Evaluate the integral using area formula: L(3 – x)dx.

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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**Evaluate the Integral Using Area Formula:**

\[ \int_{2}^{3} (3 - x) \, dx. \]

**Explanation:**

This problem involves evaluating a definite integral, \(\int_{2}^{3} (3 - x) \, dx\), using the area formula. 

To solve this, consider the function \(f(x) = 3 - x\), which is a linear function representing a straight line. The area under the curve from \(x = 2\) to \(x = 3\) forms a right triangle on the coordinate plane. 

- **Base of the Triangle:** The length along the x-axis from 2 to 3 is 1 unit.
- **Height of the Triangle:** At \(x = 2\), the function value is \(f(2) = 3 - 2 = 1\).
- **Area of the Triangle:** The area \(A\) can be calculated using the formula for the area of a triangle:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}. \]

Therefore, the value of the definite integral is \(\frac{1}{2}\).
Transcribed Image Text:**Evaluate the Integral Using Area Formula:** \[ \int_{2}^{3} (3 - x) \, dx. \] **Explanation:** This problem involves evaluating a definite integral, \(\int_{2}^{3} (3 - x) \, dx\), using the area formula. To solve this, consider the function \(f(x) = 3 - x\), which is a linear function representing a straight line. The area under the curve from \(x = 2\) to \(x = 3\) forms a right triangle on the coordinate plane. - **Base of the Triangle:** The length along the x-axis from 2 to 3 is 1 unit. - **Height of the Triangle:** At \(x = 2\), the function value is \(f(2) = 3 - 2 = 1\). - **Area of the Triangle:** The area \(A\) can be calculated using the formula for the area of a triangle: \[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}. \] Therefore, the value of the definite integral is \(\frac{1}{2}\).
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