Homework 11
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(25 points)
Show that the following functions are integrable on \(R = [0, 1] \times [0, 1]\) and find \(\int_R f dA\).(a) \(f(x,y) = \begin{cases} 0 & \text{if } 0 \le y \le \frac{1}{2} \\ 1 & \text{if } \frac{1}{2} < y \le 1. \end{cases}\)
(a) Take \(P=P_1 \times P_2\) where \(P_1 = \{x_0=0, \dots, x_N=1\}\), \(P_2 = \{y_0=0, \dots, y_N=1\}\), and \(x_i = y_i = \frac{i}{N}\).
If \(N\) is odd, \(U(f,P)=\sum_{i,j}M_{ij}a(R_{ij})=N\cdot\frac{N+1}{2}\cdot\frac{1}{N^{2}}=\frac{N+1}{2N}\)
\(L(f,P) = \sum_{i,j} m_{ij} a(R_{ij}) = N \cdot \frac{N-1}{2} \cdot \frac{1}{N^2} = \frac{N-1}{2N}\)
\(U(f,P) - L(f,P) = \frac{2}{2N} = \frac{1}{N}\). Let \(N > \frac{1}{\epsilon}\), we get \(U(f,P) - L(f,P) < \epsilon\).
Thus \(f\) is integrable, \(\frac{N-1}{2N}\le I \le \frac{N+1}{2N}\). When \(N \to \infty\), \(I = \frac{1}{2} = \int_R f dA\).If \(N\) is even, \(U(f,P) = \sum_{i,j} M_{ij} a(R_{ij}) = N \frac{N}{2} \cdot \frac{1}{N^2} = \frac{1}{2}\)
\(L(f,P) = \sum_{i,j} m_{ij} a(R_{ij}) = N (\frac{N}{2}-1) \cdot \frac{1}{N^2} = \frac{N-2}{2N}\)
\(U(f,P) - L(f,P) = \frac{N-(N-2)}{2N} = \frac{2}{2N} = \frac{1}{N}\)
Let \(N > \frac{1}{\epsilon}\), we get \(U(f,P) - L(f,P) < \epsilon\).
Thus \(f\) is integrable, \(\frac{N-2}{2N} \le I \le \frac{1}{2}\) when \(N \to \infty\), \(I = \frac{1}{2} = \int_R f dA\).
Thus \(f\) is integrable and \(\int_R f dA = \frac{1}{2}\).(b) \(g(x,y) = \begin{cases} 1 & \text{if } x = \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots \\ 0 & \text{otherwise.} \end{cases}\)
Take \(P = P_1 \times P_2\), where \(P_1 = \{x_0 = 0, \dots, x_N = 1\}\), \(P_2 = \{y_0 = 0, \dots, y_N = 1\}\), and \(x_i = y_i = \frac{i}{N}\).
Consider \(k = \{\frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{\lfloor\sqrt{N}\rfloor}\}\). \(|k| = \lfloor\sqrt{N}\rfloor - 1\).
\(U(g, P) = \sum_{i,j} M_{ij} a(R_{ij}) = \sum_{i,j: R_{ij} \cap k \neq \emptyset} M_{ij} a(R_{ij}) + \sum_{i,j: R_{ij} \cap k = \emptyset} M_{ij} a(R_{ij})\) \(\leq 2N(\lfloor\sqrt{N}\rfloor - 1) \cdot 1 \cdot \frac{1}{N^2} + N^2 \cdot \frac{1}{N+1} \cdot \frac{1}{N^2} = 2 \frac{\lfloor\sqrt{N}\rfloor - 1}{N} + \frac{1}{N+1}\).\(L(g,P) = \sum_{i,j}^N m_{ij} a(R_{ij}) = 0\)
Then \(U(g,P) - L(g,P) = 2 \frac{\lfloor\sqrt{N}\rfloor-1}{N} + \frac{1}{N+1} < \frac{2}{\sqrt{N}}.\)
Take \(N > \frac{4}{\epsilon^2}\). We get \(U(g,P) - L(g,P) < \epsilon\). Thus \(g\) is integrable.Thus \(L(g,P) = 0 \le \int \le U(g,P) = 2 \frac{\lfloor\sqrt{N}\rfloor-1}{N} + \frac{1}{N+1} \quad \text{when } N \to \infty, U(g,P) \to 0.\)
Thus \(\int_R g \, dA = 0\) -
(25 points)
Show that the function \(f(x,y) = \begin{cases}1 & \text{if } y \in \mathbb{Q} \\2x & \text{if } y \notin \mathbb{Q}\end{cases}\) is not integrable on \(R = [0, 1] \times [0, 1]\), but \(\int_{0}^{1} \int_{0}^{1} f(x,y) \, dxdy\) exists.Take \(P = P_1 \times P_2\) where \(P_1 = \{ x_0 = 0, \ldots, x_N = 1 \}\), \(P_2 = \{ y_0 = 0, \ldots, y_N = 1 \}\), and \(x_{i} = y_{i} = \frac{i}{N}\).
If \(N\) is odd, \(U(f,P)=\sum_{i,j}^{N}M_{ij}(R_{ij})=\frac{N-1}{2}N\cdot1\cdot\frac{1}{N^{2}}+N\cdot \left(\frac{1}{2}+\frac{1}{2N}\right)\cdot\frac{1}{N^{2}}+N\cdot\left(\frac{1}{2} +\frac{1}{2N}+\frac{1}{N}\right)+\ldots+N\cdot\left(\frac{1}{2}+\frac{1}{2N}+\frac{1}{N} \cdot\frac{N-1}{2}\right)\cdot\frac{1}{N^{2}}\)
\(= \frac{N-1}{2N} + \frac{N+1}{2N^2} + \frac{N+3}{2N^2} + \cdots + \frac{2N}{2N^2}\)
\(= \frac{N-1}{2N} + \frac{1}{2N^2} \times \frac{(3N+1)(N+1)}{4}\)\(L(f,P) = \sum_{i,j} m_{ij} a(R_{ij}) = \frac{N}{2} \cdot N \cdot 1 \cdot \frac{1}{N^2} + N \cdot (\frac{1}{2} - \frac{1}{2N}) \cdot \frac{1}{N^2} + N \cdot (\frac{1}{2} - \frac{1}{2N} - \frac{1}{N^2}) \cdot \frac{1}{N^2} + \dots + N(\frac{1}{2} - \frac{1}{2N} - \frac{1}{N}) \cdot \frac{1}{N}\)
\(= \frac{N-1}{2N} + \frac{N-1}{2N^2} + \frac{N-3}{2N^2} + \dots + \frac{N-N}{2N^2}\) \(= \frac{N-1}{2N} + \frac{1}{2N^2} \times \frac{N^2-1}{4}\)
Then \(U(f,P) - L(f,P) = \frac{1}{2N^2} (\frac{3N^2+4N+1-N^2+1}{4}) = \frac{1}{2N^2} \times \frac{2N^2+4N+1}{4}\) \(= \frac{1}{4}(1 + \frac{2}{N} + \frac{1}{2N^2}) > \frac{1}{4}\)
Thus \(f\) is not integrable.
\(\int_0^1 \int_0^1 f(x,y) dx dy\). For \(\int_0^1 f(x,y) dx\): If \(y \in Q\), \(= \int_0^1 1 \cdot dx = 1\). If \(y \notin Q\), \(= \int_0^1 2x dx = 1\)
Then \(\int_0^1 1 dy = 1\) exists. -
(25 points) Evaluate the iterated integral \(\int_0^1 \int_y^1 \frac{\sin x}{x} \, dx \, dy.\)
Since \(\frac{\sin x}{x}\) is continuous in \([0,1]\), then we change the order \(= \int_0^1 \int_0^x \frac{\sin x}{x} \, dy \, dx = \int_0^1 \frac{\sin x}{x} (x - 0) \, dx = \int_0^1 \sin x \, dx\)
\(= -(\cos x) \bigg|_0^1 = -\cos(1) + \cos(0) = 1 - \cos(1).\)
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(25 points) Compute the following iterated integrals in the proposed order and then in the reverse order, sketch the region of integration.
(a) \(\int_0^{\sqrt{2}} \int_0^{2-x^2} x e^{x^2} \, dy \, dx.\)
(b) \(\int_{-2}^3 \int_{y^2 - 3}^{y + 3} \left(3x^2 + y^2 \right) \, dx \, dy.\)