\section{The Meaning of the Definite Integral}
The \textbf{definite integral}\index{integral(s)!definite} of the function $f(x)$ between $x=a$ and $x=b$ is written: 
\[\int_a^b f(x)\,dx\]
Geometrically it equals the \textbf{area}\index{area} $A$ between the curve $y=f(x)$ and the $x$-axis between the vertical lines $x=a$ and $x=b$:
\begin{center}
\includegraphics{graphics/integration_review1}%
\end{center}
More precisely, assuming $a<b$, the definite integral is the net sum of the \textbf{signed} areas between the curve $y=f(x)$ and the $x$-axis where areas below the $x$-axis (i.e. where $f(x)$ dips below the $x$-axis) are counted \textbf{negatively}.

The notation used for the definite integral, $\int_a^b f(x)\,dx$, is elegant and intuitive.  We are $\int$umming ($\int dA$) the (infinitesimally) small differential rectangular areas $dA=f(x)\cdot dx$ of height $f(x)$ and width $dx$ at each value $x$ between $x=a$ and $x=b$:
\begin{center}
\includegraphics{graphics/integration_review2}%
\end{center}
We will see soon how viewing integrals as sums of differentials can be used to come up with formulas for calculations aside from just area.

\section{The Fundamental Theorem of Calculus}
As seen in a previous calculus course, the definite integral can be written as a limiting sum (Riemann Sum) of $N$ rectangles of finite width $\Delta x=(b-a)/N$ where we let the number of rectangles ($N$) go to infinity (and consequently the width $\Delta x\rightarrow 0$).  This method of evaluating a definite integral is hard or impossible to compute exactly yfor most functions.  An easy way to evaluate definite integrals is due to the \textbf{Fundamental Theorem of Calculus}\index{Fundamental Theorem of Calculus} which relates the calculation of a definite integral with the evaluation of the \textbf{antiderivative}\index{antiderivative} $F(x)$ of $f(x)$:

\theorem{\textbf{The Fundamental Theorem of Calculus:}

If $f$ is continuous on $[a,b]$ then\
\[\int_a^bf(x)\, dx=F(b)-F(a)\]
for any $F$ an antiderivative of $f$, i.e. $F'(x)=f(x)$.
}

Notationally we write $F(b)-F(a)$ with the shorthand $\displaystyle \left. F(x)\right|_a^b$, i.e.
\[\left. F(x)\right|_a^b=F(b)-F(a)\ ,\]
where, unlike the integral sign, the bar is placed \emph{on the right}.
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\section{Indefinite Integrals}
Because of the intimate relationship between the antiderivative and the definite integral, we define the \textbf{indefinite integral}\index{integral(s)!indefinite} of $f(x)$ (with no limits $a$ or $b$) to just be the antiderivative, i.e. 
\[\int f(x)\;dx=F(x)+C\]
where $F(x)$ is an antiderivate of $f(x)$ (so $F'(x)=f(x)$) and $C$ is an arbitrary constant.  The latter is required since the antiderivative of a function is not unique as $\frac{d}{dx} C=0$ implies we can always add a constant to an antiderivative to get another antiderivative of the same function.

Using our notation for indefinite integrals and our knowledge of derivatives gives the following.
\subsection*{Table of Indefinite Integrals} \index{indefinite integrals! table of}
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle\int x^n\,dx=\frac{1}{n+1}x^{n+1}+C\ \ \ \ (n\neq-1)$
\item $\displaystyle\int \cos x\,dx=\sin x+C$
\item $\displaystyle\int \sin x\,dx=-\cos x+C$
\item $\displaystyle\int \sec^2x\,dx=\tan x+C$
\item $\displaystyle\int \sec x\tan x\,dx=\sec x+C$
\item $\displaystyle\int \csc^2x \,dx=-\cot x+C$
\item $\displaystyle\int \csc x\cot x\,dx=-\csc x+C$
\item $\displaystyle\int c f(x)\,dx=c\int f(x)\,dx$
\item $\displaystyle\int \left[f(x)\pm g(x)\right]\,dx=\int f(x)\,dx\pm \int g(x)\,dx$
\end{enumerate}
~%ensures ragged bottom of second column
\end{multicols}

In the last two integration formulae $f(x)$ and $g(x)$ are functions while $c$ is a constant.  For indefinite integrals we say, for example, that $\frac{1}{n+1}x^{n+1}+C$ is the \textbf{(indefinite) integral} of $x^n$ where $x^n$ is the \textbf{integrand}\index{integrand}.  The \textbf{process} of finding the integral is called \textbf{integration}\index{integration}. Each of these indefinite integrals may be verified by differentiating the right hand side and verifying that the integrand is the result.

\section{Integration by Substitution} \index{integration! by substitution}
The last two general integral results allow us to break up an integral of sums or differences into integrals of the individual pieces and to pull out any constant multipliers.  Another useful way of solving an integral is to use the \textbf{Substitution Rule}\index{Substitution Rule} which arises by working the differentiation Chain Rule in reverse.


\theorem{\textbf{Substitution Rule (Indefinite Integrals):}\index{Substitution Rule!for indefinite integrals} Suppose $u=g(x)$ is a differentiable function whose range of values is an interval $I$ upon which a further function $f$ is continuous, then
\[\int f(g(x))g'(x)\,dx=\int f(u)\,du\ .\]
where the right hand integral is to be evaluated at $u=g(x)$ after integration.
}

Here the $du$ appearing on the right side is the differential:
\[du=g'(x)dx\]
which, recall, can be remembered by thinking $\frac{du}{dx}=g'(x)$ and multiplying both sides by $dx$.

When using the Substitution Rule with definite integrals we can avoid the final back-substitution of $u=g(x)$ of the indefinite case by instead just changing the limits of the integral appropriately to the $u$-values corresponding to the $x$-limits:
\theorem{\textbf{Substitution Rule (Definite Integrals):}\index{Substitution Rule!for definite integrals} Suppose
  $u=g(x)$ is a differentiable function whose derivative $g'$ is
  continuous on $[a,b]$ and a further function $f$ is continuous on
  the range of $u=g(x)$ (evaluated on $[a,b]$), then
\[\int_a^b f(g(x))g'(x)\,dx=\int_{g(a)}^{g(b)} f(u)\,du\ .\]
}

\section{Integration Examples}
\begin{example}
Evaluate the following integrals:
\begin{multicols}{2}
\begin{enumerate}
\item $\displaystyle \int\left(x^2+ \frac{1}{x^2}-3\cos x\right)dx $
\item $\displaystyle \int_0^{\pi/2}\left(\sqrt{x}+\sin x\right)dx $ 
\item $\displaystyle \int\left(\sqrt[3]{x^2} +\sec x\tan x\right)dx $ 
\item $\displaystyle \int\frac{(x+2)^2}{\sqrt{x}}dx $ 
\item $\displaystyle \int\frac{x}{\sqrt{ x^2 +1}}dx $ 
\item $\displaystyle \int t^2 \sec (t^3 + 5) \tan (t^3 + 5)\,dt $ 
\item $\displaystyle \int\sqrt{1+\tan \theta}\sec^2\!\theta\,d\theta $ 
\item $\displaystyle \int_{-1}^{0}x \sqrt{3x + 4}\,dx $ 
\end{enumerate}
\end{multicols}
 \begin{solution*}
   \begin{enumerate}
   \item Using basic integration formula and recalling that $\displaystyle\frac{1}{x^n}=x^{-n}$ one has:
    \begin{align*}
      \int\left(x^2 + \frac{1}{x^2}-3\cos x\right)dx
	&=\int\left(x^2+x^{-2}-3\cos x\right)dx\\
    &=\frac{1}{3}\ x^3 -x^{-1}-3\sin x + C
      =\frac{1}{3}\ x^3 -\frac{1}{x}-3\sin x + C
    \end{align*}
	\item Recalling that $\displaystyle\sqrt[n]{x}=x^\frac{1}{n}$ we evaluate the definite integral to get:
	\begin{align*}
		\int_0^{\pi/2}\left(\sqrt{x}+\sin x\right)dx
		&= \int_0^{\pi/2}\left(x^{\frac{1}{2}}+\sin x\right)dx
		= \left[\frac{2}{3}x^{\frac{3}{2}}-\cos x\right|_0^{\frac{\pi}{2}}\\
		&=\left[\frac{2}{3}\left(\frac{\pi}{2}\right)^{\frac{3}{2}}-\cos\left(\frac{\pi}{2}\right)\right]-\left[\frac{2}{3}(0)^{\frac{3}{2}}-\cos(0)\right]
		=\frac{\sqrt{2}\pi^{\frac{3}{2}}}{6}+1
	\end{align*}	

  	\item Recalling that and $\left(x^m\right)^n=x^{mn}$ gives:
    \begin{align*}
     \int\left(\sqrt[3]{x^2} +\sec x\tan x\right)dx
      &= \int(x^{2/3} +\sec x\tan x)dx  \\
      &= \frac {3}{5}x^{5/3}+ \sec x + C
      \end{align*}
      
    \item Expanding the numerator, dividing through by the denominator and simplifying $\displaystyle \frac{x^m}{x^n}=x^{m-n}$ before integrating gives:
      \begin{align*}
        \int \frac{(x+2)^2}{\sqrt{x}} \,dx \
        &= \int\frac{ x^2 + 4x + 4}{\sqrt{x}}dx \\
        &= \int\left(x^{3/2} + 4x^{1/2} + 4x^{-1/2}\right)dx \\
      &=\frac{2}{5}x^{5/2} + 4\left(\frac{2}{3}\right)x^{3/2} + 4\left(2\right)x^{1/2} + C \\
      &=\frac{2}{5}x^{5/2} + \frac{8}{3}x^{3/2} + 8x^{1/2} + C        
        \end{align*}

      \item Using the substitution $\displaystyle u= x^2 +1$ so $du= 2x \,dx \implies \frac{1}{2} \,du = x \,dx $ one has:
        \begin{align*}
          \int \frac{x}{\sqrt{x^2 +1}} \,dx
          = \int \frac{1}{2}\cdot\frac{1}{\sqrt{u}} \,du
          &=\frac{1}{2} \int u^{-1/2} \,du \\
          &=u^{1/2}+ C  = \sqrt{x^2 +1} \ + C
          \end{align*}
        \item Using the substitution $\displaystyle  u= t^3 + 5$ so $du= 3t^2 \,dt \implies \frac{1}{3} \,du = t^2 \,dt $ one has:
          \begin{align*} 
            \int t^2\sec (t^3+ 5)\tan (t^3 + 5)\,dt
            &= \frac{1}{3} \int \sec u \tan u \,du \\
            &= \frac{1}{3}\sec u + C = \frac{1}{3} \sec (t^3 + 5) + C
            \end{align*}
          \item Using the substitution $\displaystyle u= 1+ \tan \theta$ so $du = \sec^2\!\theta\,d\theta$ one has:
            \begin{align*}
              \int \sqrt{1+ \tan \theta}\sec^2\!\theta\,d\theta
              &= \int \sqrt{u}\,du = \int u^{1/2} \,du \\
              & = \frac{2}{3}\ u^{2/3} + C %\\ 
              = \frac{2}{3} (1 + \tan \theta)^{3/2}+ C
              \end{align*}
            \item Make the following substitution $u= 3x+4$ so $du= 3dx \implies \frac{1}{3} \, du= dx$.  Solving for $x$ gives $x= \frac{1}{3}(u-4)$ allowing its replacement in the integral.  Finally we change to limits in $u$:
            \begin{align*}
           x= 0 &\implies u= 0+4 =4 \\	
           x= -1 &\implies u= -3 +4 =1
           \end{align*}
           \begin{align*}
             \int_{-1}^{0}x \sqrt{3x + 4} \,dx
             &= \int_{1}^{4}\frac{1}{3}(u-4)\sqrt{u}\,\frac{1}{3}\,du \\
             &= \frac{1}{9} \int_{1}^{4}(u^{3/2}-4u^{1/2})\, du 
             = \frac{1}{9} \left[\frac{2}{5}u^{5/2}-4\left(\frac{2}{3}\right)u^{3/2}\right|_1^4 \\
             & = \frac{1}{9} \left[\frac{2}{5}(4)^{5/2}-\frac{8}{3}(4)^{3/2}\right]- \frac{1}{9}\left[\frac{2}{5}(1)^{5/2}-\frac{8}{3}(1)^{3/2}\right] \\
             & = \frac{1}{9} \left[\frac{64}{5}- \frac{64}{3}\right] -\frac{1}{9} \left[\frac{2}{5}-\frac{8}{3}\right]
             = \frac{1}{9}\left[\frac{64}{5}-\frac{64}{3}-\frac{2}{5} +\frac{8}{3}\right]\\
             &= \frac{1}{9}\left [\frac{62}{5}-\frac{56}{3}\right]
             = \frac{1}{9}\left [-\frac{94}{15}\right] \\
             &= -\frac{94}{135}
              \end{align*}
    \end{enumerate}
  \end{solution*}    

\begin{furtherquestions}
%insert:1-05.tex
  Evaluate the following integrals:
  \begin{multicols}{2}
    \begin{enumerate}
    \item $\displaystyle\int \left(t^2+\sqrt{t}-\frac{2}{t^2}\right)\,dt$
    \item $\displaystyle\int_0^1 \left(t^2+1\right)^2\,dt$
    \item $\displaystyle\int x^2\left(x^3+2\right)^{\frac{1}{3}}\,dx$
    \item $\displaystyle\int_0^{\frac{\pi}{4}}\left(\sec x-\tan x\right)\sec x\,dx$
    \item $\displaystyle\int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx$
    \item $\displaystyle\int_1^2x\sqrt{x-1}\,dx$
    \item $\displaystyle\int\sin(5\theta)\,d\theta$
    \item $\displaystyle\int_2^3\frac{3x^2-1}{\left(x^3-x\right)^2}\,dx$
    \item $\displaystyle\int t^2\sin\left(1-t^3\right)dt$
    \item $\displaystyle\int \frac{x-\sqrt{3x}}{\sqrt{2x}}\,dx$
      % \item $\displaystyle\int\frac{\sin\left(\frac{3}{x}\right)}{3x^2}\,dx$
    \item $\displaystyle\int_0^4\left(4x+9\right)^{\frac{3}{2}}\,dx$
      % \item $\displaystyle\int t\sec^2\left(t^2-1\right)\,dt$
    \item $\displaystyle\int \left(\cos \theta+\sin \theta\right)\left(\cos \theta-\sin \theta\right)^4d\theta$
    \end{enumerate}
    \end{multicols}
\end{furtherquestions}
\end{example}

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\end{reviewexercise}

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