Mathematics 242 – Principles of Analysis Solutions – Discussion 3 November 29, 2005

A) For the regular partition P_{n} of [0, 2] with n smaller intervals we have Δx = \frac{2}{n} and x_{i} = \frac{2i}{n}. The function f(x) = 1 − 6x^{2} is decreasing on [0, 2] (since f′(x) = −12x ≤ 0 on the interval). Hence

U(f, P_{n}) = ∑_{i=1}^{n} (1 − 6 (\frac{2(i − 1)}{n})^{2}) \frac{2}{n}

= \frac{2}{n} ∑_{i=1}^{n} 1 − \frac{48}{n^{3}} ∑_{i=1}^{n−1} i^{2}

= 2 − \frac{48}{n^{3}} ∑_{i=1}^{n−1} i^{2}

Similarly

\begin{aligned}L(f, P_{n}) & = ∑_{i=1}^{n} (1 − 6 (\frac{2i}{n})^{2}) \frac{2}{n} \\ & = 2 − \frac{48}{n^{3}} ∑_{i=1}^{n} i^{2} \end{aligned}

Hence,

U(f, P_{n}) − L(f, P_{n}) = \frac{48}{n^{3}} n^{2} = \frac{48}{n}

If we are given ε > 0, then \frac{48}{n} < ε whenever n > \frac{48}{ε}. So we have that f is integrable. The value of the integral is found by taking

lim_{n→∞} L(f, P_{n}) = lim_{n→∞} (2 − \frac{48}{n^{3}} ∑_{i=1}^{n} i^{2})

= lim_{n→∞} (2 − \frac{48}{n^{3}} · \frac{n(n + 1)(2n + 1)}{6})

= 2 − 16 = −14

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) = ∑_{i=1}^{n} M_{i}Δx_{i} = ∑_{i=1}^{n} Δx_{i} = 1,

while

L(f, P) = ∑_{i=1}^{n} m_{i}Δx_{i} = 0.

Given ε < 1, there is no partition P such that U(f, P) − L(f, P) < ε. Hence f is not integrable.

C) Consider the regular partition P_{n} of [a, b] into n equal subintervals: P_{n} = {x_{i} : 0 ≤ i ≤ 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}) = \frac{(b − a)}{n} ∑_{i=1}^{n} f(x_{i})

Similarly, m_{i} = f(x_{i−1}) for each i and

L(f, P_{n}) = \frac{(b − a)}{n} ∑_{i=1}^{n} f(x_{i−1}) = \frac{(b − a)}{n} ∑_{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)) · \frac{1}{n}

(note x_{n} = b and x_{0} = a). Since (b−a)(f(b)−f(a)) is constant, we have lim_{n→∞} U(f, P_{n})− L(f, P_{n}) = 0. Therefore by Theorem 7.2.8, f is integrable on [a, b].

D) Let ε > 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}) < ε/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}) < ε/2. Let P_{ε} = P_{1} ∪ P_{2}. This gives a partition of [a, b]. From the definitions, U(f, P_{ε}) = U(f, P_{1}) + U(f, P_{2}) and L(f, P_{ε}) = L(f, P_{1}) + L(f, P_{2}). This shows that

U(f, P_{ε}) − L(f, P_{ε}) = (U(f, P_{1}) − L(f, P_{1})) + (U(f, P_{2}) − L(f, P_{2})) < ε/2 + ε/2 = ε

Therefore, f is integrable on [a, b]. To see the relation ∫_{a}^{b} f = ∫_{a}^{c} f + ∫_{c}^{b} f, note for all ε > 0, there is a partition P_{ε} = P_{1} ∪ P_{2} as above, such that

|∫_{a}^{b} f − (∫_{a}^{c} f + ∫_{c}^{b} f)| = |∫_{a}^{b} f − L(f, P_{ε}) − (∫_{a}^{c} f − L(f, P_{1})) − (∫_{c}^{b} f − L(f, P_{2}))|

≤ |∫_{a}^{b} f − L(f, P_{ε})| + |∫_{a}^{c} f − L(f, P_{1})| + |∫_{c}^{b} f − L(f, P_{2})|

< ε + ε/2 + ε/2.

(The last inequalities follow since L(f, P) < ∫_{a}^{b} f < U(f, P).) Since ε > 0 is arbitrary, this

shows ∫_{a}^{b} f − (∫_{a}^{c} f + ∫_{c}^{b} f) = 0