EML 5060Analysis in Mechanical Engineering Fall 2009
Test 1 Van Dommelen (http://www.eng.famu.fsu.edu/$\sim$dommelen) Due M 08/31/09

Hand in the solution to this test on the date stated above (5% of your final grade). Read carefully. And look it up. Answer questions in order from left to right, top to bottom. You must work alone. You probably want to consult a math handbook.

Neatly draw the graph of the following functions, showing the locations of 0 and $\pm 1$ on each axis. Give the derivative. Indicate non-principal values as a broken line. Make sure that you give enough of the curves to clearly demonstrate all features. Make sure that you have answered all parts, including derivatives.

\begin{displaymath}
2x-2 \qquad \qquad \qquad \qquad
x^2 + 1 \qquad \qquad \qquad \qquad
x^4 - x^2
\end{displaymath}


\begin{displaymath}
\sin(x)\qquad \qquad \qquad \qquad
\arcsin(x)\qquad \qquad \qquad \qquad
\sinh(x)
\end{displaymath}


\begin{displaymath}
\cos(x)\qquad \qquad \qquad \qquad
\arccos(x)\qquad \qquad \qquad \qquad
\cosh(x)
\end{displaymath}


\begin{displaymath}
\tan(x)\qquad \qquad \qquad \qquad
\arctan(x)\qquad \qquad \qquad \qquad
\tanh(x)
\end{displaymath}


\begin{displaymath}
\ln(x)\qquad \qquad \qquad \qquad
e^x\qquad \qquad \qquad \qquad
\tan(x^2)
\end{displaymath}

Find (include any integration constants and absolute signs):

\begin{displaymath}
\int x^{-2} {\rm d} x= \qquad \qquad \qquad
\int_1^2 x^{-2} {\rm d} x = \qquad \qquad \qquad
\int_1^x \xi^{-2} {\rm d} \xi =
\end{displaymath}


\begin{displaymath}
\int {{\rm d} x \over x} = \qquad \qquad \qquad
\int {1\over...
... d} x = \qquad \qquad \qquad
\int {1\over 1 + x^2} {\rm d} x =
\end{displaymath}


\begin{displaymath}
\int \ln(x) {\rm d} x = \qquad \qquad \qquad
\int x e^x {\rm d} x = \qquad \qquad \qquad
\int x e^{x^2} {\rm d} x =
\end{displaymath}


\begin{displaymath}\left\vert \matrix{
1 & 2 & 3\cr
2 & 3 & 4\cr
3 & 4 & 5}
\rig...
...\qquad
{{\rm d} \over {\rm d} x} \int_x^2x f(\xi) {\rm d}\xi =
\end{displaymath}


\begin{displaymath}2 + 1 + 0 - 1 -2 -3 -4 \ldots -99 -100 = \qquad \qquad\qquad\...
...+ e^{1} + e^{0} + e^{-1} + e^{-2} + e^{-3} + e^{-4} + \ldots = \end{displaymath}


\begin{displaymath}{\rm Solve: }\quad{{\rm d} y\over {\rm d} x} = y \qquad y(1)=1\end{displaymath}