Math 317: Homework 12

Due: Friday, December 13, 2019


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


  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

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


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

    (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.