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\noindent {\bf Question}

\noindent State both parts of the Fundamental Theorem of Calculus.
Also, determine whether the following argument is correct: By the
Fundamental Theorem of Calculus,
\[ \int_{-1}^1 {1\over x^2} {\rm d}x = \left[ -{1\over x}\right]_{-1}^1 =
-2,\] and so the integral of a positive function can be negative.

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\noindent {\bf Answer}

\noindent {\bf Fundamental theorem of calculus:} Let $f$ be a
continuous function on the closed interval $[a,b]$.
\begin{itemize}
\item Consider the function on $[a,b]$ defined by
\[ F(x) =\int_a^x f(t) {\rm d}t. \]
Then, $F'(x) =f(x)$ for every $x$ in $(a,b)$.  In shorthand,
\[ f(x) =\frac{{\rm d}}{{\rm d}x} \int_a^x f(t) {\rm d}t. \]
\item If $G$ is any function on $[a,b]$ satisfying $G'(x) =f(x)$, then
\[ \int_a^b f(x) {\rm d}x = G(b) -G(a). \]
In shorthand,
\[ \int_a^b G'(x) {\rm d}x = G(b) -G(a). \]
\end{itemize}
The proof is false: the integrad is not continuous on $[-1,1]$,
and so the fundamental theorem of calculus does not apply.


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