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\begin{document}

\centerline{\bf Homework }
\centerline{\bf
UTK -- M148 -- Honors Calculus II -- Spring 2015}
\centerline{\bf
Jochen Denzler}



\begin{enumerate} \relax\setcounter{enumi}{0}\relax
\item
Find antiderivatives of the following functions:
$$
(a) \quad
f(x) = \cos (2x+4) \qquad\qquad
(b)\quad 
g(x) = 3e^x + e^{3x} + x^{3e} 
$$
$$
(c) \quad
h(x) = \frac1{5x} \mbox{ for } x>0 \qquad\qquad
(d) \quad k(x) = 1/x^5
$$
%
\item %[\underline{\underline{\bf 2.}} ]
Find the area below the curve $y=x^2$, above the $x$-axis, between $x=0$ 
and $x=b$ as a limit of a sum of rectangle areas (as done for $y=ax$ in 
class). To this end, divide the interval $[0,b]$ into $n$ pieces each of 
equal length $\Delta x=\frac bn$. Do two calculations, one by evaluating 
the function at the left endpoint, one at the right endpoint. 

Draw a picture. 

{\bf Hints: }
You may need the following formula (which is likely new to you, and you 
may just take my word for its validity): 
$1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
You may also need a similar formula ending with $(n-1)^2$, which 
you can obtain from the one I gave you by simple algebra. 

\item
(a) Evaluate $\sum_{j=0}^6 j(6-j)$. 
\\
(b) Simplify the expression $\sum_{j=1}^n f(j) - \sum_{k=2}^{n-1}
f(k)$ (where you may assume that $n$ is an integer $\ge3$).   
\\
(c) Evaluate $\sum_{j=1}^n \frac{1}{j(j+1)}$ for $n=1,2,\ldots,7$
respectively and conjecture a formula for general $n$. Then write
$\frac{1}{j(j+1)}$ as a difference of two terms, based on the
conjectured formula, and use this to prove the formula as a
telescoping sum. 

\item Estimate $\int_0^5 \frac{dx}{1+x^2}$ from above and below by a
  Riemann sum each; make the Riemann sums consist of 10 terms (using
  an equidistant partition). 

\item 
We'll get a 4-digit precise result for $\int_0^3\frac{dx}{x^2+3}$ here. 
You will like a programmable calculator for this, but the main work is still 
analytic work on the paper. Here is the philosophy of this problem: While
Riemann sums are good for theoretical purposes, a practical calculation will
give more precise results by using trapezoids rather than rectangles. 

(a) Calling $f(x):= 1/(x^2+3)$,  find out on which subinterval of $[0,3]$  we
have $f''(x)>0$ and $f''(x)<0$ respectively. 
Answer should be in the form ``$f''(x)<0$'' if $x<??$ and $f''(x)>0$
if $x>??$''. 

(b) Instead of nesting a small slice of area beneath the graph of $f$, from
$x_j$ to $x_{j+1}$, between rectangles, nest it between trapezoids: 
One trapezoid will have the oblique line connecting $(x_j,f(x_j))$ and 
$(x_{j+1},f(x_{j+1}))$, the other trapezoid will have the oblique line being a
tangent to the graph of $f$ at $x=\frac{1}{2}(x_j+x_{j+1})$. Which trapezoid
gives a lower bound for the area, which gives an upper bound, and how does the
answer depend on issues discussed previously?

(c) Write out a formula for the areas of the individual trapezoids involved in
the calculation, then sum up the appropriate areas to get an upper and a lower
bound for $\int_0^3\frac{dx}{x^2+3}$: choose 150 subintervals of
equal length in $[0,3]$.
Make sure that you select the midpoint rule for some intervals and the
trapezoidal for others, as discussed in (a) to get either an upper or
a lower bound for the integral. Now you may want to use technology and
get actual numerical values. 

\item
Evaluate the following integrals (among the expected answers, there may be: 
``cannot do it with tools available'', or, ``integral may not exist
since integrand isn't piecewise continuous''):  
$$
\begin{array}{l@{\kern3cm}l} 
   \Dst(a) \quad        \int \cos x\,dx
& \Dst(b) \quad         \int \left(3x^3 -4\sin(2x)\right)\,dx
\\[2ex] \Dst(c) \quad   \int_2^7 \frac{1}{x^2}\,dx
& \Dst(d) \quad         \int_1^5 \frac{\sin x}{x} \,dx
\\[2ex] \Dst(e) \quad   \int_{-2}^2 \frac{1}{x^2}\,dx
& \Dst(f) \quad         \lim_{x\to\infty} \int_1^x t^{-3/2}\,dt
\\[2ex] \Dst(g) \quad   \int_0^1 x^2\, dx
& \Dst(h) \quad         \int_0^1 t^2\, dx
\end{array} 
$$

\item %---------------------------------------------------------------------
Let 
$$
f(x):= \left\{ 
\begin{array}{ll} 
x^2 &\mbox{ if }|x|<1
\\
2|x|-1 &\mbox{ if } |x|\ge1
\end{array} 
\right.
$$
Find $\int_{-1}^3 f(x)\,dx$

\item %---------------------------------------------------------------------
Take the same function $f$ as before.
Calculate 
$F(x) := \int_0^x f(t)\,dt$. The answer should be a formula that also involves
if's. 
Graph both $f$ and $F$ on the interval $[-2,2]$ in the same coordinate system.
(In this graphing job, a halfway decent handmade figure,  like what
I'd draw on the blackboard, is preferable to a print-quality
technology-generated figure.)
\item %---------------------------------------------------------------------
Find the following derivatives without attempting to evaluate the integrals
first. 
$$
\begin{array}{l@{\kern3cm}l} 
   \Dst(a) \quad       \frac{d}{dx} \int_0^x \sin^2 t\,dt
&  \Dst(b) \quad       \frac{d}{dx} \int_x^{50} (\sin^2 t)/t\,dt
\\[2ex] \Dst(c) \quad  \frac{d}{dx} \int_x^{2x} \frac{dt}{1+t^4}
& \Dst(d) \quad        \frac{d}{dx} \int_x^{2x} \frac{dt}{t}
\\[2ex] \Dst(e) \quad  \frac{d}{dx} \int_0^{\sin x} \frac{\sin t}{t^2+1}\,dt
& \Dst(f) \quad        \frac{d}{dx} \int_0^{x^2} g((t+1)^2)\,dt
\end{array} 
$$

\item %----------------------------------------------------------------
Of the following three integrals, two are prohibitively difficult, one
is easy. Select the  easy one and calculate it:
$$
(a) \quad \int \sqrt{2+\sin x}\,dx\,,  \qquad
(b) \quad \int \cos x \sqrt{2+\sin x}\, dx\,, \qquad
(c) \quad \int \sin x \sqrt{2+\sin x}\, dx
$$

\item %----------------------------------------------------------------
Calculate $\Dst \int_1^2 \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$. 

\item %----------------------------------------------------------------
Calculate (a) $\Dst\int_1^2\frac{\ln x}{x}\,dx$, \quad
(b) $\Dst \int_2^4\frac{1}{x\,\ln x}\,dx$.

\item %----------------------------------------------------------------
Evaluate $\Dst\int_0^8 \sqrt{1+\sqrt{1+x}}\,dx$.

\item %----------------------------------------------------------------
Find $\int\tan x\,dx$. If, after giving it some thought, you need a hint: 
\rotate[f]{tangent is sine divided by cosine}

\item %----------------------------------------------------------------
Find $\Dst\int\frac{2dx}{e^x + e^{-x}}\,dx$,  which is the same as
$\Dst\int\frac{dx}{\cosh x}$.  


\item %----------------------------------------------------------------
(a) Calculate $\int x^3 e^{-x}\,dx$. 
\\
(b) Given any positive integer $n$, express $\int_0^N x^n e^{-x}\,dx$ in 
terms of $\int_0^N x^{n-1} e^{-x}\,dx$. Then express $\lim_{N\to\infty} 
\int_0^N x^n e^{-x}\,dx$ in terms of $\lim_{N\to\infty}\int_0^N x^{n-1} 
e^{-x}\,dx$.
\\  
(c) Evaluate $\lim_{N\to\infty}\int_0^N x^{100} e^{-x}\,dx$. Rather than writing 
the result of a huge integer with over 150 digits, write it as a product of many 
1- and 2-digit numbers without further evaluation. 

\item %----------------------------------------------------------------
Evaluate $\int e^{2x}\,\cos 3x\, dx$. 

\item %----------------------------------------------------------------
Calculate $\int_a^1 x^2(\ln x)^2\,dx$ for $a>0$. 
Then calculate $\lim_{a\to0+} \int_a^1 x^2(\ln x)^2\,dx$ from it.

\item %----------------------------------------------------------------
Calculate $\int_0^4 e^{-\sqrt{x}}\,dx$. 

\item %----------------------------------------------------------------
Calculate $\int\arcsin x\,dx$

\item %----------------------------------------------------------------
Calculate $\int\frac{\ln(\ln x)\, \ln x}{x}\, dx$

\item %----------------------------------------------------------------
Calculate $\int_1^4\sqrt{x}\,\ln x\,dx$

\item %----------------------------------------------------------------
{\sf
There once was a student named Mike Ivan Shap (as a Southerner, 
he went by his initials M.I.) who had forgotten the antiderivative of
$\frac1{x^2+1}$.  So he attempted integration by parts, using $u'=1$
and $v=\frac{1}{x^2+1}$.}

{\bf Task 1: } Carry out M.I.~Shap's calculation and see what integral
he obtained. 

{\sf Of course he couldn't do the new integral, so he went to his
  teacher for help and received the reminder that integration by parts
  wasn't needed and in actuality, he should have remembered that\ldots }

{\bf Task 2: }(too easy)
$$\int \frac{dx}{x^2+1} = ???$$

{\sf He felt frustrated and embarrassed about his forgetfulness, but
  fortunately his girl friend Miss Happy would always encourage
  him. She pointed out that he could use his unsuccessful attempt for
  finding a {\em new\/} integral; one that he would have been hard
  pressed to solve from scratch, had it been given without hint.

Namely she pointed out a simple piece of algebra:}
$$
\frac{x^2}{(x^2+1)^2} = \frac{x^2+1}{(x^2+1)^2} - \frac{1}{(x^2+1)^2} 
$$

{\it Comment from your professor: I call a piece of algebra a
  $\int$implifiation, if it simplifies the purpose of doing an
  integral. Note that sometimes, a $\int$implification goes in the
  opposite direction of what you'd have been asked to do in an algebra
  course. The task in $\int$implification is to get the expression in a
  form that is  convenient for doing some integral, whereas a
  simplification in a simple college algebra course wants you to get
  the expression in a form  that (kind of) uses the least amount of
  ink to write down.

On another note, I remind you of the trick of going around in circles
(and be known as a big wheel for it) as exemplified in the lecture and
in the textbook.}

{\bf Task 3: }Evaluate $\Dst\int\frac{dx}{(x^2+1)^2}$ according to
Miss Happy's advice. 

{\bf Task 4: }Now evaluate $\Dst\int\frac{dx}{(x^2+a^2)^2}$. 

{\sf M.I.~Shap and Miss Happy went to the movies to watch the 1968
  movie `The Thomas Crown Affair' (they are too classy to bother with
  the late 90's remake). Without spoiling much, a hinge point is
  Thomas Crown's quote: ``I have done it once, I can do it again''
  with the implied challenge `Watch me do it'. After this quote from 
  the movie, they
  were inspired to tackle the following math problem together: }

{\bf Task 5: }Evaluate $\Dst\int\frac{dx}{(x^2+1)^3}$

\item
Sometimes, definite integrals can be determined without finding an antiderivative: 
(a) Find $\int_{-7}^7 \frac{\sin x\,dx}{x^4+1}$ without attempting to obtain 
an antiderivative. Give a quick sketch of the graph of  the function
under the integral sign to illustrate your reasoning.
--- (b) Use the substitution $y=-x$ on the integral to confirm your
conclusion algebraically. 

\item 
(a) Describe the graph $y=\sqrt{25-x^2}$ in geometric terms. Based on this
deliberation, what should $\int_{-5}^5\sqrt{25-x^2}\,dx$ be? 
---
(b) Use an appropriate trig substitution to verify your conclusion
calculationally.

\item
You sure know that $\cos x\le 1$ for all $x$, and probably also that
$\sin x \le x$ for $x>0$. But you may not know yet that $\cos x \ge
1-\frac12x^2$ for all $x$, and that $\sin x\ge x-\frac16 x^3$ for $x>0$. In this
homework, you'll see why this is the case (and a bit more). 

Recall: If $f(x)\le g(x)$ for $x\in[a,b]$, then $\int_a^b f(x)\,dx \le
\int_a^b g(x)\,dx$. 

(a) Integrate the inequality $\cos t\le 1$ for all $t$ over the
interval $[0,x]$, with $x>0$. What inequality do you obtain?

(b) Next integrate the inequality so obtained (renaming the variable
into $t$) over the interval $[0,x]$ again and solve for the trig
function. What inequality do you now obtain?

(c) Repeat the procedure three more times, obtaining new inequalities
for $\sin x$ or $\cos x$ in each step. 

(d) Use the best of these inequalities to nest the values of
$\cos\frac12$ and $\sin\frac12$ between values easily calculated by
hand; i.e., find rational numbers for the `?' to make 
$?_1 \le \cos \frac12 \le {?_2}$  and $?_3 \le \sin \frac12 \le {?_4}$  
true. None of your calc's will require a pocket calculator (except possibly 
for converting fractions to decimals in the very end). 

\item
Reduce the following integrals to integrals without a square root, by
means of a trig substitution. In the problems with a $^*$, you don't
need to further deal with the thus obtained integrals (that will be a
task we do later).  In the other ones, finish up the integration
$$
(a) \quad \int \frac{dx}{(4-x^2)^{3/2}}
\,,\qquad
(b) \quad \int (4-x^2)^{3/2}\,dx
\,,\qquad
(c^*) \quad \int \frac{x^2}{\sqrt{x^2+1}}\,dx
$$
$$
(d^*) \quad \int \sqrt{x^2+x+1}\,dx
\,,\qquad
(e^*) \quad \int \sqrt{(x-1)(x-3)}\,dx
\,,\qquad
(f^*) \quad \int_1^4 \frac{x}{1+\sqrt{x^4+1}}\,dx
$$
$$
(g) \quad \int\frac{\arcsin x}{x^2}\,dx
\,,\; \qquad
(h) \quad \int_0^1\sqrt{\frac{1-x}{1+x}}\,dx
$$
In (e) you may assume $x>3$. 
Hint for (f),(g): do something else (in each case: what?) before trying the trig
substitution.
Hint for (h): $\frac{1-x}{1+x} = \frac{(1-x)^2}{(1-x)(1+x)}$. %and also
%     $\frac{1-x}{1+x} = \frac{(1-x)(1+x)}{(1+x)^2}$ 
Also to finish up (g) in the end, use that $\frac1{1-y^2} =
\frac12(\frac1{1-y}+\frac1{1+y})$. 
\item 
In one of the examples of the previous problem, the trig substitution could be avoided, and
a simple power substitution would work instead. Find which of them it
is and also carry out the power substitution for comparison. 


\item
Redo  all problems of the previous list (\#27 a--h), using a hyperbolic
substitution instead of the trig substitution. Subsequently get rid of
the hyperbolics by expressing them as exponentials and substituting
$e^t = y$ (assuming $t$ is what you named the new variable in the hyp substitution.)
If you end up with a rational expression in $y$ that is {\em not\/}
routine to integrate, you may stop here; we'll return to this task
later. Otherwise, finish them up. If possible, draw a resume whether
you feel the hyperbolic or the trigonometric substitution was easier.

\item
  (a) Reduce the integral $\int\frac{\sin^2 u}{\cos^3 u}\,du$ to the
  integral of a rational function in two ways:  Either using the
  substitution $\sin u=y$ (taking advantage of the fact that one 
  $\cos u$ can go with the differential, leaving only even powers of
  $\sin u$ and of $\cos u$). Or use $\tan\frac u2 = t$, which is a
  universal tool working for all rational expressions of $\sin u$ and
  $\cos u$. Which of the two methods leads to the `easier' rational
  integral (where `easyness' is judged by the degree of the
  denominator)? 

  (b) Convert the integrals obtained in 27c-e into integrals of
  rational expressions. 

  (c) Next you will use the hint that 
  $\frac{1}{1-y^2}=  \frac12(\frac1{1-y}+\frac1{1+y})$ and 
  $\frac{y}{1-y^2}=  \frac12(\frac1{1-y}-\frac1{1+y})$,
  which is a piece of algebra that  you will be able to invent
  yourself in a few weeks; and also a similar piece of algebra that you can obtain by
  squaring the given hints, like 
  $\frac{1}{(1-y^2)^2}=  \frac14(\frac1{(1-y)^2}+\frac1{(1+y)^2}+
  \frac{2}{1-y^2})$ (then re-use the first hint for the last term!)

  With these hints you should be able to finish up the integrals
  obtained in 27c,d,e. (Don't forget to undo the substitutions in the
  end to get answers in terms of~$x$.)

\item
  Finish up \#27f, using the universally helpful substitution for
  rational expressions in $\sin u$, $\cos u$ mentioned in the previous
  problem. (And I am not aware of a simpler procedure by means of a
  more specialized substitution). Make sure to get an exact
  expression first,  in which all occurrences of trig(arctrig(number))
  are simplified; then you may use technology to get a numeric result
  and check it for plausibility in view of the original integral. 

\item
  Find the PFD of $\Dst\frac{5x^2+4x-13}{x^3+2x^2-5x-6}$ 
  and calculate $\Dst\int_0^1 \frac{5x^2+4x-13}{x^3+2x^2-5x-6}\,dx$. 

\item 
  (a) Find the PFD of $\Dst\frac{(x+2)(x^2-2)}{(x+1)^2(x^2+1)}$ ---
  (b) Find the PFD of $\Dst\frac{x^6+76}{(x+2)^3(x-3)^2}$

\item
  Find the PFD of $\Dst\frac{25}{(x-1)^2(x^2+4x+5)}$ in two ways: one using 
  complex numbers to deal with the quadratic, one without use of complex numbers. 

\item
  Use PFD to evaluate $\Dst I(a,b;x):= 
  \int_0^x \frac{dt}{(t^2+a^2)(t^2+b^2)}$, assuming $a,b>0$ and 
  $b\neq a$. Finally, calculate the limit $\lim_{b\to a} I(a,b;x)$ (which may 
  require l'Hopital). --- {\sf The purpose of this problem is to obtain $I(a,a;x) =
  \int_0^x dt/(t^2+a^2)^2$ by a different method; we trust that 
  $I(a,a;x) = \lim_{b\to a} I(a,b;x)$, i.e., that the integral depends continuously 
  on $b$.} --- Compare your result with Hwk \#23 (Task 4).

\item
  Use the trig substitution $x=a \tan u$ on the integral $\int dx/(x^2+a^2)^2$ and 
  evaluate it this way. 
\end{enumerate}
\end{document}
