[problem: Riemann Sum Computation | 1] Compute $\int_0^1 x^3 \, d\mathcal{L}^1$ directly from the definition as a limit of Riemann sums, using the partition $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ and right endpoints. [/problem]

[solution] **Step 1 (Write the Riemann sum).** The partition $P_n$ divides $[0, 1]$ into $n$ subintervals $[(k-1)/n, k/n]$, each of width $\Delta x = 1/n$. With right endpoints $x_k^* = k/n$: \begin{align*} R_n = \sum_{k=1}^n f(x_k^*) \Delta x = \sum_{k=1}^n \left(\frac{k}{n}\right)^3 \cdot \frac{1}{n} = \frac{1}{n^4} \sum_{k=1}^n k^3. \end{align*} **Step 2 (Apply the sum formula).** The identity $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$ (provable by induction) gives: \begin{align*} R_n = \frac{1}{n^4} \cdot \frac{n^2(n+1)^2}{4} = \frac{(n+1)^2}{4n^2} = \frac{1}{4}\left(1 + \frac{1}{n}\right)^2. \end{align*} **Step 3 (Take the limit).** As $n \to \infty$: \begin{align*} \int_0^1 x^3 \, d\mathcal{L}^1 = \lim_{n \to \infty} R_n = \frac{1}{4}(1 + 0)^2 = \frac{1}{4}. \end{align*} **Step 4 (Verify via FTC).** The antiderivative of $x^3$ is $x^4/4$, so $\int_0^1 x^3 \, d\mathcal{L}^1 = 1/4 - 0 = 1/4$. $\blacksquare$ [/solution]

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[problem: Linearity And Monotonicity From The Definition | 1] Let $f, g: [a, b] \to \mathbb{R}$ be Riemann integrable and $\alpha \in \mathbb{R}$. Prove from the definition (using upper and lower sums) that $\int_a^b \alpha f \, d\mathcal{L}^1 = \alpha \int_a^b f \, d\mathcal{L}^1$. [/problem]

[solution] **Step 1 (Case $\alpha \ge 0$).** For any partition $P$: $\sup_{[x_{i-1}, x_i]} (\alpha f) = \alpha \sup_{[x_{i-1}, x_i]} f$ (since $\alpha \ge 0$ preserves the supremum). Therefore $U(\alpha f, P) = \alpha U(f, P)$. Similarly $L(\alpha f, P) = \alpha L(f, P)$. Taking the infimum and supremum over all partitions: \begin{align*} \overline{\int_a^b} \alpha f = \inf_P U(\alpha f, P) = \alpha \inf_P U(f, P) = \alpha \overline{\int_a^b} f. \end{align*} Similarly for the lower integral. Since $f$ is Riemann integrable, $\overline{\int} f = \underline{\int} f = \int f$, so $\overline{\int} \alpha f = \underline{\int} \alpha f = \alpha \int f$, giving $\int \alpha f = \alpha \int f$. **Step 2 (Case $\alpha = -1$).** On any subinterval, $\sup(-f) = -\inf(f)$ and $\inf(-f) = -\sup(f)$. Therefore $U(-f, P) = -L(f, P)$ and $L(-f, P) = -U(f, P)$. Taking extrema: \begin{align*} \overline{\int_a^b} (-f) = \inf_P U(-f, P) = \inf_P (-L(f, P)) = -\sup_P L(f, P) = -\underline{\int_a^b} f = -\int_a^b f. \end{align*} So $\int (-f) = -\int f$. **Step 3 (General $\alpha$).** Write $\alpha = |\alpha| \cdot \operatorname{sgn}(\alpha)$. If $\alpha > 0$, Step 1 applies. If $\alpha < 0$, combine Step 1 (for $|\alpha|$) with Step 2 (for the sign): $\int \alpha f = \int |\alpha|(-f) \cdot (-1) = |\alpha| \int(-f) \cdot (-1)$... more directly: if $\alpha < 0$, write $\alpha f = (-|\alpha|)f = |\alpha|(-f)$, so $\int \alpha f = |\alpha| \int(-f) = |\alpha|(-\int f) = \alpha \int f$. If $\alpha = 0$, $\int 0 = 0 = 0 \cdot \int f$. $\blacksquare$ [/solution]

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[problem: The Integral Triangle Inequality | 2] Let $(X, \Sigma, \mu)$ be a measure space and $f \in L^1(X, \mu)$. Prove the integral triangle inequality: \begin{align*} \left|\int_X f \, d\mu\right| \le \int_X |f| \, d\mu. \end{align*} [/problem]

[solution] **Step 1 (Decompose $f$).** Write $f = f^+ - f^-$ where $f^+ = \max(f, 0) \ge 0$ and $f^- = \max(-f, 0) \ge 0$. Then $|f| = f^+ + f^-$ and $\int f = \int f^+ - \int f^-$. **Step 2 (Bound using the ordinary triangle inequality).** By the triangle inequality for real numbers: \begin{align*} \left|\int_X f \, d\mu\right| = \left|\int_X f^+ \, d\mu - \int_X f^- \, d\mu\right| \le \int_X f^+ \, d\mu + \int_X f^- \, d\mu. \end{align*} **Step 3 (Identify the right side).** By linearity (applied to nonneg functions) and additivity: \begin{align*} \int_X f^+ \, d\mu + \int_X f^- \, d\mu = \int_X (f^+ + f^-) \, d\mu = \int_X |f| \, d\mu. \end{align*} **Step 4 (Conclusion).** $\left|\int_X f \, d\mu\right| \le \int_X |f| \, d\mu$. $\blacksquare$ [/solution]

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[problem: Dominated Convergence Application | 2] Let $f_n: [0, \infty) \to \mathbb{R}$ be defined by $f_n(x) = \frac{n \sin(x/n)}{x(1 + x^2)}$ for $x > 0$. Compute $\lim_{n \to \infty} \int_0^\infty f_n(x) \, d\mathcal{L}^1(x)$. [/problem]

[solution] **Step 1 (Find the pointwise limit).** For fixed $x > 0$, $n \sin(x/n) = n \cdot (x/n - (x/n)^3/6 + \cdots) \to x$ as $n \to \infty$ (since $\sin(t)/t \to 1$ as $t \to 0$). Therefore: \begin{align*} f_n(x) = \frac{n\sin(x/n)}{x(1 + x^2)} \to \frac{x}{x(1 + x^2)} = \frac{1}{1 + x^2} \quad \text{pointwise for } x > 0. \end{align*} **Step 2 (Find a dominating function).** Since $|\sin(t)| \le |t|$ for all $t$, we have $|n\sin(x/n)| \le n \cdot x/n = x$. Therefore: \begin{align*} |f_n(x)| \le \frac{x}{x(1 + x^2)} = \frac{1}{1 + x^2} =: g(x). \end{align*} The function $g$ is integrable: $\int_0^\infty \frac{1}{1 + x^2} \, d\mathcal{L}^1 = [\arctan(x)]_0^\infty = \pi/2 < \infty$. **Step 3 (Apply DCT).** By the Dominated Convergence Theorem ($f_n \to 1/(1+x^2)$ pointwise, $|f_n| \le g \in L^1$): \begin{align*} \lim_{n \to \infty} \int_0^\infty f_n \, d\mathcal{L}^1 = \int_0^\infty \frac{1}{1 + x^2} \, d\mathcal{L}^1 = \frac{\pi}{2}. \quad \blacksquare \end{align*} [/solution]

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[problem: Fubini Versus Tonelli | 3] Let $f: (0, 1) \times (0, 1) \to \mathbb{R}$ be defined by $f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}$. **(a)** Compute both iterated integrals $\int_0^1 \left(\int_0^1 f(x,y) \, dy\right) dx$ and $\int_0^1 \left(\int_0^1 f(x,y) \, dx\right) dy$. **(b)** Explain why the results differ and why this does not contradict Fubini's theorem. [/problem]

[solution] **(a) First iterated integral.** **Step 1 (Inner integral in $y$).** Observe that $\frac{x^2 - y^2}{(x^2 + y^2)^2} = -\frac{\partial}{\partial y}\left(\frac{y}{x^2 + y^2}\right)$. Verification: $\frac{\partial}{\partial y}\frac{y}{x^2 + y^2} = \frac{(x^2 + y^2) - y \cdot 2y}{(x^2 + y^2)^2} = \frac{x^2 - y^2}{(x^2 + y^2)^2}$. Therefore: \begin{align*} \int_0^1 \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy = \left[\frac{y}{x^2 + y^2}\right]_{y=0}^{y=1} \cdot (-1) \quad \text{...} \end{align*} Wait — we have $f = \frac{\partial}{\partial y}\frac{y}{x^2 + y^2}$, not $-\frac{\partial}{\partial y}$. So: \begin{align*} \int_0^1 f(x,y) \, dy = \left[\frac{y}{x^2 + y^2}\right]_0^1 = \frac{1}{x^2 + 1} - 0 = \frac{1}{x^2 + 1}. \end{align*} **Step 2 (Outer integral in $x$).** \begin{align*} \int_0^1 \frac{1}{x^2 + 1} \, dx = [\arctan(x)]_0^1 = \frac{\pi}{4}. \end{align*} **(b) Second iterated integral.** **Step 3 (By antisymmetry).** Observe $f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2} = -f(y, x)$. Therefore: \begin{align*} \int_0^1\left(\int_0^1 f(x,y) \, dx\right) dy = \int_0^1\left(-\int_0^1 f(y, x) \, dx\right) dy = -\int_0^1 \frac{1}{y^2 + 1} \, dy = -\frac{\pi}{4}. \end{align*} The two iterated integrals are $\pi/4$ and $-\pi/4$ — they differ. **(c) Why no contradiction with Fubini.** **Step 4 (Check integrability of $|f|$).** By Tonelli's theorem, if $\int \int |f| < \infty$, then Fubini applies and the iterated integrals must agree. Since they differ, $\int_0^1 \int_0^1 |f(x,y)| \, dx \, dy = \infty$. Indeed, near $(0, 0)$: $|f(x, y)| \approx |x^2 - y^2|/(x^2 + y^2)^2$. In polar coordinates $x = r\cos\theta$, $y = r\sin\theta$: $|f| \approx |\cos(2\theta)|/r^2$, and $\int_0^\varepsilon r^{-2} \cdot r \, dr = \int_0^\varepsilon r^{-1} \, dr = \infty$. Fubini's theorem requires $f \in L^1$. Since $|f| \notin L^1((0,1)^2)$, the hypotheses fail and unequal iterated integrals are permitted. $\blacksquare$ [/solution]