Actions

Fundamental theorem of calculus

The fundamental theorem of calculus is a link between the concept of the derivative of a function and the concept of integral. There are two parts to the theorem.


First part[edit]

This part is also known as the first fundamental theorem of calculus.

The theorem[edit]

Let $f$ be a continuous real-valued function which is continuous on the closed interval $\left[a,b\right]$. Let $F$ be the function defined, for all $x$ in $\left[a,b\right]$, by the following equation:

[math]F(x) = \int_a^x\!f(t)\, dt.[/math]

Then, $F$ is a primitive of $f$ on $\left[a,b\right]$.

Prerequisite information[edit]

The proof[edit]

We want to find the primitive of $F$, denoted $F'$. This is why we use the defintion of the derivative to get started:

[math]\begin{eqnarray} F'(x) &=& \lim_{h \to 0}\frac{F(x+h)-F(x)}{h} \\ &=& \lim_{h \to 0}\frac{1}{h}\left(\int_{a}^{x+h}f(t)\mathop{dt}-\int_ {a}^{x}f(t)\mathop{dt}\right)\\ &=& \lim_{h \to 0}\frac{1}{h}\left(\int_{a}^{x+h}f(t)\mathop{dt}+\int_ {x}^{a}f(t)\mathop{dt}\right)\\ &=& \lim_{h \to 0}\frac{1}{h}\int_{x}^{x+h}f(t)\mathop{dt}\,\,\,\,\textbf{(1)}\end{eqnarray}[/math]

We know that $f$ is continuous on $\left[a,b\right]$, this is why it is continuous on $\left[x,x+h\right]$[math]\left[x;x+h\right]\subset\left[a;b\right][/math]. From the mean value theorem, we know that there exist a point $c\in \left[x,x+h\right]$ such that:

[math]\begin{eqnarray} f(c)&=&\frac{1}{h}\int_x^{x+h}f(t)\mathop{dt}\,\,\,\,\textbf{(2)}\end{eqnarray}[/math]

Then, we deduce from the results $\textbf{(1)}$ and $\textbf{(2)}$ that:

[math]\begin{eqnarray}F'(x)&=&\lim_{h\to o}f(c)\end{eqnarray}[/math]

When $h \to 0$, $c\to x$, this is why we have:

[math]\begin{eqnarray}F'(x)&=&f(x)\,\,\,\#\end{eqnarray}[/math]


Second part[edit]

This part is also known as the second fundamental theorem of calculus. Sometimes it is referred to as the evaluation theorem (in context of integral calculus).

The theorem[edit]

Let $F$ be a primitive of $f$ on $\left[a,b\right]$. If $f$ is continuous on the interval $\left[a,b\right]$, then:

[math]\int_{a}^{b}f(x)\mathop{dx}=F(b)-F(a)[/math]

Prerequisite information[edit]

  • Know the first part of the fundamental theorem of calculus.

The proof[edit]

From the first part of the fundamental theorem of calculus, we have:

[math]F(x)=\int_{a}^{x}f(t)\mathop{dt}[/math]

Then, we deduce that:

[math]\begin{eqnarray}F(b)-F(a)&=&\int_{a}^{b}f(t)\mathop{dt}-\int_{a}^{a}f(t)\mathop{dt}\\&=&\int_{a}^{b}f(t)\mathop{dt}\,\,\#\end{eqnarray}[/math]