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\title{Math 166: Honors Calculus II\\
Exam I Solutions}
\date{March 2, 2000}

\begin{document}

\maketitle

\bigskip
1. a) $\log(x) = \int_1^x \frac{1}{t}\,dt$; domain $x>0$, range all $y\in\R$.

b) $\exp(x)$ is the inverse of $\log(x)$: $y=\exp(x)$ if and only if
$\log(y) = x$; domain all $x\in\R$, range $y>0$.

c) $a^x = \exp(x\log(a))$; domain all $x\in\R$, range $y>0$.

d) $\sinh(x) = (e^x - e^{-x})/2$; domain all $x\in\R$, range all $y\in\R$.
\newline
$\cosh(x) = (e^x + e^{-x})/2$; domain all $x\in\R$, range all $y\ge1$.

e) $T_nf(x;a) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$

\nextproblem

2. a) Let $f$ be integrable on $[a,b]$ and let $A(x) = \int_1^x
f(t)\,dt$ for $x\in[a,b]$. If $f(x)$ is continuous at $x$, then $A(x)$
is differentiable at $x$ and $A'(x) = f(x)$.

b) $A'(x) = \lim_{h\to0} \frac{A(x+h)-A(x)}{h} = \lim_{h\to0}
\frac{1}{h}\int_x^{x+h}f(t)\,dt$.\newline
Since $f$ is continuous on the interval from $x$ to $x+h$, the Mean
Value Theorem implies there exists a $c_h$ between $x$ and $x+h$ such
that $f(c_h) =\frac{1}{h}\int_x^{x+h}f(t)\,dt$. Therefore, using the
fact that $f$ is continuous at $x$ and $c_h\to x$ as $h\to 0$, we get
$A'(x) = \lim_{h\to0} f(c_h) = f(\lim_{h\to0}c_h) = f(x)$.

\nextproblem

3. a) Assume that $f$ has continuous derivatives to order $n+1$ on an
interval conatining~$a$. Then,
\[ f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k +
\frac{1}{n!}\int_a^xf^{(n+1)}(t)(x-t)^n\,dt\]

b) If $F(t) = T_nf(x;t) = \sum_{k=0}^n \frac{f^{(k)}(t)}{k!}(x-t)^k$,
then $F(x) = f(x)$ and $F(a) = T_nf(x;a)$, so $E_nf(x) = f(x)
- T_n(x;a) = F(x) - F(a) = \int_a^x F'(t)\,dt$.  Now, by the
Product Rule, $F'(t) = f'(t) + \sum_{k=1}^n(a_{k+1} - a_k)$
where $a_k =\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}$. Since this is
a telescoping sum, we get $F'(t) = f'(t) + a_{n+1} - a_1 =
\frac{f^{(n+1)}(t)}{n!}(x-t)^n$,
so that $E_nf(x) = \frac{1}{n!}\int_a^xf^{(n+1)}(t)(x-t)^n\,dt$.

%b) By the Weighted Mean Value Theorem, there is a $c_x$ between $a$ and $x$
%such that
%\[ f^{(n+1)}(c_x) = \frac{\int_a^x f^{(n+1)}(t)(x-t)^n\,dt}
%{\int_a^x(x-t)^n\,dt}
%= \frac{n! E_nf(x)}{\frac{1}{n+1}(x-a)^{n+1}}\]
%Therefore,
%\[E_nf(x) = \frac{1}{n!}\int_a^xf^{(n+1)}(t)(x-t)^n\,dt =
%\frac{f^{(n+1)}(c_x)}{(n+1)!}(x-a)^{n+1}\]

4. a) Let $A(x) = \int_a^x \log(\cos(\sqrt{t}))\,dt$. Then by the
Fundamental Theorem of Calculus, $A'(x) = \log(\cos(\sqrt{x}))$. Since
$\int_{1/x^2}^{x^2} \log(\cos(\sqrt{t}))\,dt = A(x^2) - A(1/x^2)$,
the Chain Rule implies
\begin{eqnarray*}
\ddx\int_{1/x^2}^{x^2} \log(\cos(\sqrt{t}))\,dt &=&
A'(x^2)2x - A'(1/x^2)(-2/x^3) \\
&=& 2x \log(\cos(|x|)) + \frac{2}{x^3}\log(\cos(1/|x|)) =
\log\Big(\cos(|x|)^{2x}\cos(1/|x|)^{2/x^3}\Big)
\end{eqnarray*}

\newpage

b)
\begin{eqnarray*}
\ddx \Big(1+x^3\Big)^{(1+x^5)} &=& \ddx \exp[(1+x^5)\log(1+x^3)]\\
&=& \exp[(1+x^5)\log(1+x^3)]\Big(5x^4\log(1+x^3)+\frac{1+x^5}{1+x^3}3x^2\Big)\\
&=& \Big(1+x^3\big)^{x^5}\Big(5x^4(1+x^3)\log(1+x^3) + 3x^2(1+x^5)\Big)
\end{eqnarray*}

c) $f^{-1}(1) = x$ if and only if $1 = f(x) = x + e^x$, and this equations
holds only for $x = 0$. By the Theorem on the Derivatives of Inverses,
\[ (f^{-1})'(1) = \frac{1}{f'(0)} = \frac{1}{1 + e^0} = \frac{1}{2} \]

\nextproblem

5 a) Substitute $u = \sqrt{x-1}$, $u^2+1 = x$, $2u\,du = dx$ to get
\begin{eqnarray*}
\int \frac{\sqrt{x-1}}{x}\,dx &=& \int \frac{2u^2}{u^2+1}\,du\\
&=& \int 2 - \frac{2}{u^2+1}\,du= 2u - 2\arctan(u) + C\\
&=& 2\sqrt{x-1} - 2\arctan(\sqrt{x-1}) + C
\end{eqnarray*}

b)
$\int (x\cos(x))^2\,dx = \int x^2(\frac{1}{2}+\frac{1}{2}\cos(2x))\,dx
=  \frac{1}{6}x^3+\frac{1}{2}\int x^2\cos(2x)\,dx$. Use integration by
parts with $u = x^2$, $dv = \cos(2x)\,dx$, $du = 2x\,dx$, $v =
\frac{1}{2}\sin(2x)\,dx$ to get
\[ \int x^2\cos(2x))\,dx = \frac{1}{2}x^2\sin(2x) - \int x\sin(2x)\,dx \]
Integration by parts again gives $\int x\sin(2x)\,dx =
-\frac{1}{2}x\cos(2x) +\frac{1}{4}\sin(2x) + C$, so
\[ \int (x\cos(x))^2\,dx =  \frac{1}{6}x^3+\frac{1}{4}x^2\sin(2x)
+\frac{1}{4}x\cos(2x) -\frac{1}{8}\sin(2x) + C \]

c) Complete the square to get $\sqrt{x^2 - x} = \sqrt{(x-1/2)^2 - (1/2)^2}$.
Then make a trignonometric subsitution, $(1/2)\tan(u) = \sqrt{(x-1/2)^2 -
(1/2)^2}$, $(1/2)\sec(u) = x - 1/2$, $(1/2)\sec(u)\tan(u)\,du = dx$, $x =
(1/2)(\sec(u) + 1)$, to get
\begin{eqnarray*}
\int \frac{x}{\sqrt{x^2-x}}\,dx
&=& \frac{1}{2}\int \frac{\sec(u)+1}{\tan(u)}\sec(u)\tan(u)\,du\\
&=& \frac{1}{2}\int \sec^2(u)+\sec(u)\,du\\
&=& \frac{1}{2}\tan(u) + \frac{1}{2}\log|\sec(u)+\tan(u)| + C\\
&=& \sqrt{x^2-x} + \frac{1}{2}\log|2x - 1 + 2\sqrt{x^2-x}| + C
\end{eqnarray*}

\nextproblem

6.  a) $\frac{1}{x^3-1}= \frac{1}{(x-1)(x^2+x+1)} = \frac{A}{x-1} +
\frac{Bx+C}{x^2+x+1}$. Clear the denominators to get
\[ 1 = A(x^2 + x + 1) + (Bx+C)(x-1) = (A+B)x^2 + (A-B+C)x + (A-C) \]
This equation must hold for all $x$, so
\[ \begin{array}{ccccccc}
A &+& B & &   &=& 0\\
A &-& B &+& C &=& 0\\
A & &   &-& C &=& 1
\end{array} \]
Solving these equations yields $A = 1/3$, $B=-1/3$ and $C = -2/3$.

b) Use the substitution $u = \tan(x/2)$, $dx = 2du/(1+u^2)$,
$\sin(x)=2u/(1+u^2)$ and $\cos(x) = (1-u^2)/(1+u^2)$ to get
\[ \int \frac{(\sin(x)+1)^2}{\cos(x)+3}\,dx =
\int \frac{(2u/(1+u^2) + 1)^2}{((1-u^2)/(1+u^2) + 3)}\frac{2}{(1+u^2)}\,du 
= \int \frac{(u+1)^4}{(u^2+2)(u^2+1)^2}\,du \]

\nextproblem

7. a) $\tan(0) = 0$, $\tan'(0) = \sec^2(0) = 1$, $\tan''(0) =
2\sec^2(0)\tan(0) = 0$, $\tan'''(0) = 4\sec^2(0)\tan(0) + 2\sec^4(0) = 2$,
so $T_3\tan(x) = x + \frac{2}{3!}x^3  = x + \frac{1}{3}x^3$.

b) First recall that
$T_n\log(1-x) = T_n\Big(-\int_0^x\frac{1}{1-t}\,dt\Big) =
-\int_0^xT_{n-1}\Big(\frac{1}{1-t}\Big)\,dt =-\int_0^x 1 + t + \dots +
t^{n-1}\,dt = -\big(x + \frac{1}{2}x^2 + \dots + \frac{1}{n}x^n\big)$.
Then by the Substitution Property,
$T_n\log(1+x) = x - \frac{1}{2}x^2 + \dots + (-1)^{n+1}\frac{1}{n}x^n$.
Therfore, by the Linearity Property, 
\begin{eqnarray*}
T_{2n}\log(1-x^2) &=& T_{2n}(\log(1-x)+\log(1+x)) = T_{2n}\log(1-x)
+ T_{2n}\log(1+x)\\
&=& -\big(x + \frac{1}{2}x^2 + \dots + \frac{1}{2n}x^{2n}\big)
+ x - \frac{1}{2}x^2 + \dots + (-1)^{2n+1}\frac{1}{2n}x^{2n}\\
&=& - x^2 - \frac{1}{2}x^4 - \dots - \frac{1}{n}x^{2n} 
\end{eqnarray*}

c)
\begin{eqnarray*}
\frac{1+x^n}{1+x^{2n}} &=& \frac{1}{1+x^n} + x^n\frac{1}{1+x^{2n}}\\
&=& 1 - x^{2n} + \frac{x^{4n}}{1+x^{2n}} + x^n\Big(1 - x^{2n} +
\frac{x^{4n}}{1+x^{2n}}\Big)\\
&=& 1 + x^n - x^{2n} - x^{3n} + x^{3n}\frac{x^n+x^{2n}}{1+x^{2n}}
\end{eqnarray*}
By the Uniqueness Theorem for Taylor Polynomials,
$T_{3n}\Big(\frac{1+x^n}{1+x^{2n}}\Big)= 1 + x^n - x^{2n} - x^{3n}$.

\nextproblem

8. By Taylor's Formula,
$e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 +
\frac{e^{c_x}}{4!}x^4$ for some $c_x$ between 0 and~$x$.
Then
\begin{eqnarray*}
\int_{0.5}^{1.0} \frac{e^x}{x}\,dx &=& 
\int_{0.5}^{1.0} \frac{1}{x} + 1 + \frac{1}{2}x + \frac{1}{3!}x^2 +
\frac{e^{c_x}}{4!}x^3\,dx\\
&=& \log(x) + x + \frac{1}{4}x^2 + \frac{1}{18}x^3 \Big|_{0.5}^{1.0} + 
\int_{0.5}^{1.0}\frac{e^{c_x}}{4!}x^3\,dx\\
&=& \log(2) + \frac{1}{2} + \frac{3}{16} + \frac{7}{144} + E\\
&\cong& 1.429258 + E
\end{eqnarray*}
Here $E$ is the error term and can be estimated using the inequality
$e^{c_x} < e^{1.0} = e$:
\[ E = \int_{0.5}^{1.0} \frac{e^{c_x}}{4!}x^3\,dx < \int_{0.5}^{1.0}
\frac{e}{4!}x^3\,dx = \frac{e}{4\cdot4!}x^4\Big|_{0.5}^{1.0} = \frac{5e}{512}
\cong 0.026546 \]
Since $E$ is positive, the value of the original integral lies between 
$1.429258$ and $1.429258+0.026546 = 1.455804$. Another way to organize this
information is to take the approximation to be the midpoint, $1.443531$, and
say that the error is $\pm 0.026545/2 = \pm 0.013273$. (The actual value of
the integral is approximately $1.440898$.)

\end{document}
