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12. Evaluate (2+ r+3|) dx by interpreting it in terms of areas. Y-3+ 1-4 +3 %3D 2. a dy X dr t 3 dk 11-1+e V =3 4 2. A, +Az A, e 을 .bb .4.3 .12 Az =żbih

12. Evaluate (2+ r+3|) dx by interpreting it in terms of areas. Y-3+ 1-4 +3 %3D 2. a dy X dr t 3 dk 11-1+e V =3 4 2. A, +Az A, e 을 .bb .4.3 .12 Az =żbih

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Calculating Definite Integral Using Area Interpretation **Problem Statement:** Evaluate the integral \(\int_{-4}^{2} (2 + |x + 3|) \, dx\) by interpreting it in terms of areas. --- **Explanation and Solution:** **Graph Description:** - The graph is a piecewise function consisting of linear segments, reflecting the absolute value function \(|x + 3|\). - The function can be broken down into: - \(y = 2 + |x + 3|\) **Analysis of the Function:** 1. **Function Analysis:** - The equation can be expressed as two separate lines: - For \(x < -3\), \(y = 2 - (x + 3) = -x - 1\) - For \(x \geq -3\), \(y = 2 + (x + 3) = x + 5\) 2. **Graphical Representation:** - The graph shows two regions (areas \(A_1\) and \(A_2\)) under the line and above the x-axis within the interval \([-4, 2]\). - There’s a diagonal line intersecting at \(y = 3\), where the function changes at \(x = -3\). **Calculation of Areas:** - **Area \(A_1\) (Triangle from \(x = -4\) to \(x = -3\)):** - Triangle with base = 4, height = 3 - \(A_1 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 4 \cdot 3 = 6\) - **Area \(A_2\) (Triangle from \(x = -3\) to \(x = 2\)):** - Triangle with base = 2, height = 2 - \(A_2 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 2 = 2\) **Total Area:** - The total area under the curve is the sum of areas \(A_1\) and \(A_
Transcribed Image Text:### Calculating Definite Integral Using Area Interpretation **Problem Statement:** Evaluate the integral \(\int_{-4}^{2} (2 + |x + 3|) \, dx\) by interpreting it in terms of areas. --- **Explanation and Solution:** **Graph Description:** - The graph is a piecewise function consisting of linear segments, reflecting the absolute value function \(|x + 3|\). - The function can be broken down into: - \(y = 2 + |x + 3|\) **Analysis of the Function:** 1. **Function Analysis:** - The equation can be expressed as two separate lines: - For \(x < -3\), \(y = 2 - (x + 3) = -x - 1\) - For \(x \geq -3\), \(y = 2 + (x + 3) = x + 5\) 2. **Graphical Representation:** - The graph shows two regions (areas \(A_1\) and \(A_2\)) under the line and above the x-axis within the interval \([-4, 2]\). - There’s a diagonal line intersecting at \(y = 3\), where the function changes at \(x = -3\). **Calculation of Areas:** - **Area \(A_1\) (Triangle from \(x = -4\) to \(x = -3\)):** - Triangle with base = 4, height = 3 - \(A_1 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 4 \cdot 3 = 6\) - **Area \(A_2\) (Triangle from \(x = -3\) to \(x = 2\)):** - Triangle with base = 2, height = 2 - \(A_2 = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 2 = 2\) **Total Area:** - The total area under the curve is the sum of areas \(A_1\) and \(A_
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