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\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Final Exam Solutions}
\centerline{\it May 7, 2002}
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\noindent
I. A) The time interval $\Delta t = 10$ minutes, or $1/6$ hours.  Since
the speeds are given in miles per hour, we will use $\Delta t = 1/6$.
The left-hand sum approximation to the total distance traveled is
$$LHS = (35 + 36 + 71 + 23 + 27 + 40) \doteq 38.7\ {\rm miles}$$
The right-hand sum approximation is
$$RHS = (36 + 71 + 23 + 27 + 40 + 35) \doteq 38.7\ {\rm miles}$$
also.  Our best estimate is the average, $38.7$ miles.
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\noindent
B)  This was a ``trick question''.  Ordinarily, we would use the 
relation $|RHS - LHS| = |v(1) - v(0)|\Delta t$ ({\it Note: mentioning
this fact earned full credit})  to answer a question
like this.  However, since the velocity is not steadily increasing or
decreasing, it is not possible to say that the actual distance traveled
lies between the LHS and RHS values.  {\it It is not possible to determine
the information asked for from the information given.}
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\noindent
II. A)  $\int_0^9 f(x)\ dx = $ area above $x$-axis, minus area below $x$-axis.
Using the area formulas for circles and trapezoids (or triangles and rectangles),
we get
$$\int_0^9 f(x)\ dx = 4 + {1\over 4} \pi(3)^2 - {1\over 4}\pi(4)^4 = 
4 - {7\pi\over 4} \doteq -1.5$$
(Note: the integral only goes from $x = 0$ to $x = 9$.  The triangle
below the $x$-axis from $x = 9$ to $x = 12$ should be ignored.)
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\noindent
B)  Let $f$ be continuous on the interval $[a,b]$, and let $F$ be
any antiderivative of $f$.  Then $\int_a^b f(x)\ dx = F(b) - F(a)$.
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\noindent
III.  A)  Use the power rule.  Note: ${5\over \pi}$ is a constant!
$$\int x^7 - 3x^{-1/2} + {5\over \pi}\ dx = {x^8\over 8} - 6 x^{1/2} + {5\over \pi}x + C$$
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\noindent
B)  Use substitution:  $w = x^2 + 1$, $dw = 2x\ dx$.  Then
$$\int x e^{x^2+1}\ dx = {1\over 2}\int e^w\ dw = {1\over 2}e^w + C = 
{1\over 2}e^{x^2 + 1} + C$$
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\noindent
C)  Use integration by parts, twice, with $u = $ power of $x$ (or the table)
$$\eqalign{\int x^2 \cos(3x)\ dx &= {x^2\over 3}\sin(3x) - {2\over 3}\int x \sin(3x)\ dx\cr
&={x^2\over 3}\sin(3x) - {2\over 3}\left( {-x\over 3}\cos(3x) + {1\over 3}\int \cos(3x)\ dx \right)\cr
&= {x^2\over 3}\sin(3x) + {2x\over 9}\cos(3x) - {2\over 27}\sin(3x) + C\cr}$$
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\noindent
D)  This has the form of \# 28 in the table with $x \to 2x$, $a = 1$:
$$\int {1\over \sqrt{1 - 4x^2}}\ dx = 
{1\over 2}\int {2\over \sqrt{1 - (2x)^2}}\ dx
= {1\over 2}\arcsin(2x) + C$$
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\noindent
E)  This one is slightly tricky.  The most direct way to do it is 
to add and subtract a term on the top to break the fraction up into
two things that are easy to integrate.  The second integral is $du/u$:
$$\eqalign{\int {1\over 1 + e^x}\ dx &= \int{1 + e^x - e^x\over 1 + e^x}\ dx\cr
&= \int {1+e^x\over 1+e^x} - {e^x\over 1 + e^x}\ dx \cr
&= \int 1\ dx - \int {e^x\over 1 + e^x}\ dx \cr
&= x - \ln(1 + e^x) + C\cr}$$
(Another way is to use a substitution $u = e^x$.  After some simplification,
the form is \# 26 in the table with $a = -1$, $b = 0$.)
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\noindent
IV.  A)  Rotating about the $x$-axis, the cross-sections of the solid
are disks, radius $4 - x^2$, thickness $dx$.  So the volume is 
$$V = \int_0^1 \pi(4 - x^2)^2\ dx = \pi\int_0^1 16 - 8x^2 + x^4\ dx = 
\pi(16x - {8x^3\over 3} + {x^5\over 5})|_0^1 = {203\pi\over 15}$$ 
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\noindent
B)  Subdivide the plate into strips with long edges parallel to the $y$-axis.
If $\Delta x$ = width is small enough, then the density is nearly constant
on each strip and the mass of the strip is approximately (density)(area):
$$\eqalign{{\rm Total Mass} &\doteq \sum (1+x)(4 - x^2)\Delta x\cr
&= \int_0^1 (1+x)(4 - x^2)\ dx\cr
&= \int_0^1 4 + 4x - x^2 - x^3\ dx\cr
&= (4x + 2x^2 - {x^3\over 3} - {x^4\over 4})|_0^1 = {65\over 12}\cr}$$
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\noindent
C)  The $x$-coordinate of the center of mass is
$$\overline{x} = {\int_0^1 x(1+x)(4-x^2)\ dx\over \int_0^1 (1+x)(4 - x^2)\ dx }$$
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\noindent
V. A)  The probability is 
$$\int_1^2 {2\over 9}x(3-x)\ dx = {2\over 9}\left({3x^2\over 2} - {x^3\over 3}\right)|_1^2 =  {13\over 27}$$
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\noindent
B)  The mean snowfall is 
$$\int_0^3 x\cdot {2\over 9}x(3-x)\ dx = {3\over 2}$$
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\noindent
VI.  With $f(x) = x\cos(3x)$, we differentiate using the product and
chain rules, and simplify as we go along:
$$\eqalign{f(x) = x\cos(3x) &\qquad f(0) = 0\cr
f'(x) = \cos(3x) - 3 x\sin(3x) &\qquad f'(0) = 1\cr
f''(x) = -6\sin(3x) - 9 x\cos(3x) &\qquad f''(0) = 0\cr
f'''(x) = -27\cos(3x) + 27 x\sin(3x) &\qquad f'''(0) = -27\cr
f''''(x) = 108\sin(3x) + 81x\cos(3x) &\qquad f''''(0) = 0\cr
f^{(5)}(x) = 405\cos(3x) - 243x\cos(3x) &\qquad f^{(5)}(0) = 405\cr}$$
So 
$$p_5(x) = x - {27\over 3!}x^3 + {405\over 5!}x^5 = 
x - {9\over 2}x^3 + {27\over 8}x^5$$
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\noindent
B)  Using the Taylor series for $\cos(x)$:
$$x\cos(3x) = x\left(1 - {(3x)^2\over 2!} + {(3x)^4\over 4!} + \cdots \right)
= x - {9\over 2}x^3 + {27\over 8}x^5 + \cdots$$
This agrees with part A.
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\noindent
VII. Let $H$ be the horseshoe temperature.  The surrounding 
constant temperature is the water temperature $30$.
Newton's Law says:
$${dH\over dt} = k(H - 30)$$
for some constant of proportionality $k$.  So 
$$\eqalign{\int {dH\over H - 30} &= \int k\ dt\cr
\ln|H - 30| &= kt + c\cr
H - 30 &= b e^{kt} \qquad (b = \pm e^c)\cr 
H &= 30 + be^{kt}\cr}$$
We are given $H(0) = 100$, so $b = 70$.  When $t = 10$, $T(10) = 70$,
so 
$$70 = 30 + 70e^{10k} \Rightarrow k = {\ln(4/7)\over 10} \doteq -.005596$$
Then we must solve
$$35=30 + 70e^{-.005596 t}\Rightarrow t = {\ln(5/70)\over -.005596} \doteq 47$$
(47 minutes).

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