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\leftline{Mathematics 134 -- Intensive Calculus for Science 2}
\leftline{Practice Final Exam Solutions}
\bigskip
\noindent
I.  A)  The statement of the Fundamental Theorem of Calculus is:
If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative 
of $f$, then 
$$\int_a^b f(x)\ dx = F(b) - F(a)$$
\noindent
B)  The function $F$ is decreasing for $-2 < x < -1$ (since 
$f$ is $< 0$ there), increasing for $-1 < x < 1$ (since $f$
is $ > 0$ on that interval, constant for $2 < x < 3$,  and 
decreasing for $3 < x < 4$.  For the intervals where $f$ is 
horizontal, the slope of $F$ is constant (so that part of the
graph of $F$ is a part of a straight line).  
The values of $F$ at the points $-2,-1,0,1,2,3,4$
can be found from the Fundamental Theorem.  Since $F(0) = 0$ is
given, by considering areas we see:
$$\eqalign{F(0) - F(-2) &= \int_{-2}^0 f(x)\ dx = 0\ {\rm so}\ F(-2) = 0\cr
           F(0) - F(-1) &= \int_{-2}^0 f(x)\ dx = 1\ {\rm so}\ F(-1) = -1\cr
           F(1) - F(0) &= \int_{0}^1 f(x)\ dx = 1/2\ {\rm so}\ F(1) = 1/2\cr
           F(2) - F(0) &= \int_{0}^2 f(x)\ dx = 1/2\ {\rm so}\ F(2) = 1/2\cr
           F(3) - F(0) &= \int_{0}^3 f(x)\ dx = 3/2\ {\rm so}\ F(3) = 3/2\cr
           F(4) - F(0) &= \int_{0}^4 f(x)\ dx = 1/2\ {\rm so}\ F(4) = 1/2\cr}$$
Here is a sketch of the plot:
\vglue 1.5truein
\noindent
II.  
\item{A)}  By the power (and sum and constant multiple) rules
$$\int 5x^4 - \sqrt{x} + {3\over x}\ dx = x^5 - {2\over 3}x^{3/2} + 3\ln|x| + c$$
\item{B)}  Let $u = x^4 + 1$, so $du = 4x^3 + 1$.  The given integral 
is 
$$\eqalign{\int x^3 \sin(x^4 + 1)\ dx &= {1\over 4}\int \sin(u)\ du\cr
&= {-1\over 4}\cos(u) + c\cr
&= {-1\over 4}\cos(x^4 + 1) + c\cr}$$
\item{C)}  Integrate by parts twice, letting $u$ be the power
of $x$ and $dv = e^{-5x}dx$:
$$\eqalign{\int x^2 e^{-5x}\ dx 
&= {-x^2\over 5}e^{-5x} + {2\over 5}\int xe^{-5x}\ dx\cr
&= {-x^2\over 5}e^{-5x} + {2\over 5}\left({-x\over 5}e^{-5x} + {1\over 5}\int e^{-5x}\ dx\right)\cr
&= {-x^2\over 5}e^{-5x} + {2\over 5}\left({-x\over 5}e^{-5x} - {1\over 25}e^{-5x}\right) + c\cr
&= {-x^2\over 5}e^{-5x} - {2x\over 25}e^{-5x} - {2\over 125}e^{-5x} + c\cr}$$
(This can also be derived from entry 14 in the table of integrals,
taking $a = -5$.)
\item{D)}  Use entry 24 in the table, with $a = 3$, $u = 2x$.  We have
$$\int {1\over u^2 + a^2} = {1\over a}\arctan\left({u\over a}\right) + c$$
so 
$$\eqalign{
\int {1\over (2x)^2 + 3^2}\ dx &= {1\over 2}\int{1\over u^2 + 3^2}\ du\cr
&={1\over 2}\cdot {1\over 3}\arctan\left({x\over 3}\right) + c\cr
&={1\over 6}\arctan\left({x\over 3}\right) + c\cr}$$
\item{E)}  Let $u = \ln(x)$.  Then $du = {1\over x}dx$, so the given
integral is
$$\eqalign{\int {(\ln(x))^4 \over x}\ dx &= \int u^4\ du\cr
                                         &= {u^5\over 5} + c\cr
                                         &= {(\ln(x))^5\over 5} + c\cr}$$
\bigskip
\noindent
III.  A)  Rotating about the $x$-axis, we get disk cross-sections, so 
the volume is 
$$V = \int_0^4 \pi (x^{3/2})^2\ dx = \int_0^4 \pi x^3\ dx = \pi{x^4\over 4}|_0^4 = 64\pi$$
\item{B)}  Rotating about the line $y = -4$, we get ``washer'' cross-sections
with inner radius 4 and outer radius $4 + x^{3/2}$.  The volume
is
$$\eqalign{V &= \int_0^4 \pi (x^{3/2} + 4)^2 - \pi 4^2\ dx\cr
&= \pi \int_0^4 x^3 + 8x^{3/2} + 16 - 16\ dx\cr
&= \pi \left({x^4\over 4} + {16\over 5}x^{5/2}\right)|_0^4\cr
&= {832\pi\over 5} \doteq 522.76\cr}$$
\bigskip
\noindent
IV.  A)  Since the density is 0 for all $x < 0$, we have
$$\int_{-\infty}^\infty p(x)\ dx = \int_0^\infty .1e^{-.1x}\ dx = 
\lim_{b->\infty} -e^{-.1x}|_0^b = \lim_{b->\infty}(1 - e^{-.1b}) = 1$$
\item{B)}  The median time between cars is the solution $T$ of the 
equation $\int_{-\infty}^T p(x) = 1/2$.  As in part A, since $p(x) = 0$
if $x < 0$, this says
$$1/2 = \int_0^T .1e^{-.1x}\ dx = 1 - e^{-.1T}.$$
Hence $e^{-.1T} = 1/2$, so $T = \ln(1/2)/(-.1) \doteq 6.93$.
\bigskip
\noindent
V.  A) To compute the derivatives here, you need to recall the product
rule from last semester!  $f(x) = xe^x$, so $f'(x) = xe^x + e^x$,
$f''(x) = xe^x + e^x + e^x = xe^x + 2e^x$, and 
$f''(x) = xe^x + e^x + e^x + e^x = xe^x + 3e^x$.  Therefore the
3rd degree Taylor polynomial at $a = 0$ is
$$p_3(x) = f(0) + f'(0)x + {f''(0)\over 2} x^2 + {f''(0)\over 6} x^3 =
x + (2/2)x^2 + (3/6)x^3 = x + x^2 + x^3/2.$$
\item{B)}  Both $p_3(.1)$ and $(.1)e^{.1}$ equal $.1105$ to four decimal
places, so the agreement is close.
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\noindent
VI.  A) From the Taylor series for ${1\over 1-u}$ at $a = 0$, we have
${1\over 1 - u} = 1 + u + u^2 + u^3 + \cdots$.  Letting $u = -x^2$, 
we get
$${1\over 1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots$$
\item{B)}  We know $\int{1\over 1 + x^2}\ dx = \arctan(x)$.  Hence
to get the Taylor series for arctan we can integrate
series for ${1\over 1 + x^2}$ term by term.  The result is 
$$\arctan(x) = c + x - {x^3\over 3} + {x^5\over 5} - {x^7\over 7} + \cdots$$
for some constant $c$.  Since $\arctan(0) = 0$, $c$ must be zero, and 
$$\arctan(x) = x - {x^3\over 3} + {x^5\over 5} - {x^7\over 7} + \cdots$$
\item{C)}  The series for $g(x) = {\arctan(x) - x\over x^3}$ can be found
by algebraic manipulations of the series from part B:
$${\arctan(x) - x\over x^3} = {x - {x^3\over 3} + {x^5\over 5} + \cdots - x\over x^3}
= {-1\over 3} + {1\over 5}x^2 + \cdots$$
The coefficient of $x^2$ is the second derivative of $g(x)$ at zero, divided
by 2, so $g''(0) = {2\over 5} > 0$.  Hence $g$ is concave up at $0$.
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\noindent
VII.  A)  The number of calories needed per day to maintain weight is $20W$, so
if $I$ is the constant calorie intake per day, 
$${dW\over dt} = k(20W - I)$$
for some constant of proportionality $k$.
\item{B)}  Separating variables and integrating:
$$\eqalign{ \int {dW\over 20W - I} &= \int k\ dt\cr
            {1\over 20} \ln|20W - I| &= kt + c\cr
            20W - I &= d e^{20kt}\cr
            W = {I+ d e^{20kt} \over 20}\cr}$$
\item{C)}  Given $I = 3000$, and $k = {-1\over 3500}$.  When $t = 0$, $W(0) = 160$, so we have
$160 = {3000 + d\over 20}$, so $d = 3200 - 3000 = 200$.  Then the 
person's weight at the end of one year is 
$$W(365) = {3000 + 200e^{-365/3500}\over 20} \doteq 151.2$$
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\noindent
Extra Credit:  The idea here is to integrate the equations of 
the parabolas to find the areas of the given cross-sections, then
use those to derive a right-hand sum approximation of the integral
$\int_0^{20} A(x)\ dx$ with $\Delta x = 5$ to approximate the volume
enclosed by the hull.  For example, the area of the cross-section
at $x = 5$ is $\int_{-1}^1 0 - (2x^2 - 2)\ dx = 8/3$.  Doing the 
same for the other given cross-sections, the volume is estimated by
$$(8/3)\cdot 5 + 4\cdot 5 + (16/3)\cdot 5 + {32\over 3\sqrt{5}}\cdot 5\doteq 83.9 {\rm ft}^3$$



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