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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions -- Discussion 3}
\centerline{November 29, 2005}
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\noindent
A)  For the regular partition $P_n$ of $[0,2]$ with $n$ smaller
intervals we have $\Delta x = {2\over n}$ and $x_i = {2i\over n}$.  
The function $f(x) = 1 - 6x^2$ is decreasing on $[0,2]$ 
(since $f'(x) = -12x \le 0$ on the interval).  Hence
$$\eqalign{U(f,P_n) &= \sum_{i=1}^n 
\left(1 - 6\left({2(i-1)\over n}\right)^2\right) {2\over n}\cr
&= {2\over n}\sum_{i=1}^n 1 - {48\over n^3} \sum_{i=1}^{n-1} i^2\cr
&= 2 - {48\over n^3} \sum_{i=1}^{n-1} i^2\cr}$$
Similarly 
$$\eqalign{L(f,P_n) &= \sum_{i=1}^n 
\left(1 - 6\left({2i\over n}\right)^2\right) {2\over n}\cr
&= 2 - {48\over n^3} \sum_{i=1}^n i^2\cr}$$
Hence,
$$U(f,P_n) - L(f,P_n) = {48\over n^3} n^2 = {48 \over n}$$
If we are given $\varepsilon > 0$, then ${48 \over n} < \varepsilon$
whenever $n > {48 \over\varepsilon}$.  So we have
that $f$ is integrable.  
The value of the integral is found by taking 
$$\eqalign{\lim_{n\to\infty} L(f,P_n) 
&= \lim_{n\to\infty} \left(2 - {48\over n^3} \sum_{i=1}^n i^2\right)\cr
&= \lim_{n\to\infty} \left(2 - {48\over n^3} \cdot {n(n+1)(2n+1)\over 6}\right)\cr
&= 2 - 16 = -14\cr}$$ 

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\noindent
B)  The answer is NO:  $f(x)$ is not integrable on $[0,1]$ (or on
any other interval $[a,b]$ with $b > a$).  The reason is that since every
such interval contains both rational and irrational numbers.  So
for every partition $P$, we have $M_i = 1$ all $i$, and $m_i = 0$
all $i$.  Hence
$$U(f,P) = \sum_{i=1}^n M_i \Delta x_i = \sum_{i=1}^n \Delta x_i = 1,$$
while 
$$L(f,P) = \sum_{i=1}^n m_i \Delta x_i = 0.$$
Given $\varepsilon < 1$, there is no partition $P$
such that $U(f,P) - L(f,P) < \varepsilon$.  Hence
$f$ is not integrable.  

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\noindent
C)  Consider the regular partition $P_n$ 
of $[a,b]$ into $n$ equal subintervals: $P_n = \{x_i : 0 \le i \le n\}$
with $x_i - x_{i-1} = (b-a)/n$ for all $i$.  Then since $f$ is increasing
$M_i = f(x_i)$ for all $i$ and 
$$U(f,P_n) = {(b - a)\over n} \sum_{i = 1}^n f(x_i)$$
Similarly, $m_i = f(x_{i-1})$ for each $i$ and 
$$L(f,P_n) = {(b - a)\over n} \sum_{i = 1}^{n} f(x_{i-1}) = {(b - a)\over n} \sum_{i = 0}^{n-1} f(x_i).$$
In the difference $U(f,P_n) - L(f,P_n)$, all the terms
except the last one in $U$ and the first one in $L$ cancel (i.e. 
the sum ``telescopes''), leaving
$$U(f,P_n) - L(f,P_n) = (b-a)(f(b) - f(a))\cdot {1\over n}$$
(note $x_n = b$ and $x_0 = a$).  Since $(b - a)(f(b) - f(a))$ is constant,
we have $\lim_{n\to\infty} U(f,P_n) - L(f,P_n) = 0$.  Therefore by 
Theorem 7.2.8, $f$ is integrable on $[a,b]$. 


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\noindent
D) Let $\varepsilon > 0$.
Since $f$ is integrable on $[a,c]$, there is a partition $P_1$
of $[a,c]$ such that $U(f,P_1) - L(f,P_1) < \varepsilon/2$.
Similarly, since $f$ is integrable on $[c,b]$
there is a partition $P_2$ of $[c,b]$
such that $U(f,P_2) - L(f,P_2) < \varepsilon/2$.
Let $P_\varepsilon = P_1 \cup P_2$.  This gives a partition of $[a,b]$.
From the definitions, $U(f,P_\varepsilon) = U(f,P_1) + U(f,P_2)$ 
and $L(f,P_\varepsilon) = L(f,P_1) + L(f,P_2)$.  This shows
that 
$$U(f,P_\varepsilon) - L(f,P_\varepsilon) = (U(f,P_1) - L(f,P_1)) + (U(f,P_2) - L(f,P_2)) 
< \varepsilon/2 + \varepsilon/2 = \varepsilon$$
Therefore, $f$ is integrable on $[a,b]$.  To see the
relation 
$\int_a^b f = \int_a^c f + \int_c^b f$, note
for all $\varepsilon > 0$, there is a partition $P_\varepsilon = P_1 \cup P_2$ as above,
such that 
$$\eqalign{\left|\int_a^b f - \left(\int_a^c f + \int_c^b f\right)\right| &=
\left|\int_a^b f - L(f,P_\varepsilon) - \left(\int_a^c f - L(f,P_1)\right) - \left(\int_c^b f - L(f,P_2)\right)\right|\cr
&\le  \left|\int_a^b f - L(f,P_\varepsilon)\right| +  \left|\int_a^c f - L(f,P_1)\right| + \left|\int_c^b f - L(f,P_2)\right|\cr
&< \varepsilon + \varepsilon/2 + \varepsilon/2.\cr}$$
(The last inequalities follow since $L(f,P) < \int_a^b f < U(f,P)$.)
Since $\varepsilon > 0$ is arbitrary, this shows 
$\int_a^b f - \left(\int_a^c f + \int_c^b f\right) = 0$
\bye