Fundamental Theorem of Calculus

Differentiation and Integration are actually inverse operations!

First, we prove that integration yields continuous functions.

\begin{proof} Let

$$A(x)={\int }_a^{x}f(t)dt.$$

Then, $A$ is continuous iff ${\lim }_{h\to 0}A(c+h)=A(c)$. By definition,

$$A(c+h)-A(c)=\{\begin{array}{ll}{\int }_c^{c+h}f(t)dt& \text{if }h\ge 0\\ {\int }_{c+h}^{c}f(t)dt& \text{if }h<0.\end{array}$$

Since $f$ is bounded and $h\to 0$, both tend to $0$. \end{proof}

Now, we have

The First Fundamental Theorem of Calculus §

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“First Fundamental Theorem of Calculus”,“label”:“first-fundamental-theorem-of-calculus”,“_index”:0}] Theorem 1 (First Fundamental Theorem of Calculus). Let $f$ be integrable on $[a,b]$ and $A$ its integral. If $c\in (a,b)$ and $f$ is continuous at $c$, then $A$ is differentiable at $c$ and $A^{\prime }(c)=f(c)$.

\begin{proof} To show that $A(x)$ is differentiable at $c$, we need to evaluate

$$L=\underset{h\to 0}{\lim }\frac{A(c+h)-A(c)}{h}.$$

Letting

$$S=\{\begin{array}{ll}[c,c+h],& \text{if }h\ge 0\\ [c+h,c],& \text{if }h<0,\end{array}$$

we get

$$A(c+h)-A(c)={\int }_{S}f(t)dt.$$

Letting $s={\text{sup}}_f(S)$ and $i={\text{inf}}_f(S)$,

$$hi\le {\int }_{S}f(t)dt\le hs.$$

Therefore,

$$i\le \frac{{\int }_{S}f(t)dt}{h}\le s,$$

so, taking the limit, we get

$$\underset{h\to 0}{\lim }i\le L\le \underset{h\to 0}{\lim }s.$$

Since $f$ is continuous at $c$, $f(c)={\lim }_{h\to 0}i={\lim }_{h\to 0}s$. Hence, $L=f(c)$. \end{proof}

Second Fundamental Theorem of Calculus §

First, we define

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Primitive”,“label”:“primitive”,“_index”:1}] Definition 2 (Primitive). Let $f$ be a function on an open interval $I$. A primitive $\varphi$ of $f$ on $I$ is a differentiable function on $I$ with ${\varphi }^{\prime }(x)=f(x)$ for all $x\in I$. Not unique: $\varphi \to \varphi +c$.

Now,

[!math|{“type”:“theorem”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Second Fundamental Theorem of Calculus”,“label”:“second-fundamental-theorem-of-calculus”,“_index”:2}] Theorem 3 (Second Fundamental Theorem of Calculus). Suppose $f,\varphi$ are functions on $[a,b]$ with $f$ integrable and $\varphi$ is a primitive, defined and continuous. Then,

$$\varphi (b)-\varphi (a)={\int }_a^{b}f(x)dx$$

\begin{proof} Let $P$ be a partition of $[a,b]$ with

$$P:a=t_0<t_1<\cdots <t_n=b$$

and set $M_j={\text{sup}}_f([t_{j-1},t])$ and $m_j=\text{inf}([t_{j-1},t_j])$. By definition,

$$L(f,P)=\sum _{j=1}^n{m}_j(t_j-t_{j-1})\le {\int }_a^{b}f(x)dx\le \sum _{j=1}^m{M}_j(t_j-t_{j-1})=U(f,P).$$

On the other hand, by MVT, there is $c_j\in [t_{j-1},t_j]$ s.t.

$${\varphi }^{\prime }(c_j)=\frac{\varphi (t_j)-\varphi (t_{j-1})}{t_j-t_{j-1}}.$$

But $f(c_j)={\varphi }^{\prime }(c_j)$ and $m_j\le f(c_j)\le M_j$, so

$$m_j(t_j-t_{j-1})\le \varphi (t_j)-\varphi (t_{j-1})\le M_j(t_j-t_{j-1}).$$

Adding over all $j$, we get

$$L(f,P)\le \sum _{j=1}^n(\varphi (t_j)-\varphi (t_{j-1}))=\varphi (b)-\varphi (a)\le U(f,P).$$

Since $f$ is integrable, we get that

$$L(f,P)=U(f,P)=\varphi (b)-\varphi (a)={\int }_a^{b}f(x)dx.$$

\end{proof}