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\def\sol{\bigskip\item{} {\it Solution:\ }}
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\leftline{Mathematics 134 -- Intensive Calculus for Science 2}
\leftline{Exam I Solutions -- February 17, 2006}
\bigskip
\noindent
I.  The given graph shows $y = f(x)$. 
\vglue 1.5truein
\item{A)} (10)  Use the graph to determine $\int_0^4 f(x)\ dx$.
\sol  Using the interpretation of the integral as 
signed area, we see there is a triangle below the $x$-axis.
The region above the $x$-axis can be split into a small triangle,
a trapezoid ($2 \le x \le 3$) and another larger triangle:
$$\int_0^4 f(x)\ dx = -{1\over 2}\cdot {3\over 2}\cdot 1 + 
{1\over 2}\cdot {1\over 2}\cdot 1 + {1 + 2\over 2}\cdot 1 + {1\over 2}\cdot 1 \cdot 2 = 
2.$$
\item{B)}  (5) Let $F(x)$ be the antiderivative of $f(x)$
with $F(0) = -2$.  What is $F(4)$?
\sol By the Fundamental Theorem of Calculus
$$2 = \int_0^4 f(x)\ dx = F(4) - F(0) = F(4) + 2 \Rightarrow F(4) = 0$$ 
\item{C)}  (15) Give a qualitative sketch of $y = F(x)$
from part B.  Your graph should show where $F$ is 
increasing/decreasing, and where $F$ is concave up/concave down.
\sol  Proceeding as in part B we can get the following values:
$F(0) = -2$, $F(1) = -5/2$, $F(2) = - 5/2$, $F(3) = -1$ and $F(4) = 0$.
The graph of $F$ is decreasing on $0 < x < 3/2$, and increasing the 
rest of the way to $x = 4$.  It is concave down for $0 < x < 1$
and again for $3 < x < 4$.  It is concave up for $1 < x < 3$.  The 
graph looks like this:
\vglue 1.5 truein
\item{D)} (5) The function $G(x)$ is $G(x) = \int_0^x f(t)\ dt$
using the function $f$ in the graph above.  What is $G'(2)$?
\sol By the second part of the Fundamental Theorem,
$G'(2) = f(2) = 1$, from the given graph.  
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\noindent
II.  On the Moon, the acceleration due to gravity is about -1.6 m/sec${}^2$
(versus about -9.8 m/sec${}^2$ on Earth).  
\medskip
\item{A)} (10) A rock is tossed upward with initial velocity $v(0) = 1$
 m/sec.  
Solve the equation ${dv\over dt} = -1.6$ to find 
the velocity of the ball as a function of time. 
\sol  $v(t) = \int -1.6\ dt = -1.6 t + C$. From the initial condition
$v(0) = 1$, $C = 1$, so $v(t) = -1.6 t + 1$.
\item{B)} (10) Use your answer $v(t)$ from part A.  If the 
rock's initial height is $h(0) = 1$ m,  solve the equation 
${dh\over dt} = v(t)$ to find the height of the rock as a function 
of time.   
\sol The height is the integral of the velocity: 
$$h(t) = \int v(t)\ dt = \int  -1.6 t + 1\ dt = -0.8 t^2 + t + C$$
From the initial height $h(0) = 1$, we get $C = 1$, so 
$h(t) = -0.8 t^2 + t + 1$.
\item{C)}  (10) How long does the rock take to fall back 
to ground level on the Moon?  Would the corresponding time be longer
or shorter on the Earth?
\sol We solve the equation $h(t) = 0$ by the quadratic formula:
$$t = {1 \pm \sqrt{1 - 4(-.8)(1)}\over 2(-.8)} \doteq -.66, 1.91$$
We take the positive value $1.91$ sec.  This is {\it longer}
than the corresponding time on Earth.  (Intuitively, the Moon's 
gravity is ``weaker'' and pulls the rock down more slowly.)
\bigskip
\noindent
III.  Compute each of the integrals below using basic rules
and/or substitution.  You must show all work
for full credit. 
\medskip
\item{A)} (10) 
$$\int 3x^7 - {4\over x^3} + 5\sqrt{x} + {7\over x}\ dx$$
\sol  By the power rule (including the special case for $\int x^{-1}\ dx$)
$$ = {3\over 8} x^8 + 2 x^{-2} + {10\over 3}x^{3/2} + 7\ln|x| + C$$ 
\item{B)} (15)
$$\int_0^1 \sin(\pi x) + \cos(\pi x)\ dx$$
\sol  We apply the substitution $u = \pi x$, so $du = \pi dx$.  The integral
becomes
$$\eqalign{{1\over \pi} \int_{u=0}^{u=\pi} \sin(u) + \cos(u)\ du
&= {1\over \pi} \left(-\cos(u) + \sin(u)\right|_0^\pi \cr
&= {1\over \pi} (-\cos(\pi) + \sin(\pi) - (-\cos(0) + \sin(0)))\cr
&= {2\over \pi}\cr}$$
\item{C)} (10) 
$$\int x e^{5x^2 + 7}\ dx$$
\sol Use $u = 5x^2 + 7$, so $du = 10x dx$.  The integral is 
$${1\over 10}\int e^u\ du = {1\over 10} e^u + C = {1\over 10} e^{5x^2 + 7} + C$$
\bye