15. Find the area below the function fAx) = sin(x) + 2 on the interval [0, T]. %3D

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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Question 15

**Problem 15: Find the area below the function \( f(x) = \sin(x) + 2 \) on the interval \([0, \pi]\).**

### Detailed Explanation

1. **Graph Description:**
   - The graph shows the function \( f(x) = \sin(x) + 2 \) plotted over the interval \([0, \pi]\).
   - The sine function \( \sin(x) \) oscillates above and below the x-axis, however, due to the +2, it is shifted vertically upwards by 2 units.
   - At \( x = 0 \), the value of the function is \( f(0) = \sin(0) + 2 = 2 \).
   - At \( x = \pi \), the value of the function is \( f(\pi) = \sin(\pi) + 2 = 2 \).
   - The graph features a single peak where \( \sin(x) \) reaches its maximum, which is 1. At this peak, \( f(x) = \sin(x) + 2 = 3 \).

2. **Mathematical Solution:**
   - To find the area under the curve \( f(x) \) over the interval \([0, \pi]\), we need to compute the definite integral:
   
     \[
     \text{Area} = \int_{0}^{\pi} [\sin(x) + 2] \, dx
     \]

   - This integral can be split into two separate integrals for ease of calculation:

     \[
     \int_{0}^{\pi} \sin(x) \, dx + \int_{0}^{\pi} 2 \, dx
     \]

   - The first part is:

     \[
     \int_{0}^{\pi} \sin(x) \, dx = [-\cos(x)]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 - (-1) = 1 + 1 = 2
     \]

   - The second part is:

     \[
     \int_{0}^{\pi} 2 \, dx = 2 \left[ x \right]_{0}^{\pi} = 2(\pi - 0) =
Transcribed Image Text:**Problem 15: Find the area below the function \( f(x) = \sin(x) + 2 \) on the interval \([0, \pi]\).** ### Detailed Explanation 1. **Graph Description:** - The graph shows the function \( f(x) = \sin(x) + 2 \) plotted over the interval \([0, \pi]\). - The sine function \( \sin(x) \) oscillates above and below the x-axis, however, due to the +2, it is shifted vertically upwards by 2 units. - At \( x = 0 \), the value of the function is \( f(0) = \sin(0) + 2 = 2 \). - At \( x = \pi \), the value of the function is \( f(\pi) = \sin(\pi) + 2 = 2 \). - The graph features a single peak where \( \sin(x) \) reaches its maximum, which is 1. At this peak, \( f(x) = \sin(x) + 2 = 3 \). 2. **Mathematical Solution:** - To find the area under the curve \( f(x) \) over the interval \([0, \pi]\), we need to compute the definite integral: \[ \text{Area} = \int_{0}^{\pi} [\sin(x) + 2] \, dx \] - This integral can be split into two separate integrals for ease of calculation: \[ \int_{0}^{\pi} \sin(x) \, dx + \int_{0}^{\pi} 2 \, dx \] - The first part is: \[ \int_{0}^{\pi} \sin(x) \, dx = [-\cos(x)]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 - (-1) = 1 + 1 = 2 \] - The second part is: \[ \int_{0}^{\pi} 2 \, dx = 2 \left[ x \right]_{0}^{\pi} = 2(\pi - 0) =
Expert Solution
Step 1: Set up the integral to calculate the area under the curve

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