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\begin{center}
{\bf EML 5060\hfill Analysis in Mechanical Engineering \hfill Fall 1996 }\\
{\it Test 1 \hfill Van Dommelen (dommelen@eng.fsu.edu) \hfill Due 8/30/96 }
\end{center}

Hand in the solution to this test on 8/30 (5\% of your final grade).
If your performance is insufficient, you will need to hand in a
corrected version; however, only your initial grade counts.
{\it This test must have been {\bf accepted} before exam 1, or you
also receive a 0 grade for exam 1.}

Neatly draw the graph for {\it all applicable} $x$ (showing the locations
of 0 and 1 on each axis) and give the derivative:
$$
x-1 \qquad \qquad \qquad \qquad
x^2 - 1 \qquad \qquad \qquad \qquad
x^3 - x
$$
\vfill
$$
\sin(x)\qquad \qquad \qquad \qquad
\arcsin(x)\qquad \qquad \qquad \qquad
\sinh(x)
$$
\vfill
$$
\cos(x)\qquad \qquad \qquad \qquad
\arccos(x)\qquad \qquad \qquad \qquad
\cosh(x)
$$
\vfill
$$
\tan(x)\qquad \qquad \qquad \qquad
\arctan(x)\qquad \qquad \qquad \qquad
\tanh(x)
$$
\vfill
$$
\ln(x)\qquad \qquad \qquad \qquad
e^x\qquad \qquad \qquad \qquad
\sin(x^2)
$$
\vfill

Find (include any integration constants and absolute signs):
$$
\int x^5 {\rm d} x= \qquad \qquad \qquad
\int_0^1 x^5 {\rm d} x = \qquad \qquad \qquad
\int_0^x \xi^5 {\rm d} \xi =
$$
$$
\int {{\rm d} x \over x} = \qquad \qquad \qquad
\int {1\over 1 - x^2} {\rm d} x = \qquad \qquad \qquad
\int {1\over 1 + x^2} {\rm d} x =
$$
$$
\int \ln(x) {\rm d} x = \qquad \qquad \qquad
\int x \cos(x) {\rm d} x = \qquad \qquad \qquad
\int x \cos(x^2) {\rm d} x =
$$

$$\left| \matrix{
1 & 2 & 3\cr
1 & 2 & 4\cr
1 & 4 & 7}
\right| = \qquad \qquad \qquad
\lim_{x\to 0} {\sinh(x) \over x} = \qquad \qquad \qquad
{{\rm d} \over {\rm d} x} \int_a^x f(\xi) {\rm d}\xi =
$$

$$ 3 + 6 + 9 + 12 + \ldots + 297 + 300 = \qquad \qquad\qquad\qquad\qquad
1 + {1\over 3} + {1\over 9} + {1\over 27} + \ldots = $$

$$ {\rm Solve: }\quad{{\rm d} y\over {\rm d} x} = -y \qquad y(1)=2$$

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