Homework 9

Solution.

Since $f$, $g$, and $e^x$ are differentiable on $I$, the product rule and chain rule imply that $h$ is also differentiable on $I$ with

Moreover, since $f\p{a} = f\p{b} = 0$, we immediately see that $h\p{a} = h\p{b} = 0$ as well. Thus, by Rolle's theorem, there exists $x \in \p{a, b}$ such that

where in the last step, we used that $e^t \neq 0$ for all $t \in \R$ to divide through by $e^{g\p{x}}$.

Solution.

Since $f$ is bounded, $U\p{f}$ and $L\p{f}$ are well-defined. Moreover, Theorem 32.4 and the definitions of $U\p{f}$ and $L\p{f}$ imply

Thus, $U\p{f} = L\p{f}$, so $f$ is integrable. Next,

By the squeeze theorem, $\lim_{n\to\infty} \p{\int_a^b f - L_n} = 0$, which also implies that $\lim_{n\to\infty} L_n = \int_a^b f$ (since $\int_a^b f$ is a constant!). Finally,

as well (by limit laws).

Common Mistakes

The most common big mistake I saw was something like this:

Since $\lim_{n\to\infty} \p{U_n - L_n} = 0$, we have $\lim_{n\to\infty} U_n = \lim_{n\to\infty} L_n$. Thus, ...

Remember that limits are additive when both limits exist--take another look at the assumptions of Theorem 9.3.

In the same vein, many students who made this mistake also made the following one:

We must have $\lim_{n\to\infty} U_n = U\p{f}$ and $\lim_{n\to\infty} L_n = L\p{f}$, so ...

Even if you did know that $\p{U_n}$ and $\p{L_n}$ both converge to the same number, more work needs to be done to show that $\lim_{n\to\infty} U_n = U\p{f}$ and $\lim_{n\to\infty} L_n = L\p{f}$. To prove this, you would need to use Theorem 32.4 and the definitions of $U\p{f}$ and $L\p{f}$ (that they're the supremum and infimum of some set, respectively).

Lastly, the most common small mistake was this:

Since $\lim_{n\to\infty} \p{U_n - L_n} = 0$, given any $\varepsilon > 0$ we have a partition $P_n$ such that

$$U\p{f, P_n} - L\p{f, P_n} = U_n - L_n < \varepsilon$$

for sufficiently large $n$. Thus, $f$ is integrable by Theorem 32.5.

The issue here is that you're not given that $U_n$ and $L_n$ correspond to the same partition, so you can't invoke Theorem 32.5 immediately. However, there is an easy fix: you do know that $U_n = U\p{f, P_n}$ and $L_n = L\p{f, Q_n}$ for some partitions $P_n, Q_n$, so by Lemma 32.2 applied to the common refinement $P_n, Q_n \subseteq P_n \cup Q_n$,