# Harrison Chapman

## Math 317: Homework 12

Due: Friday, December 13, 2019

Note:

This homework assignment is just an example, for practice with our end-of-semester material. It will not be collected or graded.

Exercises:

1. Let $$f$$ be integrable on $$[a,b]$$. Prove that, for any subset $$S \subseteq [a, b]$$ we have

$M(|f|, S) - m(|f|, S) \le M(f, S) - m(f, S)$

Hint. For $$x_0, y_0 \in S$$, we have $$\vert f(x_0)\vert - \vert f(y_0)\vert \le \vert f(x_0) - f(y_0)\vert \le M(f, S) - m(f, S)$$.

Note: This is the primary step in proving that $$|f|$$ is integrable whenever $$f$$ is.

Proof. The first part of the hint follows from the triangle inequality, as

$\vert f(x_0)\vert = \vert f(x_0) - f(y_0) + f(y_0) \vert \le \vert f(x_0) - f(y_0) \vert + \vert f(y_0) \vert.$

The second part of the hint comes from how $$M(f, S) \ge f(x_0)$$ and $$m(f, S) \le f(y_0)$$ for all $$x_0, y_0 \in S$$.

On the other hand, we have that for any $$\epsilon > 0$$, properties of the supremum and infimum guarantee that there exists $$x_0, y_0 \in S$$ so that,

$M(\vert f\vert, S) < \vert f(x_0)\vert + \epsilon/2 \quad\textrm{and} \quad m(\vert f\vert, S) > \vert f(y_0) \vert - \epsilon/2.$

So we have that,

$M(\vert f\vert, S) - m(\vert f\vert, S) < \vert f(x_0)\vert - \vert f(y_0) \vert + \epsilon \le M(f, S) - m(f, S) + \epsilon.$

This holds for all $$\epsilon > 0$$, so we conclude that the desired inequality holds. $$\Box$$

2. Show that $$\left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert \le \frac{16\pi^3}{3}$$.

Proof. Using Theorems 33.4 and 33.5, together with the FTC I (for example), we have that (as $$\left\vert \sin^8(e^x )\right\vert \le 1$$)

\begin{align*} \left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \sin^8(e^x) \right\vert\; dx \\ &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \right\vert\; dx = \int_{-2\pi}^{2\pi} x^2 \; dx \\ &\le \frac{16\pi^3}3. \Box \end{align*}
3. Let $$f$$ be the function,

$f(x) = \left\{\begin{array}{lr}\sin(\frac 1x) & x \ne 0\\0 & x=0\end{array}\right.$

Prove that $$f$$ is integrable on $$[-1, 1]$$. Hint. See Excercise 33.11(c) and its solution in the textbook. (Why can we not apply the Dominated Convergence Theorem to prove that $$f$$ is integrable?)

Proof. Let $$\epsilon > 0$$. Notice that $$f$$ is continuous on each of $$[-1, -\epsilon/8]$$ and $$[\epsilon/8, 1]$$, and is hence integrable. This means that there exist partitions $$Q_1$$ of $$[-1, -\epsilon/8]$$ and $$Q_2$$ of $$[\epsilon/8, 1]$$ so that

$U(f, Q_1) - L(f, Q_1) < \epsilon/4 \quad\text{and}\quad U(f, Q_2) - L(f, Q_2) < \epsilon/4.$

Notice that $$P = Q_1 \cup Q_2$$ is a partition of $$[-1, 1]$$ and furthermore that

$U(f, P) = U(f, Q_1) + \epsilon/4 + U(f, Q_2)$

and,

$L(f, P) = L(f, Q_1) - \epsilon/4 + L(f, Q_2).$

(Why? The only terms in the two sums are those contributed by the interval $$[-\epsilon/8, \epsilon/8]$$ but we know that the supremum and infimum of $$f$$ on any such interval are $$1$$ and $$-1$$ respectively and that the width of this interval is $$\epsilon/4$$) So,

\begin{align*} U(f, P) - L(f, P) &= (U(f, Q_1) + \epsilon/4 + U(f, Q_2)) - (L(f, Q_1) - \epsilon/4 + L(f, Q_2)) \\ &= (U(f, Q_1) - L(f, Q_1)) + (U(f, Q_2) - L(f, Q_2)) + \epsilon/2 \\ &< \epsilon/4 + \epsilon/4 + \epsilon/2 = \epsilon. \end{align*}

We can find any such $$P$$ for any $$\epsilon > 0$$, so we conclude that $$f$$ is integrable on $$[-1, 1]$$.

4. Let $$f$$ be a continuous function on $$\mathbb R$$ and define

$F(x) = \int_0^{\sin x}{f(t)\; dt}\quad\textrm{for } x \in \mathbb R.$

Show that $$F$$ is differentiable on $$\mathbb R$$ and compute $$F'$$.

Let $$G(x) = \int_0^x{f(t)\; dt}$$. Then $$G(x)$$ is differentiable and $$G'(x) = f(x)$$ by the FTC II as $$f$$ is continuous. Notice that $$F(x) = G(\sin x)$$. So by the chain rule, $$F$$ is differentiable and $$F'(x) = G'(\sin x)\cos x = f(\sin x)\cos x$$.

5. Let $$f$$ be the function,

$f(t) = \left\{\begin{array}{lr} t & t < 0\\ t^2+1 & 0 \le t \le 2\\ 0 & t > 2 \end {array}\right.$
1. Determine the function $$F(x) = \int_0^x f(t)\; dt$$.

2. Sketch $$F$$. Where is $$F$$ continuous?

3. Where is $$F$$ differentiable? Calculate $$F'$$ at the points of differentiability.

1. $F(x) = \left\{\begin{array}{lr} t^2 & t < 0\\ t^3/3+t & 0 \le t \le 2\\ 14/3 & t > 2 \end {array}\right.$
2. By the FTC II, $$F(x)$$ is continuous everywhere.

3. By the FTC II, $$F(x)$$ is differentiable everywhere that $$f(x)$$ is continuous, hence at all $$x$$ except possibly $$x=0$$ and $$x=2$$.

We can prove that $$F(x)$$ is not differentiable at $$x=0$$ and $$x=2$$ by showing that the left-hand and right-hand difference quotient limits disagree.