
%rev151.tex: ``Answers to Prof. Rong's Review Problems (Math151, Fall 2003,
%Rutgers Univ.)
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\bf
\centerline
{
The Right Answers to The  Rong Questions:
}
\centerline
{Solutions to the Review Questions for Final Exam Math 151, Fall 2003,
Rutgers Univ}

\rm
\bigskip
\centerline{ Doron Zeilberger (``Dr. Z.'')}
 
{\bf Disclaimer:} I believe the answers to be correct, but I assume
no responsibility to any damage due to possible errors and/or typos.

{\bf 1.} For what values of $B$ is the following function
continuous at $x=1$?
$$
f(x)=\cases{ 3x^3-x^2+Bx, & if  $x>1$; \cr
             Bx-2, & if $x\leq 1$.\cr}
$$
{\bf Sol to 1.:} The limit from the left (at $x=1$), 
$\lim_{x \rightarrow 1^{-}}f(x)$ is
$3\cdot(1)^3-1^2+B\cdot 1=2+B$, whereas the limit from the right
$\lim_{x \rightarrow 1^{+}}f(x)$ is $B\cdot 1-2=B-2$.
In order for $f(x)$ to be continuous at $x=1$ we set these equal
and solve. $2+B=B-2$ getting $4=0$. This has no solutions.

{\bf Ans. to 1.:} None.

{\bf 2.} Find the equation of the tangent line to the curve
$y={{x-1} \over {x+1}}$ at each point where the tangent is
parallel  to the line $x-2y=2$.

{\bf Sol. to 2.} The line $x-2y=2$, can be written as $y=x/2-1$, hence
its slope is $1/2$. Now the slope of the tangent to the curve
$y={{x-1} \over {x+1}}$ is
$$
{{dy} \over {dx}}={{(x-1)'(x+1)-(x-1)(x+1)'} \over {(x+1)^2}}=
{{1\cdot(x+1)-(x-1)\cdot 1} \over {(x+1)^2}}=
{{2} \over {(x+1)^2}} \quad .
$$
Setting this equal to $1/2$ yields that the points we are interested in
satisfy
$$
{{2} \over {(x+1)^2}}={{1} \over {2}} \quad .
$$
Solving this we get
$(x+1)^2-4=0$ hence, $x^2+2x-3=(x+3)(x-1)=0$,
whose roots are $x=-3$ and $x=1$. Plugging-in into 
$y={{x-1} \over {x+1}}$, gives, 
when $x=-3$ $y=2$, and when $x=1$, $y=0$. So the points we need are
$(-3,2)$ and $(1,0)$.

{\bf Ans. to 2:} The tangent at the point $(-3,2)$ is
$(y-2)=(1/2)(x+3)$, which is $x-2y=-7$, and the
tangent at the point $(1,0)$ is $(y-0)=(1/2)(x-1)$,
which is $x-2y=1$.

{\bf 3.} Find the equation of the tangent line to the curve defined
by the equation $\ln(xy)+2x-y+1=0$ at the point 
$({{1} \over {2}}, 2)$.

{\bf Sol. to 3.} First simplify to get
$$
\ln x + \ln y +2x-y+1=0 \quad.
$$
By implicit differentiation
$$
{{1} \over {x}}+{{y'} \over {y}}+2-y'=0 \quad.
$$
Now plug-in $x=1/2$ and $y=2$, to get
$$
{{1} \over {1/2}}+{{y'} \over {2}}+2-y'=0 \quad,
$$
getting $4=y'/2$, and so the slope $y'$ (at the designated point)
is $8$. The equation of the tangent line at the designated
point $({{1} \over {2}}, 2)$ is
$(y-2)=8(x-{{1} \over {2}})$, which is $y=8x-2$.

{\bf Ans.:} $y=8x-2$.

{\bf 4.} Find the following limits
$$
a) \lim_{x \rightarrow -1} (x^2-2x+1)
\quad
b) \lim_{x \rightarrow \infty} {{3x^2-7} \over {\sqrt{x^2+2}}}
\quad
c) \lim_{x \rightarrow 0} {{\sin 3x} \over {2 \sin 5x}}
$$
$$
d) \lim_{x \rightarrow \infty} {{3e^x+4e^{-x}} \over {5e^x+4e^{-x}}}
\quad
e) \lim_{x \rightarrow 8^{-}} {{\vert x-8 \vert} \over {x-8}} \quad.
$$
{\bf Sol. to 4.}

{\bf (a)} Plugging in we get $(-1)^2-2(-1)+1=4$. 

{\bf Ans. to 4a):} $4$.

{\bf (b)} ``forgetting about the little ones'' we get
$$
\lim_{x \rightarrow \infty} {{3x^2-7} \over {\sqrt{x^2+2}}}=
\lim_{x \rightarrow \infty} {{3x^2} \over {\sqrt{x^2}}}
=\lim_{x \rightarrow \infty} {{3x^2} \over {x}}
=\lim_{x \rightarrow \infty} 3x= \infty \quad .
$$
{\bf Ans. to 4b):} $\infty$.

{\bf (c)} When the argument of $\sin$ goes to $0$, you can replace
$\sin w$ by $w$, hence
$$
\lim_{x \rightarrow 0} {{\sin 3x} \over {2 \sin 5x}}
=
\lim_{x \rightarrow 0} {{3x} \over {2 (5x)}}
=\lim_{x \rightarrow 0} {{3} \over {10}}
={{3} \over {10}} \quad.
$$
{\bf Ans to 4c)}: $3/10$.

{\bf d)} It is better to replace $e^x$ by another variable, let's call it
$y$.
$$
\lim_{x \rightarrow \infty} {{3e^x+4e^{-x}} \over {5e^x+4e^{-x}}}
=
\lim_{y \rightarrow \infty} {{3y+4/y} \over {5y+4/y}}
$$
Now `forgetting about the little ones' this equals
$$
\lim_{y \rightarrow \infty} {{3y} \over {5y}}=
{{3} \over {5}} \quad .
$$
{\bf Ans to 4d)}: $3/5$.

{\bf e)}  Recall that $\vert z\vert=-z$ for $z<0$, so
$\vert x-8 \vert=-(x-8)$ for $x<8$. We have
$$
\lim_{x \rightarrow 8^{-}} {{\vert x-8 \vert} \over {x-8}}=
\lim_{x \rightarrow 8^{-}} { {-(x-8)} \over {x-8}}=
\lim_{x \rightarrow 8^{-}} (-1)= -1 \quad.
$$
{\bf Ans. to 4e):} $-1$.

{\bf 5.} Find the following limits
$$
a)\quad \lim_{x \rightarrow 0^{+}} \sqrt{x} \ln x
\quad\quad
b) \quad \lim_{x \rightarrow 0^{+}} x^{\sin x}
\quad\quad
c) \quad \lim_{x \rightarrow \infty} {{ (\ln x)^2 } \over {x}} \quad .
$$

{\bf Sol. to 5} 
{\bf a)} We use L'H\^opital:
$$
\lim_{x \rightarrow 0^{+}} \sqrt{x} \ln x=
\lim_{x \rightarrow 0^{+}}  {{\ln x} \over {x^{-1/2}}  }=
\lim_{x \rightarrow 0^{+}}  {{(\ln x)'} \over {(x^{-1/2})'}  }=
\lim_{x \rightarrow 0^{+}}  {{1/x} \over {(-1/2)x^{-3/2}}  }=
$$
$$
\lim_{x \rightarrow 0^{+}}  {{x^{1/2}} \over {(-1/2)}  }=
\lim_{x \rightarrow 0^{+}}  -2\sqrt{x}  =
\lim_{x \rightarrow 0^{+}}  -2\sqrt{0}  =0\quad .
$$
{\bf Ans. to 5a)}: $0$.

b) We must first take the $\ln$, getting
$\ln (x^{\sin x})=\sin x \ln x$, and writing it as a quotient
we get ${{\ln x} \over {\csc x}}$. Taking the
limit of this we have:
$$
\lim_{x \rightarrow 0^{+}} {{\ln x} \over {\csc x}}=
\lim_{x \rightarrow 0^{+}} {{(\ln x)'} \over {(\csc x)'}}=
\lim_{x \rightarrow 0^{+}} {{1/x} \over {-\cot x \csc x}}=
$$
$$
\lim_{x \rightarrow 0^{+}} {{\tan x \sin x} \over {x}}=
\lim_{x \rightarrow 0^{+}} {{\sin x \sin x} \over {(\cos x) x}}=
\lim_{x \rightarrow 0^{+}} {{x^2} \over { x}}=
\lim_{x \rightarrow 0^{+}} x=0 \quad .
$$
Now the limit of the original is the exponential of this

{\bf Ans. to 5b)}: $e^0=1$.

{\bf Shortcut:} When $x$ is near $0$, $\sin x$ can be replaced by
$x$, so we get the simpler limit $\lim_{x \rightarrow 0^{+}} x^x$
whose $\ln$ is $x \ln x $, and the limit
$$
\lim_{x \rightarrow 0^{+}} x \ln x=
\lim_{x \rightarrow 0^{+}}  {{\ln x} \over {x^{-1}} }=
\lim_{x \rightarrow 0^{+}}  {{(\ln x)'} \over {(x^{-1})'}}=
$$
$$
\lim_{x \rightarrow 0^{+}}  {{1/x} \over {-x^{-2}}}=
\lim_{x \rightarrow 0^{+}}  (-x) =0 \quad .
$$
and we get quicker {\bf Ans.}: $e^0=1$.

{\bf c)} By L'H\^opital again:
$$
\lim_{x \rightarrow \infty} {{ (\ln x)^2 } \over {x}}=
\lim_{x \rightarrow \infty} {{ ((\ln x)^2)' } \over {x'}}=
\lim_{x \rightarrow \infty} {{ 2(\ln x)(1/x) } \over {1}}=
\lim_{x \rightarrow \infty} {{ 2\ln x } \over {x}} \quad.
$$
Using L'H\^opital once again, we have that this equals
$$
2\lim_{x \rightarrow \infty} {{ (\ln x)' } \over {x'}}=
2\lim_{x \rightarrow \infty} {{ 1/x } \over {1}}=
2\lim_{x \rightarrow \infty} {{ 1 } \over {x}}= 0 \quad.
$$
{\bf Ans to 5c)}: $0$.

{\bf 6.} Find $y'$ in each case. Do not simplify your answer.
$$
a) \quad y=x^7-3x+6-{{1} \over {x^4}} \quad
b) \quad y=x^{5x} \sin (x^2) \quad
c) \quad y={{2 \tan x} \over {\sqrt{1-x^2}}} 
$$
$$
d) \quad x^4 y+ 5y^6x^3=8 \quad 
e) \quad xe^{xy+3y}=y \quad f) \quad f(x)=(1+2x)^{1/x} \quad .
$$
{\bf Sol. to 6.}

{\bf a)}.
$$
y'=(x^7-3x+6-{{1} \over {x^4}})'=
(x^7-3x+6-x^{-4})'=
7x^6-3-(-4)x^{-5}=
7x^6-3+{{4} \over {x^5}} \quad .
$$
{\bf Ans. to 6a)}: $7x^6-3+{{4} \over {x^5}}$.

{\bf b)}
$$
y'=(x^{5x} \sin (x^2))'=
(x^{5x})' \sin (x^2)+x^{5x} (\sin (x^2))'=
(e^{5x\ln x})' \sin (x^2)+x^{5x} (\sin (x^2))'=
$$
$$
(5x\ln x)'(e^{5x\ln x}) \sin (x^2)+x^{5x} (2x) \cos (x^2)=
$$
$$
(5\ln x+ 5)(e^{5x\ln x}) \sin (x^2)+x^{5x} (2x) \cos (x^2)=
$$
$$
(5\ln x+ 5)x^{5x} \sin (x^2)+x^{5x} (2x) \cos (x^2) \quad .
$$

{\bf Ans. to 6b)}:
$(5\ln x+ 5)x^{5x} \sin (x^2)+x^{5x} (2x) \cos (x^2)$.

{\bf c)}
$$
y'=({{2 \tan x} \over {\sqrt{1-x^2}}})'=
{{(2 \tan x)' \sqrt{1-x^2}- 
(2 \tan x) ((1-x^2)^{1/2})'
} \over {(\sqrt{1-x^2})^2}}=
$$
$$
{{2 \sqrt{1-x^2}\sec^2 x - 
(2 \tan x) (1/2)(1-x^2)^{-1/2}\cdot(-2x)
} \over {1-x^2}}=
$$
$$
{{2 \sqrt{1-x^2} \sec^2 x 
+2x(1-x^2)^{-1/2}\tan x 
} \over {1-x^2}}=
{
{2\sec^2 x } \over {\sqrt{1-x^2}}}
 +{{2x \tan x} \over {\sqrt{1-x^2}^3}}
$$

{\bf Ans, to 6c)}:
${{2\sec^2 x } \over {\sqrt{1-x^2}}}+{{2x \tan x} \over {\sqrt{1-x^2}^3}}$.

{\bf d)}
$$
( x^4 y+ 5y^6x^3)'=0 \quad,
$$
so
$$
(x^4)'y+x^4y'+5y^6(x^3)'+5(y^6)'x^3=0 \quad,
$$
$$
4x^3y+x^4y'+5y^6(3x^2)+5(6y^5y')x^3=0 \quad,
$$
$$
4x^3y+x^4y'+15y^6x^2+30y^5y'x^3=0 \quad.
$$
Moving all the terms not involving $y'$ to the right gives
$$
x^4y'+30x^3y^5y'=-4x^3y-15y^6x^2 \quad.
$$
$$
(x^4+30x^3y^5)y'=-4x^3y-15y^6x^2 \quad.
$$
Hence
$$
y'={{-4x^3y-15y^6x^2} \over {x^4+30x^3y^5} }
$$
Since we are not supposed to simplify, we have

{\bf Ans. to 6d):} $y'={{-4x^3y-15y^6x^2} \over {x^4+30x^3y^5} }$.

{\bf e)}:
$$
(xe^{xy+3y})'=y' \quad,
$$
$$
x'e^{xy+3y} + x(e^{xy+3y})'=y' \quad,
$$
$$
e^{xy+3y} + xe^{xy+3y}\cdot (xy+3y)'=y' \quad,
$$
$$
e^{xy+3y} + xe^{xy+3y}\cdot (x'y+xy'+3y')=y' \quad,
$$
$$
e^{xy+3y} + xe^{xy+3y}\cdot(y+(x+3)y')=y' \quad,
$$
$$
e^{xy+3y} + xye^{xy+3y}+
(x+3)xy'e^{xy+3y}=y' \quad
$$
$$
(1+xy)e^{xy+3y}=
y'(1-x(x+3)e^{xy+3y}) \quad
$$
Hence
$$
y' ={{e^{xy+3y}(1+xy)} \over {1-x(x+3)e^{xy+3y}}} \quad .
$$

{\bf Ans. to 6e)}:
$y' ={{e^{xy+3y}(1+xy)} \over {1-x(x+3)e^{xy+3y}}}$.

{\bf f)}
$$
(\ln f(x))'= ({{\ln(1+2x)} \over {x}})'
$$
$$
{{f'(x)} \over {f(x)}}=
{{(\ln(1+2x))'x-\ln(1+2x) \cdot x'} \over {x^2}}
=
{{(2x/(1+2x))-\ln(1+2x)} \over {x^2}}
$$
Hence
$$
f'(x)=f(x){{(2x/(1+2x)-\ln(1+2x)} \over {x^2}}=
(1+2x)^{1/x} {{2x/(1+2x)-\ln(1+2x)} \over {x^2}} \quad .
$$

{\bf Ans. to 6f):}  $(1+2x)^{1/x} {{2x/(1+2x)-\ln(1+2x)} \over {x^2}}$.

{\bf 7.} State the formal definition of the derivative of
the function $f(x)$. Use the definition to calculate
$f'(x)$ for $f(x)=\sqrt{3-5x}$.

{\bf Sol. to 7.}:

The formal definition of the derivative of a function $f(x)$ is
$$
f'(x)=\lim_{s \rightarrow x} {{f(s)-f(x)} \over {s-x}} \quad.
$$
{\bf Note:} There is another version with $h$ and 
$\lim_{h \rightarrow 0}$, but for {\it this} problem it is more
convenient to use this one.

$$
f'(x)=\lim_{s \rightarrow x} {{\sqrt{3-5s}-\sqrt{3-5x}} \over {s-x}}
=
\lim_{s \rightarrow x} {{(\sqrt{3-5s}-\sqrt{3-5x})
(\sqrt{3-5s}+\sqrt{3-5x})
} \over {(s-x)(\sqrt{3-5s}+\sqrt{3-5x})}}
$$
$$
=
\lim_{s \rightarrow x} {{(\sqrt{3-5s})^2-(\sqrt{3-5x})^2} 
\over {(s-x)(\sqrt{3-5s}+\sqrt{3-5x})}}
=
\lim_{s \rightarrow x} {{(3-5s)-(3-5x)} 
\over {(s-x)(\sqrt{3-5s}+\sqrt{3-5x})}}
$$
$$
=
\lim_{s \rightarrow x} {{-5(s-x)} 
\over {(s-x)(\sqrt{3-5s}+\sqrt{3-5x})}}
=
$$
$$
\lim_{s \rightarrow x} {{-5} 
\over {\sqrt{3-5s}+\sqrt{3-5x}}}=
{{-5} \over {\sqrt{3-5x}+\sqrt{3-5x}}}=
{{-5} \over {2\sqrt{3-5x}}} \quad.
$$

{\bf Ans. to 7.} ${{-5} \over {2\sqrt{3-5x}}}$
(but it had to be derived from the definition!).

{\bf 8.} Suppose that $S(x)=\sqrt{x}$ for $x\geq 0$, and
let $f$ and $g$ be differentiable functions about which the
following is known
$$
f(3)=2 ,\quad f'(3)=7, \quad  g(3)=4 , \quad g'(3)=5 \quad .
$$
Compute the following
$$
(f+g)'(3) \quad (f \cdot g)'(3) ,\quad
\left ( {{f} \over {g}} \right )' (3) ,\quad
(S \circ g)'(3) ,\quad 
\left ( {{f\cdot g} \over {f-g}} \right )' (3) \quad .
$$
{\bf Sol. to 8.}
$$
(f+g)'(3)=f'(3)+g'(3)=7+5=12 \quad.
$$
$$
(f\cdot g)'(3)=f'(3)\cdot g(3)+f(3) \cdot g'(3)=
7\cdot 4+2\cdot 5=38 \quad.
$$
$$
\left ( {{f} \over {g}} \right )' (3) =
 {{f'(3)g(3)-f(3)g'(3)} \over {g(3)^2}}  =
 {{7 \cdot 4-2 \cdot 5} \over {4^2}} =
 {{18} \over {16}} = {{9} \over {8}} \quad.
$$
$$
(S\circ g)'(3)=S'(g(3))g'(3)=S'(4) \cdot 5 =5 \cdot (1/2)(4)^{-1/2}=5/4 \quad.
$$
$$
\left ( {{ f \cdot g}
\over {f-g}} \right )' (3) =
$$
$$
\left ( {{(f\cdot g)'(f-g)-(f\cdot g)(f-g)'}
\over {(f-g)^2}} \right ) (3)=
$$
$$
\left ( {{(f'g+fg')(f-g)-(f\cdot g)(f'-g') }
\over {(f-g)^2}} \right ) (3) =
$$
$$
\left ( {{f^2g'-f'g^2}
\over {(f-g)^2}} \right ) (3) =
$$
$$
 {{f(3)^2g'(3)-f'(3)g(3)^2}
\over {(f(3)-g(3))^2}} =
$$
$$
 {{2^2\cdot 5- 7 \cdot 4^2}
\over {(2-4)^2}} =-23 \quad.
$$

{\bf Ans. to 8}: $12,38,9/8,5/4,-23$.

{\bf 9.} Suppose $f''(x)=-3x+\cos (\pi x)$ and $f(1)=2$ and
$f'(1)=-1$. What is $f(5)$?

{\bf Sol. to 9.}: The general antiderivative of 
$-3x+\cos (\pi x)$ is $f'(x)=-3x^2/2+\sin(\pi x)/\pi+A$,
where $A$ is a general constant. To find $A$, we plug-in
$x=1$, and get $f'(1)=-3\cdot 1^2/2+\sin(\pi)/\pi+A=-3/2+0+A$.
On the other hand, in this specific problem,
we are told that $f'(1)=-1$, so we get the equation $-1=A-3/2$,
and solving for $A$, we get $A=1/2$.
So 
$$
f'(x)=-(3/2)x^2+\sin(\pi x)/\pi+1/2 \quad .
$$

To get $f(x)$, we take the antiderivative of the above, getting
$f(x)=-(1/2)x^3-\cos(\pi x)/\pi^2+x/2+B$, where $B$ is a constant
yet to be determined. Plugging-in $x=1$ into this yields
$f(1)=-1/2-\cos(\pi)/\pi^2+1/2+B=1/\pi^2+B$. 
On the other hand, in this specific problem,
we are told that $f(1)=2$, so we get the equation $2=1/\pi^2+B$,
so $B=2-1/\pi^2$. Hence
$$
f(x)=-(1/2)x^3-\cos(\pi x)/\pi^2+x/2+2-1/\pi^2
$$
Finally, plugging-in $x=5$, we get
$f(5)=-(1/2)5^3-\cos(5 \pi )/\pi^2+5/2+2-1/\pi^2=$
$-125/2+1/\pi^2+5/2+2-1/\pi^2=-120/2+2=-60+2=-58$.

{\bf Ans. to 9:} -58.

{\bf 10.} A farmer with $450$ feet of fencing wants to enclose the
four sides of a rectangular region into four pens with fencing
parallel to one side of the rectangle. What is the largest possible
total area of the four pens?

{\bf Sol. to 10.} Let's call the length of the side parallel to the
three internal walls $x$, and the length of the other side
(perpendicular to these walls) $y$. 

We have  to {\bf maximize} the {\bf area}. So the
{\bf Goal} function is $xy$. We also have the
{\bf constraint}: $2y+5x=450$. Solving for $y$, we get
$y=(450-5x)/2$, and plugging this into the goal function, we
express it in terms of $x$ alone, getting
$f(x)=x(450-5x)/2=(450/2)x-(5/2)x^2$. To find the
maximum, we take the derivative
$f'(x)=450/2-5x$, and then set it equal to $0$.
Solving $450/2-5x=0$, we get $x=45$. 
Plugging this into $f(x)$ we get that the maximal 
total area is $f(45)=45 \cdot (450-5\cdot 45)/2=45^2\cdot 5/2=10125/2$.

{\bf Ans to 10}: $10125/2$.

{\bf Note:} In some problems, they might ask you about
the {\bf dimensions} of the optimal rectangle. Then
the answer would be $45 \times (225/2)$.

{\bf 11.} A $15$ feet long ladder is leaning against a wall when its base
begins to slide along the floor, away from the wall. By the
time the base is $12$ ft. away from the wall, the base is moving
at the rate of $5$ ft/sec. How fast is the top of the ladder sliding
down the wall then? How fast is the area of the triangle formed by
ladder, wall, and floor changing at that time?

{\bf Sol. to 11.} This is a related rates problem. Let's call the
height of the top of the ladder $y$, and the distance of the
foot of the ladder from the wall $x$. By
Pythagoras,
$$
x^2+y^2=15^2 \quad.
$$
Taking derivatives with respect to time (using the chain rule in
its abstract setting, a.k.a. implicit differentiation), we get
$$
2x {{dx} \over {dt}}+2y {{dy} \over {dt}}=0 \quad.
$$
Right now $x=12$ and $dx/dt=5$. We also need $y$ right now.
Using the relation, $x^2+y^2=15^2$ we plug-in $x=12$ and solve for
$y$: $12^2+y^2=15^2$, getting $y=9$. Substituting these values
into the derived relation we get
$$
2\cdot 12 \cdot 5+2 \cdot 9 {{dy} \over {dt}}=0 \quad,
$$
and solving for $dy/dt$, we get $dy/dt=-20/3=-6{{2} \over {3}}$.
Since the sign is negative, it is sliding {\bf down}
(which is also obvious from common sense).

{\bf Ans. to the first part of 11:} The top of the ladder is sliding
down at a speed of $20/3=6{{2} \over {3}}$ ft/sec.

The {\bf area}, let's call it $A$, is given by the relation
$A=xy/2$. Taking derivative with respect to time 
(using the product rule), we get
$$
{{dA} \over {dt}}=
{{1} \over {2}} \left(x {{dy} \over {dt}}+ y {{dx} \over {dt}} \right ) \quad,
$$
and we plug-in the above numbers, getting
$$
{{dA} \over {dt}}=
{{1} \over {2}} \left(12 \cdot (-20/3)
+ 9\cdot 5 \right ) =-35/2\quad,
$$

{\bf Ans. to second part of 11):} The area is decreasing at a rate
of $17 {{1} \over {2}}$ square feet per second.

{\bf 12.} Let $f(x)={{3x} \over {x^2-1}}$. Find the domain of
the function, the intervals where $f(x)$ is increasing or decreasing,
maximum and minimum, the concavity and inflection points, horizontal
and vertical asymptotes of the graph. Then sketch the graph of
$f(x)$.

{\bf Sol. to 12.} Domain: $\{x \vert x \neq -1,1\}$.
Even though  the horizontal, and vertical asymptotes are mentioned
{\bf last} in the statement of the problem, it is best to
start with them, since they contain the most important information.

{\bf horizontal asymptotes:}
$$
\lim_{x \rightarrow \infty} f(x)=
\lim_{x \rightarrow \infty} {{3x} \over {x^2-1}}=
\lim_{x \rightarrow \infty} {{3x} \over {x^2}}= 0 \quad.
$$
So $y=0$ (alias the x-axis) is a horizontal asymptote.

{\bf vertical asymptotes}: Set the denominator equal to $0$,
we get $x=-1$ and $x=1$.

Let's consider each vertical asymptote on its own.
Plugging-in a value of $x$ slightly less than $-1$, say
$-1.01$, we get $-BIG$, so this means that
$\lim_{x \rightarrow -1^{-}}f(x)=-\infty$, and the
curve would drop down to ``hell'' ($-\infty)$ as it
approaches the vertical asymptote $x=-1$.
Plugging-in a value of $x$ slightly bigger than
$-1$, say $-.99$ we get that $f(-.99)=BIG$, so
the curve ``comes down from heavens'' right after $x=-1$.
Similarly the curve goes down to hell as $x$ approaches $x=1$,
and emerges from heavens right after $x=1$.

It is also a good idea to find the $x$-intercept
(solve $y=0$, getting $(0,0)$) and the $y$-intercept,
plugging-in $x=0$, getting, once again $(0,0)$. So the
curve intersects the $x$-axis only at the origin $(0,0)$.

Notice that we haven't taken derivatives yet, but we can already
pretty much draw the function. When $x$ is very negative,
it is almost coincident with the $x$-axis (the horiz. Asymptote)
but it is below the $x$-axis, since $f(x)$ is negative there,
and then it is dropping down-down right before $x=-1$.
By inspection you can see that the curve is {\bf concave down}
at the interval $-\infty <x \leq -1$

Then from $x=-1$ to $x=1$ it is a steady decrease (since 
at $x=-1$ it is coming from $y=\infty$ dropping to
$y=-\infty$ at $x=1$. Somewhere between these, there should be an inflection
point, that by symmetry is $(0,0)$. Also for $x>1$ it
dropping from $\infty$ to $0$. So you can draw the curve without any
calculus! Also it is clear that there are no local maxima and
minima.

But now we can do some calculus.

By the quotient rule
$$
f'(x)= -3{{x^2+1} \over {(x^2-1)^2}} \quad,
$$
this is always negative, so 
{\bf intervals of increase:} none.

{\bf intervals of decrease:} $(-\infty,-1)$,$(-1,1)$, $(1,\infty)$.


To find potential (local) max or min. points, solve
$f'(x)=0$, which has no roots. Hence

{\bf max and min. points}: none.

Using the quotient rule one more time, we get
$$
f''(x)= {{6x(x^2+3)} \over {(x^2-1)^3}} \quad.
$$

The inflection point(s) are where $f''(x)=0$. Setting the
numerator equal to $0$, we get $x=0$. The $y$ coordinate is
$f(0)=0$. So

{\bf inflection point}: $(0,0)$. 

The transition points in the life of this function are
$x=-1$,$x=0$, and $x=1$. Plugging-in random values in each
interval we get

The interval $(-\infty, -1)$ is {\bf concave down} (since. e.g.
$f''(2)<0$).

The interval $(-1, 0)$ is {\bf concave up} (since. e.g.
$f''(-{{1} \over {2}})>0$).

The interval $(0,1)$ is {\bf concave down} (since. e.g.
$f''({{1} \over {2}})<0$).

The interval $(1,\infty)$ is {\bf concave up} (since. e.g.
$f''(2)>0$).

{\bf Ans. to 12.}:  max. and min. points: none;
intervals of increase: none;
intervals of decrease: $(-\infty,-1),(-1,1),(1,\infty)$;
inflection point: $(0,0)$; 
intervals of concave down:$(-\infty,-1)$, $(0,1)$;
intervals of concave up:$(-1,0)$, $(1,\infty)$;
horiz. asymptote: $y=0$; vertical asymptotes: $x=-1$ and $x=1$.


{\bf 13.} Repeat the same problem for $y={{x^2} \over {x^2+3}}$.

{\bf Sol. to 13.}:

Horiz. Asymptotes: $y=1$. Vertical asymptote: none.
By the quotient rule (do it!)
$$
f'(x)={{6x} \over {x^2+3}}
$$
$$
f''(x)={{-18(x^2-1)} \over {x^2+3}}
$$
So potential max or min.: $x=0$. Plugging into the second
derivative, we get $f''(0)=6>0$ so $x=0$ is a {\bf minimum}.
Plugging $x=0$ into $f(x)$, we get that when $x=0$, $y=0$,
so the {\bf point} $(0,0)$ is a {\bf local minimum}, and that's
the only one.

To get the inflection points solve $f''(x)=0$, and
get $x=-1$ and $x=1$. When $x<-1$, $f''(x)$ is negative
so in the interval $-\infty < x < -1$ the curve is
{\bf concave down}. In the interval
$-1 < x < 1$ $f''(x)$ is positive, so the curve is {\bf concave up},
and in the interval $ 1< x < \infty$ the curve is
{\bf concave down} since $f''(x)$ is negative there.

Now draw it!

{\bf Ans. to 13}: Horiz. asymptote $y=1$; Vertical asymptotes: none.
local min. $(0,0)$; inflection points: $(-1,1/4)$ and
$(1,1/4)$; interval of decrease: $(-\infty,0)$;
interval of increase: $(0,\infty)$;
intervals of concave down:$(-\infty,-1),(1,\infty)$;
interval of concave up: $(-1,1)$.

{\bf 14.} Sketch the graph of $f(x)$ which satisfies the
following conditions $f'(1)=f'(-1)=0$, $f'(x)<0$ if
$\vert x\vert<1$, $f'(x)>0$ if $1<\vert x\vert<2$,
$f'(x)=-1$ if $\vert x \vert >2$, $f''(x)<0$ if
$-2<x<0$, inflection point $(0,1)$.

{\bf Sol. to 14.} First let's spell everything out w/o the annoying
absolute value symbol $\vert \vert$.

For $-\infty<x <-2: f'(x)=-1$.

For $-2<x <-1: f'(x)>0$.

For $-1<x <1: f'(x)<0$.

For $1<x <2: f'(x)>0$.

For $2<x <\infty: f'(x)=-1$.

Now we are almost ready to sketch, noting that at $x=-2$ and
$x=1$ we have local minima, since the derivative changes sign from
negative to positive, and at $x=-1$ and $x=2$ we have local maxima,
since $f'(x)$ changes sign from positive to negative.
We also have an inflection point at $(0,1)$ and since
$f''(x)<0$ between $x=-2$ and $x=0$, it is concave down there,
and hence between $x=0$ and $x=2$ it is concave up.

To actually draw a curve fitting all these
specifications, draw a straight line with slope
$-1$ coming from $-\infty$ to the point $(-2,0)$, then draw
a hat-like curve that peaks at $x=-1$ and inflects at
the point $(0,1)$ turning into a cup-like arc that bottoms
at $x=1$ and goes up to $x=2$ and then it suddenly becomes
a straight line with slope $-1$.

{\bf 15.} Find the linearization of $f(x)=\sqrt{x+1}$ at
$a=15$ and use it to find an approximation to
$\sqrt{15}$ and $\sqrt{17}$ . Use the second derivative  to determine
whether the estimate is greater or less than the actual value.

{\bf Sol to 15.} The linearization of a function $f(x)$ at $x=a$ is
$L(x)=f(a)+f'(a)(x-a)$. In this  problem
$f(x)=(x+1)^{1/2}$, so 
$$
f'(x)=(1/2)(x+1)^{-1/2}={{1} \over {2\sqrt{x+1}}} \quad.
$$
We have $f(a)=f(15)=\sqrt{1+15}=\sqrt{16}=4$,
and $f'(a)=f'(15)={{1} \over {2\sqrt{15+1}}} =1/8$.
Hence

{\bf Ans. to first part of 15}: $L(x)=4+{{1} \over {8}}(x-15)$ .

Now $\sqrt{15}=f(14)$, and plugging  $x=14$, we get
$L(14)=4+{{1} \over {8}}(14-15)={{31} \over {8}}=3{{7} \over {8}}$,
hence 

{\bf Ans. to second part of 15}:  
$\sqrt{15}$ is approximately $3{{7} \over {8}}.$

Also $\sqrt{17}=f(16)$, and plugging $x=16$, we get
$L(16)=4+{{1} \over {8}}(16-15)={{33} \over {8}}=4{{1} \over {8}}$,
hence 

{\bf Ans. to third part of 15}:  
$\sqrt{17}$ is approximately $4{{1} \over {8}}.$

$$
f''(x)=(-1/4)(x+1)^{-3/2}={{-1} \over {4\sqrt{x+1}^3}} \quad.
$$
Since $f''(15)=-1/256$ is {\bf negative}, the above approximations
to $\sqrt{15}$ and $\sqrt{17}$ are {\bf more} than the actual value.

{\bf Ans. to fourth part of 15}:  both are more than the actual values.

{\bf 16.} Find the derivative of the following functions:
$$
a) \quad F(x)=\int_{\pi}^{x} \tan (s^2) ds
\quad\quad
b)\quad g(x)=\int_{1}^{\cos x} \sqrt{1-t^2} dt
\quad .
$$

{\bf Sol to 16.a)}  Since the upper-limit of integration
is {\bf identical} to the argument of $F$, (both are $x$), the
derivative is simply the integrand with $s$ replaced by $x$.

{\bf Ans. to 16a)}: $F'(x)=\tan(x^2)$.

{\bf Sol to 16b)} Now the upper-limit of integration, 
$\cos x$ is {\bf not} identical with the argument of $g$, which
is $x$. We must first use the chain rule, multiplying top and
bottom by $d\cos x$ (so to speak!).
$$
g'(x)= {{d} \over {x}} \int_{1}^{\cos x} \sqrt{1-t^2} dt=
{{d \cos x} \over {dx}} \cdot
{{d} \over {\cos x}} \int_{1}^{\cos x} \sqrt{1-t^2} dt \quad .
$$

The first piece is a straightforward differentiation, while the second
is done exactly as in a): replace $t$ by $\cos x$ in the integrand.
Hence $F'(x)=-\sin x \sqrt{1-\cos x^2}$. Using the famous trig. identity
this is $-\sin^2 x$.

{\bf Ans. to 16b)}: $g'(x)=-\sin^2 x$.

{\bf 17.} Suppose that $f$ is continuous on $[0,4]$, $f(0)=1$
and $2 \leq f'(x) \leq 5$ for all $x \in (0,4)$. Show that
$9 \leq f(4) \leq  21$.

{\bf Sol. to 17.} We must use the {\bf Mean Value Theorem}.
If $f(x)$ is continuous in $[a,b]$ and differentiable in
$(a,b)$ then there exists a $c$ in $(a,b)$ such that
$$
f'(c)= {{f(b)-f(a)} \over {b-a}} \quad.
$$
In this problem $a=0$ and $b=4$ so, there exists a $c$ in $(0,4)$ such that
$$
f'(c)= {{f(4)-f(0)} \over {4-0}} =
{{f(4)-1} \over {4-0}} 
$$
But we know that $2 \leq f'(x) \leq 5$ for {\it every} $x \in (0,4)$,
in particular for the $c$, whatever it is. Hence
$$
2 \leq {{f(4)-1} \over {4}} \leq 5 \quad.
$$
Multiplying by $4$, we get (since $4$ is positive, the inequalities are
preserved) $8 \leq f(4)-1 \leq 20$, and adding $1$ everywhere,
we get $9 \leq f(4) \leq 21$.

{\bf 18.} Let $f(x)=3x^3-2x^2+x-1$. Show that $f(x)$ must have
real roots in $[0,1]$.

{\bf Sol. to 18.} The given function
 $f(x)$ is continuous (because it is a polynomial),
Plug-in the leftmost end, $0$, into $f(x)$, getting
$f(0)=-1$. Plug-in the rightmost end, $1$, into $f(x)$, getting
$f(1)=3 \cdot 1^3-2 \cdot 1^2+1-1=1$. Since these two numbers
$-1$ and $1$ have {\bf opposite signs} (one is positive and the
other is negative), it follows by the {\bf intermediate value theorem},
that there must be a $c \in (0,1)$ such that $f(c)=0$. In other
words, $f(x)=0$ must have a real root in $[0,1]$.

{\bf 19.} Show that the equation $x^{101}+x^{51}+x-1=0$ has
exactly one real root.

{\bf Sol. to 19}: Let $f(x)=x^{101}+x^{51}+x-1$.
Since $\lim_{x \rightarrow -\infty} f(x)=-\infty$ and
$\lim_{x \rightarrow \infty} f(x)=\infty$ 
(because the polynomial $f(x)$ has {\bf odd degree}), and
these have {\bf opposite signs} (like in problem 18), there
must be {\bf at least} one root in $(-\infty,\infty)$.

Taking derivative, $f'(x)=101x^{100}+51x^{50}+1$. Since
all the powers are even, and the coefficients are {\bf positive}
and the constant term ($1$) is {\bf strictly positive},
$f'(x)$ is {\bf strictly positive}. If there would have been
more than one root, by Rolle's theorem there would have been
a $c$ such that $f'(c)=0$, contradicting the fact that $f'(x)$ is
always positive. Hence  there is {\bf exactly one real root}.

{\bf 20.} Use two steps of Newton's method to approximate the root of the
equation $x^4+x-4=0$ in the interval $[1,2]$.

{\bf Sol. to 20}, Recall that {\bf Newton's method} is:
$$
x_{n}=x_o- {{f(x_o)} \over {f'(x_o)}} \quad,
$$
where $x_0$ is the {\bf old} approximation and $x_n$ is
the {\bf new} approximation.
Implementing this here, we get $f(x)=x^4+x-4$, and
$f'(x)=4x^3+1$, so for this problem we have
$$
x_{n}=x_o- {{x_o^4+x_o-4} \over {4x_o^3+1}} \quad.
$$
$f(1)=-2$ and $f(2)=14$. Since $-2$ is so much closer to $0$ then
$14$, it is reasonable to take as the {\bf initial approximation}
$x_o=1$. We get
$$
x_{n}=1- {{1^4+1-4} \over {4\cdot 1^3+1}} =
1-{{-2} \over {5}} = 7/5 \quad.
$$
Now let $x_0=7/5$ and we get
$$
x_{n}=7/5- {{(7/5)^4+7/5-4} \over {4\cdot (7/5)^3+1}}  \quad.
$$
Since you don't have a calculator, you leave it like this.
(It turns out to be 9703/7485).

{\bf Ans. to 20}: $7/5- {{(7/5)^4+7/5-4} \over {4\cdot (7/5)^3+1}}=
9703/7485$.

{\bf 21.} [Note: slightly changed from previous version].
If a stone is thrown vertically upward from the surface
of the moon with a velocity of $10 m/s$, its height after
$t$ seconds is $h(t)=10t-0.83 t^2$.
\smallskip
(a) What is the velocity of the stone after $3 s$?
\smallskip
(b) What is the velocity of the stone after it has risen $26.72 m$?

{\bf Sol. to 21.}: The velocity is the derivative of the position
(in this problem height). So
$$
v(t)=h'(t)=10-1.66 t \quad.
$$
The velocity after $3$ seconds is $v(3)=10- 1.66 \cdot 3=10-4.98=5.02$.

{\bf Ans. to 21a)}: $5.02 m/s$.

To solve b), we need to find at what time the height is $26.72$, so
we have to solve the equation
$$
10t-.83t^2=26.72 \quad,
$$
getting $t=4$ and $t=8.8048$ (you need a calculator for this) .
Since $t=4$ comes before, we have $h(4)=10-1.66 \cdot 4=3.36$

{\bf Ans. to 21b)}: $3.36$.

{\bf 22.} Find the absolute maximum and absolute minimum values
of $f(x)={{x} \over {x^2+1}}$ in $[0,2]$.

{\bf Sol. to 22.} Using the quotient rule, we have
$$
f'(x)={{x'(x^2+1)-x(x^2+1)'} \over {(x^2+1)^2}}=
{{1\cdot(x^2+1)-x \cdot 2x} \over {(x^2+1)^2}}=
{{x^2+1-2x^2} \over {(x^2+1)^2}}={{1-x^2} \over {(x^2+1)^2}} \quad.
$$
We have to solve $f'(x)=0$. Setting the {\bf numerator} equal to $0$,
we get $1-x^2=0$, so $(1-x)(1+x)=0$ and we get the two roots
$x=-1$ and $x=1$. But $x=-1$ does not satisfy the
{\bf residency requirement}, i.e. it does not belong to the interval
$[0,2]$, so we have the three
{\bf finalists}: $0,1,2$. (The endpoints of the interval always
participate in the {\bf final contest}).

Now let's plug these numbers in  $f(x)={{x} \over {x^2+1}}$:
$$
f(0)={{0} \over {0^2+1}}=0 \quad\quad
f(1)={{1} \over {1^2+1}}= {{1} \over {2}}=.5 \quad\quad
f(2)={{2} \over {2^2+1}}= {{2} \over {5}}=.4 \quad .
$$

{\bf Ans. to 22}: Absolute minimum value is $0$,
Absolute maximum value is $1/2$.

{\bf 23.} Find the most general antiderivative of the functions
$$
a) \quad f(x)=1-x^3+5x^5-3x^7 \quad 
b) \quad g(x)={{5-4x^3+3x^6} \over {x^6}}
$$
$$
c) \quad h(x)= 3e^x+7\sec^2 x+5(1-x^2)^{-{{1} \over {2}}} .
$$

{\bf Sol. to 23.} Recall the formula 
$$
\int x^n={{x^{n+1}} \over {n+1}}+C \quad .
$$

{\bf a)}
$$
\int f(x) dx =\int (1-x^3+5x^5-3x^7)dx=
x-{{x^4} \over {4}}+5{{x^6} \over {6}}-
3{{x^8} \over {8}}=
x-(1/4)x^4+(5/6)x^6-(3/8)x^8+C \quad .
$$

{\bf Ans. to 23a}:$x-(1/4)x^4+(5/6)x^6-(3/8)x^8+C$ .

{\bf b)} First {\bf break it up}, then integrate {\bf term by term}.
$$
\int g(x) dx= \int {{5-4x^3+3x^6} \over {x^6}} dx=
\int (5 \cdot x^{-6} -4x^{-3}+3) dx=
5x^{-5}/(-5)-4 (x^{-2}/(-2))+3x=
$$
$$
-x^{-5}+2x^{-2}+3x=
{{-1} \over {x^5}}+{{2} \over {x^2}}+3x+C \quad.
$$

{\bf Ans. to 23b)}: ${{-1} \over {x^5}}+{{2} \over {x^2}}+3x+C$.

{\bf c)} Recall (or look up in the formula sheet) that 
$(\tan x)'=\sec^2 x$, hence the {\bf antiderivative} of
$\sec^2 x$ is $\tan x+C$. Also $(\sin^{-1} x)'=(1-x^2)^{-{{1} \over {2}}}$,
hence the antiderivative of 
$(1-x^2)^{-{{1} \over {2}}}$ is $\sin^{-1} x+C$.
Of course the antiderivative of $e^x$ is $e^x+C$.
So
$$
\int f(x) dx=
\int (3e^x+7\sec^2 x+5(1-x^2)^{-{{1} \over {2}}})dx=
3e^x+7\tan x+5\sin^{-1}x+C \quad.
$$

{\bf Ans. to 23c)}: $3e^x+7\tan x+5\sin^{-1}x+C$.

{\bf 24.} Evaluate the integral, if it exists.
$$
a) \quad \int_{1}^{9} {{\sqrt{u}-2u^2} \over {u}} du . \quad
b) \quad \int_{0}^{2} y^2 \sqrt{1+y^3} dy .\quad
c) \quad \int_{1}^{5} {{dt} \over {(t-4)^2}} \quad.
$$
$$
d) \quad \int_{0}^{1} t^2 \cos(t^3) dt . \quad
e) \quad \int {{e^{\sqrt{x}}} \over {\sqrt{x}}} dx .
$$
{\bf Sol. to 24.}

{\bf a)}
$$
\int_{1}^{9} {{\sqrt{u}-2u^2} \over {u}} du =
\int_{1}^{9} (u^{-{{1} \over {2}}}-2u) du =
u^{{{1} \over {2}}}/({{1} \over {2}})-{{2u^2} \over {2}}\vert_{1}^{9} =
2\sqrt{u}-u^2 \big \vert_{1}^{9} =
$$
$$
[2\sqrt{9}-9^2]-
[2\sqrt{1}-1^2]=-76 \quad.
$$

{\bf Ans. to 24a)}: $-76$.

{\bf b)}: We make the substitution $u=1+y^3$ which entails
$du/dy=3y^2$ hence $dy=du/(3y^2)$. Also the limits 
of the new integral must be computed. When $y=0$, $u=1$, and
when $y=2$, $u=9$. So
$$
b) \int_{0}^{2} y^2 \sqrt{1+y^3} dy 
=
\int_{1}^{9} y^2 \sqrt{u} {{du} \over {3y^2}}= 
{{1} \over {3}}\int_{1}^{9} \sqrt{u} du= 
{{1} \over {3}}\int_{1}^{9} u^{{{1}\over {2}}} du= 
{{1} \over {3}}{{1} \over {3/2}} u^{{{3}\over {2}}}\vert_{1}^{9}= 
$$
$$
{{2} \over {9}} \sqrt{u}^3 \vert_{1}^{9}= 
{{2} \over {9}}( \sqrt{9}^3 -\sqrt{1}^3)=
{{2} \over {9}}( 3^3 -1^3)={{52} \over {9}} \quad.
$$

{\bf Ans. to 24b)}: $52/9$.

{\bf c)} Beware! The integrand blows up (and hence is undefined)
at $t=4$, that lies {\bf within} the interval of integration.

{\bf Ans. to 24c}: Undefined!

{\bf d)} Here the substitution is $u=t^3$ giving $du/dt=3t^2$, and
so $dt=du/(3t^2)$. As for the limits of integration, when
$t=0$ $u=0$ and when $t=1$, $u=1$.
$$
\int_{0}^{1} t^2 \cos(t^3) dt =
\int_{0}^{1} t^2 \cos(u) {{du} \over {3t^2}} =
{{1} \over {3}} \int_{0}^{1}  \cos(u) du=
{{1} \over {3}} \sin u\vert_{0}^{1} =
{{1} \over {3}}(\sin 1 - \sin 0)=(\sin 1)/3 \quad.
$$

{\bf Ans. to 24d):} $(\sin 1)/3$.

{\bf e)}: This is an {\it indefinite integral} since there
are no limits of integration. The answer should  be an
{\bf expression in $x$}. In the method of {\bf substitution},
we must translate back at the end.

Take $u=\sqrt{x}=x^{{{1} \over {2}}}$ , so
$du/dx=(1/2)x^{-{{1} \over {2}}}$ and
$dx=2 x^{{{1} \over {2}}}du=2\sqrt{x}du$. We have
$$
e) \int {{e^{\sqrt{x}}} \over {\sqrt{x}}} dx 
=\int {{e^u} \over {\sqrt{x}}}  2\sqrt{x}du
=2\int e^u du= 2e^u=2e^{\sqrt{x}}+C \quad.
$$

{\bf Ans. to 24e)}: $2e^{\sqrt{x}}+C$.

{\bf 25.} If $\int_{1}^{5} f(x) dx=12$ and
$\int_{4}^{5} f(x) dx=3.6$ , find $\int_{1}^{4} f(x)$.

{\bf Sol. to 25.} We use the obvious relation 
$$
\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx \quad,
$$
or rather its companion
$$
\int_{a}^{c} f(x)dx=\int_{a}^{b} f(x)dx-\int_{c}^{b} f(x)dx \quad.
$$
So
$$
\int_{1}^{4} f(x)dx=\int_{1}^{5} f(x)dx-\int_{4}^{5} f(x)dx=
12-3.6=8.4 \quad.
$$

{\bf Ans. to 25}: $8.4$.

{\bf 26.} Find the area bounded by the two curves 
(a) $y_1=2x^2$ and $y_2=8x$.
(b) $y_1=\sin x$ and $y_2=\cos x$, $0 \leq x \leq {{\pi} \over {2}}$

{\bf Sol. to 26.} Recall that the area is (if the two curves meet
only at two points)
$$
\int_{a}^{b} (TOP-BOTTOM)dx \quad,
$$
where $a$ and $b$ are the two roots of $y_1=y_2$.
Solving $2x^2=8x$ yields $2x(x-4)=0$ so $x=0$ and $x=4$, and
indeed there are only two meeting points.
To find who's on top just plug-in any $x$ in the interval $(0,4)$.
$x=1$ is a good choice. When $x=1$, $y_1=2$ and $y_2=8$, since
$2<8$ we know that $y_2=8x$ is on top and $y_1=2x^2$ is
at the bottom.
So the area is
$$
\int_{0}^{4} (8x-2x^2)dx=
4x^2-(2/3)x^3 \vert_{0}^{4}=
4\cdot 4^2-(2/3)\cdot 4^3={{64} \over {3}}=
21{{1} \over {3}} \quad.
$$

{\bf Ans. to 26a)}: ${{64} \over {3}}$.

{\bf b)}: (Note, in class I solved a different problem).

$y_1=\sin x$ and $y_2=\cos x$ meet when $\sin x=\cos x$,
and that happens when $x=\pi/4$. There are two
parts (equal by symmetry). The one bounded by
$y=\sin x$, $y=\cos x$, and the $y$-axis (alias $x=0$),
and the one bounded by $y=\sin x$, $y=\cos x$, and  $x=\pi/2$.
The first one happens between $x=0$ and $x=\pi/4$ and
the second between $x=\pi/4$ and $x=\pi/2$. In the
first half $y=\cos x$ is on top and $y=\sin x$ is at the
bottom, and at the second half it is the other way.
Hence the area is
$$
\int_{0}^{\pi/4}(\cos x-\sin x)dx+
\int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx=
(\sin x+ \cos x)\vert_{0}^{\pi/4}+
(-\cos x-\sin x)\vert_{\pi/4}^{\pi/2}=
$$
$$
(\sin (\pi/4)+ \cos (\pi/4))-(\sin (0)+ \cos (0))+
(-\cos (\pi/2)-\sin (\pi/2))-
(-\cos (\pi/4)-\sin (\pi/4))=2 \sqrt{2}-2 \quad.
$$

{\bf Ans. to 26b):} $2 \sqrt{2}-2$.

{\bf 27.} Evaluate the limit
$$
\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \sin({{i \pi} \over {n}})
{{\pi} \over {n}} \quad.
$$
This is an instance of the formal definition of the {\bf definite
integral}
$$
\int_{a}^{b} f(x) dx=
\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i) h\quad,
$$
where $h=(b-a)/n$ and $x_i=a+ih$, $i=1,2, \dots , n$.

The name of the game is to first decide on the function
$f(x)$, which in this problem is obviously $f(x)=\sin x$.
The argument of $\sin$ is $i\pi/n$ so
$x_i=i\pi/n$. To get $a$ we plug-in $i=0$ getting
$a=0$, and to get $b$ we plug in $i=n$, getting $b=\pi$.
Hence $h=\pi/n$, which is exactly as in the problem.
Hence this complicated-looking limit is nothing but
$$
\int_{0}^{\pi} \sin x=-\cos x \vert_{0}^{\pi}=
-\cos(\pi)-(-cos(0))=-(-1)+1=2 \quad.
$$
{\bf Ans. to 27.}: $2$.

{\bf 28.} a) Use the definition of the natural logarithm as an
integral and compare areas to prove that $\ln 2 <1 <\ln 3$.

b) Use a) to deduce that $2<e<3$.

{\bf Sol. to 28}: {\bf a)} Recall that
$$
\ln x=\int_{1}^{x} {{1} \over {t}} dt \quad.
$$
Hence $\ln 2$ is the area under the curve $y=1/t$ between
$t=1$ and $t=2$. To get an {\bf upper bound} for
$\ln 2$ note that it is enclosed in the square bounded
by the $t$-axis, the horizontal line $t=1$, and the
vertical lines $t=1$ and $t=2$. Its area is $1 \cdot 1=1$,
so $\ln 2<1$.

The {\bf lower bound} $1 < \ln 3$ is more tedious.
Take the inscribed rectangles with $n=8$, getting that
$\ln 3$, the area under the curve $y=1/t$ between
$t=1$ and $t=3$ is definitely more than
$$
{{1} \over {4}} \left (
{{1} \over {5/4}}+
{{1} \over {6/4}}+
{{1} \over {7/4}}+
{{1} \over {8/4}}+
{{1} \over {9/4}}+
{{1} \over {10/4}}+
{{1} \over {11/4}}+
{{1} \over {12/4}} \right )={{28271} \over {27720}} >1\quad.
$$

{\bf b)}: Since $e^x$ is an increasing function we can 
{\bf exponentiate} the inequality $\ln 2<1<\ln 3$ getting
$$
e^{\ln 2}<e^{1}<e^{\ln 3} \quad.
$$
But $e^{\ln w}=w$, so
$$
2<e<3 \quad.
$$

\bye



