\magnification=\magstep1
\centerline{Mathematics 136 -- AP Calculus}
\centerline{Practice Exam 2 Solutions}
\centerline{October 29, 2003}
\bigskip
\noindent
I.  ({\it Note:} On second thought this one is probably more
complicated than a reasonable exam question would be; although
the algebra of solving for the critical point is not complicated,
thinking through the set-up could take some time.  It's good practice
though!)
\bigskip
Here's one way to set it up.  Let $x$ be the number of chairs
sold.  The revenue will be $R(x) = x\cdot P(x)$, where $P(x) = $
price per chair.  If $x \le 300$, then $R(x) = 90x$, so it is 
increasing (no critical points).  
For $x > 300$, the price per chair depends on the number of chairs
in the following way:  
$$P(x) = 90 - (.25)(x - 300),$$
so the revenue is 
$$R(x) = x(90 - .25(x - 300)) = -.25x^2 - 165x,$$
So to find the critical points, we take 
$$R'(x) = -.5 x - 165 = 0,$$
and hence $x = 330$.  This is a local maximum by the
2nd derivative test:  $R''(x) = -.5 < 0$.  The revenue at
$x = 330$ is $R(330) = (330)(90 - (.25)(30)) = \$27225$.  The
revenues at the endpoints of the interval are $R(0) = \$0$
and $R(400) = (400)(90 - (.25)(100)) = \$26000$.  Hence
the global maximum revenue is $\$27225$, and the minimum
is $\$0$.  
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\noindent
II.  
\item{A)} If $f$ is continuous on $[a,b]$, then
the definite integral of $f$ is the limit of the Riemann
sums for $f$: 
$$\int_a^b f(x)\ dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i^*) \Delta x$$
(it would also be OK to give the LHS or RHS here).
\item{B)}  The Fundamental Theorem of Calculus says:
\itemitem{part 1:}  If $f$ is continuous on $[a,b]$ and
$F$ is an antiderivative of $f$, then
$$\int_a^b f(x)\ dx = F(b) - F(a)$$
\itemitem{part 2:}  The function 
$$F(x) = \int_a^x f(t)\ dt$$
is an antiderivative of $f$:
$${d\over dx} \int_a^x f(t)\ dt = f(x).$$
\vglue 7.0truein
\noindent
V.  
\item{A)} (Should have said IV.17 -- oops!)
Let $u = \sin^{n-1}(x)$ and $dv = \sin x\ dx$,
so $du = (n-1)\sin^{n-2}(x)\cos(x)\ dx$ and $v = -\cos(x)$.
Then the integration by parts formula says
$$
\int \sin^n(x)\ dx = -\sin^{n-1}(x)\cos(x) + 
(n-1)\int \sin^{n-2}(x)\cos^2(x)\ dx$$
In the remaining integral, use the trig identity 
$\cos^2(x) = 1 - \sin^2(x)$, so 
$$\eqalign{\int \sin^n(x)\ dx &= -\sin^{n-1}(x)\cos(x) + 
(n-1)\int \sin^{n-2}(x)(1 - \sin^2(x))\ dx\cr
&= -\sin^{n-1}(x)\cos(x) + (n-1)\int \sin^{n-2}(x)\ dx - (n-1)\int\sin^n(x)\ dx\cr}$$
This is one of the cases where we get the same integral back, up 
to a constant.  Add $(n-1)\int \sin^n(x)\ dx$ to both sides to yield
$$n\int \sin^n(x)\ dx = -\sin^{n-1}(x)\cos(x) + (n-1)\int \sin^{n-2}(x)\ dx$$
Then divide by $n$ to get the table entry:
$$\int \sin^n(x)\ dx = {-1\over n}\sin^{n-1}(x)\cos(x) +
{(n-1)\over n}\int \sin^{n-2}(x)\ dx$$
\bigskip
\item{B)}  There are several ways to do this.  I think the most direct 
is to make a substitution first for the function in the exponent.
Let $u = x^2$, then $du = 2x\ dx$.  The $x^5\ dx$ is ${1\over 2}u^2\ du$
so the integrand is
$${1\over 2}\int u^2 e^{3u}\ du$$
Now use table entry III.14 (twice):
$$\eqalign{
{1\over 2}\int u^2 e^{3u}\ du &= {1\over 2}\left(
{u^2\over 3} e^{3u} - {2u\over 9}e^{3u} + {2\over 27}e^{3u}\right)+ C \cr
&={x^4\over 6}e^{3x^2} - {x^2\over 9}e^{3x^2} + {1\over 27} e^{3x^2} + C\cr}$$
\bigskip
\item{C)}  Let $u = sin(2x) + 4$, so $du = 2\cos(2x)$.  The form
is 
$${1\over 2}\int u^{1\over 3}\ du = {1\over 2}\cdot {3\over 4} u^{4\over 3} + C
= {3\over 8}(sin(2x) + 4)^{4\over 3} + C$$
\bigskip
\item{D)} The denominator is $4x^2 + 4x - 3 = (2x+1)^2 - 4$ so there are
real roots $2x + 1 = 2, 2x + 1 = -2$, or $x = 1/2, -3/2$.
To put this in the form to use number 27 in the table, we need to multiply
top and bottom by $1/4$ to make the leading coefficient 1:
$$\int {x+3\over 4x^2 + 4x - 3}\ dx = 
\int{ x/4 + 3/4 \over x^2 + x - 3/4}\ dx = 
\int{ x/4 + 3/4 \over (x - 1/2)(x + 3/2)}\ dx$$
So we use 27 with $c = 1/4, d = 3/4, a = 1/2, b = -1/2$. The antiderivative
is 
$$F(x) = {7\over 16}\ln(x - 1/2) - {3\over 16}\ln(x + 3/2)$$
For the definite integral, we take
$F(4) - F(2) \doteq .2859$.
\bye