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\begin{document}

\title{MTH 1420, Spring 2012\\Dr. Graham-Squire}
\author{Section 5.4: Fundamental Theorem of Calculus }
\date{}
\maketitle

HW: 4, 5, 8, 14, 23

Practice: 3, 9, 13, 17, 19, 21

\section{Introduction}

As we have seen in the Evaluation Theorem of section 5.3 and Lab 2, the operations of differentiation and integration are inverses.  Thus we can use anti differentiation to calculate integrals, and if we integrate and then differentiate we end up where we started.  The lab gives an intuitive idea of the fundamental theorem, in this section we give precise statements of the theorem as well as investigate more complicated problems.

\section{Examples}

We begin with two quick examples to remind us about area functions and their derivatives:

\bexa
	If $f(t)=$
	
	\includegraphics[angle=90, height=70mm]{coordaxes.pdf}
	
	and $\d g(x)=\int^x_0f(t)dt$, calculate $g(1)=\blank{.5}, g(2)=\blank{.5}, g(3)=\blank{.5}, g(4)=\blank{.5},\\ g(6)=\blank{.5}$
\eexa
\newpage


\bexe
	Let $\d g(x)=\int^x_1 (1+\sqrt{t})dt$.  Calculate $g(x)$ and $g'(x)$.
\eexe

\vspace{2in}

\brem
	Notice that in the exercise above, you would get the same result for $g'(x)$ no matter what the lower limit of integration for $g(x)$ is.  Thus although the function $g$ might change if you adjust the lower limit of integration, the rate of change of $g$ will be no different.
\erem

\section{The Fundamental Theorem of Calculus}

\bthe(Fundamental Theorem of Calculus)
\benu
	\item If $f$ is continuous on $[a,b]$, and $\d g(x)=\int^x_a f(t)dt$ (for some $x$ between $a$ and $b$), then $g(x)$ is an antiderivative of $f$.  Notationally, $g'(x)=f(x)$ for all $a<x<b$, and we can summarize it more succinctly as
	\[ \frac{d}{dx}\int^x_a f(t)dt=f(x).
	\]   
	\item $\d\int^b_af(x)dx=F(b)-F(a)$, where $F$ is any antiderivative of $F$. (This is the same as the Evaluation Theorem from section 5.3)
	\eenu
\ethe

\bpro
	Proof of part 1 of the theorem:
\epro

\vspace{1in}

\bexe
	Let $\d h(x)=\int^x_3\frac{\cos t}{t}dt$. Use the Fundamental Theorem to find $h'(x)$.
\eexe
\vspace{1in}

\section{More tricky FTC questions}

\bexa
	Consider a question very similar to the exercise above, but with a different upper limit of integration.  Let $\d h(x)=\int^{\sqrt{x}}_3\frac{\cos t}{t}dt$. Find $h'(x)$.
\eexa
\vspace{3.5in}

\bexa
	Another method for solving the example above is:
\eexa
\vspace{2.5in}
\newpage

\bexe
	Find the derivative of the function:
	\[ \int^{\tan x}_0 \sqrt{t+\sqrt{t}}\,dt
	\]	
\eexe
\vspace{3.5in}

\bexe
	Find a function $f$ such that $f(2)=0$ and $\d f'(x)=\frac{\sin x}{x}$.  Hint: Use the fundamental theorem of calculus to find an antiderivative for $f'$, then choose the correct lower limit of integration so that $f(2)=0$.
\eexe




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