## Fundamental Theorem of Calculus

1. Review. So we introduced some technique to figure out the area between a curve $y=f(x)$ and the line $y=0$. This was mildly complicated involving sums, partitions, and other exotic names.

The question: can’t we do something simpler?

The answer: yes…kind of. We can use a function’s antiderivative to figure out the area under a curve between two points.

Fundamental Theorem of Calculus. Let $f$ be a continuous function on $[a,b]$. Then
(1)$\displaystyle F(x)=\int^{x}_{a}f(t)\,\mathrm{d}t$
is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover
(2)$\displaystyle F'(x)=f(x)$

Corollary. If $f$ is continuous on $[a,b]$, then
(3)$\displaystyle \int^{b}_{a}f(x)\,\mathrm{d}x=F(b)-F(a)$
where $F(x)$ is the antiderivative for $f(x)$.

Notational Warning: often times in practice we will write
(4)$\displaystyle \left.\int^{b}_{a}f(x)\,\mathrm{d}x=F(x)\right|^{b}_{a}$
with the understanding this vertical bar is shorthand for
(5)$\displaystyle \left.\phantom{\int}F(x)\right|^{b}_{a}=F(b)-F(a)$
(End of notational warning!)

Remark. Sometimes we refer to this corollary as the fundamental theorem of calculus. It’s all fairly loose. The intuitive notion underpinning the fundamental theorem of calculus is derivatives and integrals are “dual”, in some sense, to each other.

In a sense we can think of this as giving us a method calculating the area under a curve using antiderivatives. An incorrect but sometimes helpful intuition is to take (3) as a definition for definite integrals. The correct and rigorous approach would be to return to Riemann sums and prove everything with them. We will not do that (at least, not now…when we get to real analysis, we will prove everything rigorously).

Example 1. Consider $\displaystyle\int^{2\pi}_{0}\sin(x)\,\mathrm{d}x$. We should recall that
(6)$\displaystyle\frac{\mathrm{d}\cos(x)}{\mathrm{d}x}=-\sin(x)$
which tells us the antiderivative for sine is negative cosine. Thus
(7)$\displaystyle\left.\int^{2\pi}_{0}\sin(x)\,\mathrm{d}x=-\cos(x)\right|^{2\pi}_{0}=-[\cos(2\pi)-\cos(0)]=0$
This is a trivial answer, but not-as-trivial calculation.

Example 2. Consider $\displaystyle\int^{2}_{0}x^{5}\,\mathrm{d}x$. We see that
(8)$\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}x^{6}=6x^{5}$
thus
(9)$\displaystyle\frac{1}{6}\frac{\mathrm{d}}{\mathrm{d}x}x^{6}=x^{5}$
So the antiderivative for $f(x)=x^{5}$ is $F(x)=x^{6}/6$. Thus the integral becomes
(10)$\displaystyle\int^{2}_{0}x^{5}\,\mathrm{d}x = F(2)-F(0)=\frac{64}{6}$
This concludes our second example.

2. What Does the Theorem Mean? Well, if we examine the function $f(x)$ which is differentiable on the interval $[a,b]$, then we can note the fundamental theorem of calculus enables us to write
(11)$\displaystyle\int^{b}_{a}\frac{\mathrm{d}f(x)}{\mathrm{d}x}\,\mathrm{d}x = f(b)-f(a)$
The intuition is that we have a “differential”
(12)$\displaystyle\mathrm{d}f =\frac{\mathrm{d}f(x)}{\mathrm{d}x}\,\mathrm{d}x$
and we integrate it over the domain $[x=a, x=b]$.