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{\bf EML 5060\hfill Analysis in Mechanical Engineering \hfill Fall 1998 }\\
{\it Test 1 \hfill Van Dommelen (dommelen@eng.fsu.edu) \hfill Due 8/28/98 }
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Hand in the solution to this test on 8/28 (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.  Please
note: This test must have been {\it accepted} before exam 1, on 9/16,
or you also receive a 0 grade for exam 1. Read carefully.  And look it up.

\medskip
Neatly draw the graph of the following functions, showing the
locations of 0 and $\pm 1$ on each axis. Also give the derivative.
Indicate nonprincipal values as a broken line.  Make sure that you
draw enough of the curves to {\em clearly} demonstrate {\em all}
features.
$$
1-x \qquad \qquad \qquad \qquad
x^2 - 2 \qquad \qquad \qquad \qquad
x^3 - x
$$
\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
$$
\ln(x)\qquad \qquad \qquad \qquad
e^x\qquad \qquad \qquad \qquad
\cos(x^2)
$$
\vfill

Find (include any integration constants and absolute signs):
$$
\int x^4 {\rm d} x= \qquad \qquad \qquad
\int_0^1 x^4 {\rm d} x = \qquad \qquad \qquad
\int_0^x \xi^4 {\rm d} \xi =
$$
$$
\int {{\rm d} x \over x} = \qquad \qquad \qquad
\int {1\over 1 - x^2} {\rm d} x = \qquad \qquad \qquad
\int {1\over 1 + x^2} {\rm d} x =
$$
$$
\int \ln(x) {\rm d} x = \qquad \qquad \qquad
\int x \sin(x) {\rm d} x = \qquad \qquad \qquad
\int x \sin(x^2) {\rm d} x =
$$

$$\left| \matrix{
1 & 2 & 3\cr
1 & 2 & 6\cr
1 & 4 & 6}
\right| = \qquad \qquad \qquad
\lim_{x\to 0} {\tanh(x) \over x} = \qquad \qquad \qquad
{{\rm d} \over {\rm d} x} \int_a^x f(\xi) g(x) {\rm d}\xi =
$$

$$ 2 + 4 + 6 + 8 + \ldots + 998 + 1000 = \qquad \qquad\qquad\qquad\qquad
{1\over \sqrt2} + {1\over \sqrt2^2} + {1\over \sqrt2^3} + \ldots = $$

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

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