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\title[]{MA140-Engineering Calculus}
\author[ Lecture 18 ]{\large{ Lecture 18} 
\and %\\ \small{Supervisor: Dr. John Burns}
\\ 
 \includegraphics[scale=0.0]{Figures/lec14.jpg}}
%\date[June 2-4, 2016]{June 2-4, 2016}
%\institute[Uppsala University]{ }


\begin{document}
\begin{frame}
\titlepage
\end{frame}


\section{MA140}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}
\begin{definition}
A function $F$ is an \textit{antiderivative} of $f$ on an interval $I$ if $F'(x)=f(x)$ for all $x$ in $I$.\\
$f$ is the derivative of $F$ $\longleftrightarrow$ $F$ is an antiderivative of $f$
\end{definition}
{\textcolor {red} {\textit{Note:}}} If $F$ is an antiderivative of $f$, then the most general antiderivative of $f$ is $$F(x)+c$$
Where $c$ is an arbitrary constant. \\
For example the antiderivative of $f(x)=2x$ equals $F(x)=x^{2}+c$\\
or \\
For example the antiderivative of $f(x)=3x^{2}$ equals $F(x)=x^{3}+c$
\end{frame}
\begin{frame}
\begin{definition}
We call $$\int f(x) dx$$ an \textit{indefinite integral} if $$\int f(x) dx=F(x)+c$$ where $F(x)$ is an antiderivative of $f(x)$.
\end{definition}
\begin{example}
\begin{itemize}
\item $\int 2x dx=x^{2}+c$
\item $\int 3x^{2} dx=x^{3}+c$
\end{itemize}
\end{example}
\end{frame}
\begin{frame}
\begin{example}
\begin{itemize}
\item $$\int x dx=\frac{1}{2}x^{2}+c$$
Because the derivative of $\frac{1}{2} x^{2}+c$ is equal to $x$
\item $$\int x^{2} dx=\frac{1}{3} x^{3}+c$$
\end{itemize}
\end{example}
In general when $x\neq -1$: $$\int x^{n} dx=\frac{x^{n+1}}{n+1}+c$$ 

\end{frame}
\begin{frame}
\begin{example}
\begin{itemize}
\item $$\int \sin (x) dx=-\cos (x)+c$$
\item $$\int \cos (x) dx=\sin (x)+c$$
\item $$\int \sec^{2} (x) dx=\tan(x)+c$$
\item $$\int \csc^{2} (x) dx=-\cot (x)+c$$
\item $$\int \sec (x)\tan (x) dx=\sec (x)+c$$
\item $$\int \csc (x)\cot (x) dx=-\csc(x)+c$$
\end{itemize}
\end{example}
\end{frame}
\begin{frame}

{\textcolor {red} {\textit{Some useful integral properties:}}}
\begin{itemize}
\item[(1)] $$\int cf(x) dx=c\int f(x) dx$$
\item[(2)] $$\int [f(x)\pm g(x)] dx=\int f(x) dx\pm \int g(x) dx$$
\end{itemize}
\begin{example}
evaluate the following integral: $$\int (2x^{2}+9x^{7}) dx$$
\end{example}
$$\int (2x^{2}+9x^{7}) dx=\int 2x^{2} dx+\int 9x^{7} dx=2\int x^{2} dx+9\int x^{7} dx$$
\end{frame}
\begin{frame}
$$=[2(\frac{x^{3}}{3})+c_{1}]+[9(\frac{x^{8}}{8})+c_{2}]=2(\frac{x^{3}}{3})+9(\frac{x^{8}}{8})+(c_{1}+c_{2})$$ 
$$\int (2x^{2}+9x^{7}) dx=2(\frac{x^{3}}{3})+9(\frac{x^{8}}{8})+c$$

\begin{theorem}
If $u=g(x)$ is a differentiable function, then: $$\int f(g(x))g'(x) dx=\int f(u)du$$
\end{theorem}
\end{frame}
\begin{frame}
\begin{example}
evaluate the following integral: $$\int 3x^{2} \sin (x^{3}) dx$$
\end{example}
We see that $3x^{2}$ is the derivative of $x^{3}$. So if we make the substitution $u=x^{3}$, then $\displaystyle {\frac{du}{dx}}=3x^{2}$  or in the differential form $du=3x^{2} dx$, so 
$$\int 3x^{2} \sin (x^{3}) dx=\int \sin (u) du $$
Now we integrate with respect to $u$:
$$\int \sin (u) du=-\cos (u)+c$$
Replace $u$ by $x^{3}$, we get:
$$\int 3x^{2} \sin (x^{3}) dx=\int \sin (u) du=\int \sin (u) du=-\cos (u)+c=-\cos (x^{3})+c$$
\end{frame}
\begin{frame}
\begin{example}
Find: $$\int 2x \sqrt{1+x^{2}} dx$$
\end{example}
We define a function of $x$, called $u$.\\
Let $u=1+x^{2}$, so $\displaystyle {\frac{du}{dx}=2x}$ or $du=2x dx$
So the above integral becomes: $$\int \sqrt{u} du=\int u^{\frac{1}{2}} du=\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{2}{3}u^{\frac{3}{2}}+c=\frac{2}{3}(1+x^{2})^{\frac{3}{2}}+c$$
\end{frame}
\begin{frame}
\begin{example}
Find: $$\int \cos(4x-7) dx$$
\end{example}
Let $u=4x-7$. then $\displaystyle{\frac{du}{dx}}=4$ or $du=4dx$.\\
We can rewrite the original integral as: $$\frac{1}{4}\int 4 \cos(4x-7) dx$$  then the above integral equals: $$\frac{1}{4} \int \cos(u) du=\frac{1}{4} \sin(u)+c=\frac{1}{4} \sin(4x-7)+c$$  
\end{frame}
\begin{frame}
\begin{example}
Find: $$\int e^{-5x} dx$$
\end{example}
Let $u=-5x$,  then $\displaystyle{\frac{du}{dx}}=-5$ or $du=-5dx$.\\
We can rewrite the original integral as: $$\frac{-1}{5}\int -5e^{-5x} dx$$  then the above integral equals: $$\frac{-1}{5}\int e^{u} du=\frac{-1}{5}e^{u}+c=\frac{-1}{5}e^{-5x}+c $$
\end{frame}
\begin{frame}
\begin{example}
Find: $$\int \sin^{3}x \cos x dx$$
\end{example}
\end{frame}




\end{document}