Note:
This homework assignment is just an example, for practice with our end-of-semester material. It will not be collected or graded.
Exercises:
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. \(\Box\)
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 \(\left\vert \sin^8(e^x )\right\vert \le 1\))
\[\begin{align*} \left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \sin^8(e^x) \right\vert\; dx \\ &\le \int_{-2\pi}^{2\pi} \left\vert x^2 \right\vert\; dx = \int_{-2\pi}^{2\pi} x^2 \; dx \\ &\le \frac{16\pi^3}3. \Box \end{align*}\]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 \([-1, 1]\).
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 \(F'(x) = G'(\sin x)\cos x = f(\sin x)\cos x\).
Let \(f\) be the function,
\[f(t) = \left\{\begin{array}{lr} t & t < 0\\ t^2+1 & 0 \le t \le 2\\ 0 & t > 2 \end {array}\right.\]Determine the function \(F(x) = \int_0^x f(t)\; dt\).
Sketch \(F\). Where is \(F\) continuous?
Where is \(F\) differentiable? Calculate \(F'\) at the points of differentiability.
By the FTC II, \(F(x)\) is continuous everywhere.
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.