ |
$f(x)=x^n\sin(x)$ |
$r_f(x)=$ | $g(x)=$ |
Example