EXAMPLE 1 Use rectangles to estimate the area under the parabola y = 4x² from 0 to 1 (the parabolic region S illustrated to the left). SOLUTION We first notice that the area must be somewhere between o and 4 because S is contained in a rectangle of side lengths 1 and 4, but we can certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = x- and as in the Figure (a) below. y (1, 4) (1, 4) y= 4x S4 S3 (a) (b) We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip (as in Figure (b) above). In other words, the heights of these rectangles are the values of the function fx) = 4x? at the right endpoints of the subintenvals o. 4. [ 글 [을 퀴 and (), «()'. «(), and 4(1)°, If we let R, be the sum of the areas of these Each rectangle has width and the heights are approximating rectangles, we get We see that the area A is less than R4, s0 Instead of using the rectangles above we could use the smaller rectangles whose heights are the values of f at the left endpoints of the sub 112 114
EXAMPLE 1 Use rectangles to estimate the area under the parabola y = 4x² from 0 to 1 (the parabolic region S illustrated to the left). SOLUTION We first notice that the area must be somewhere between o and 4 because S is contained in a rectangle of side lengths 1 and 4, but we can certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = x- and as in the Figure (a) below. y (1, 4) (1, 4) y= 4x S4 S3 (a) (b) We can approximate each strip by a rectangle whose base is the same as the strip and whose height is the same as the right edge of the strip (as in Figure (b) above). In other words, the heights of these rectangles are the values of the function fx) = 4x? at the right endpoints of the subintenvals o. 4. [ 글 [을 퀴 and (), «()'. «(), and 4(1)°, If we let R, be the sum of the areas of these Each rectangle has width and the heights are approximating rectangles, we get We see that the area A is less than R4, s0 Instead of using the rectangles above we could use the smaller rectangles whose heights are the values of f at the left endpoints of the sub 112 114
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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![In this educational content, we explore a method for approximating the area under the curve \( y = 4x^2 \) between \( x = \frac{1}{8} \) and \( x = 1 \), using rectangles whose heights are determined by function values at either the left or right endpoints of subintervals.
### Calculation Process
1. **Using Left Endpoints:**
- The sum of the areas of the rectangles using the left endpoints is given by:
\[
L_4 = \frac{1}{4} \times 4(0)^2 + \frac{1}{4} \times 4\left(\frac{1}{4}\right)^2 + \frac{1}{4} \times 4\left(\frac{1}{2}\right)^2 + \frac{1}{4} \times 4\left(\frac{3}{4}\right)^2
\]
- This calculation results in an area of 1.09375.
2. **Using Right Endpoints:**
- Similarly, rectangles using the right endpoints yield a sum calculated as shown, with an area of 1.59375.
### Visual Representation
- **Graph (a):** Displays rectangles using left endpoints with their heights underestimating the curve.
- **Graph (b):** Displays rectangles using right endpoints with heights overestimating the curve.
### Estimation Insights
- We determine the area \( A \) is bounded by:
\[
1.0937500 < A < 1.5937500
\]
- Increasing the number of strips refines this approximation, as demonstrated when dividing \( S \) into eight equal-width strips.
- Using 50 strips, the area is estimated between 1.2936 and 1.3736.
- With 1000 strips, it narrows to between 1.3313340 and 1.3353340. Averaging these numbers gives a good estimate for \( A \).
This educational tool demonstrates numerical integration using approximations, particularly illustrating how increasing the number of subdivisions improves estimation accuracy.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5abad2ef-7d64-4544-b6f7-dd3dfbff74f8%2F5325a096-1030-4433-a96a-a2f7ccedcb42%2Flqglc9q_processed.png&w=3840&q=75)
Transcribed Image Text:In this educational content, we explore a method for approximating the area under the curve \( y = 4x^2 \) between \( x = \frac{1}{8} \) and \( x = 1 \), using rectangles whose heights are determined by function values at either the left or right endpoints of subintervals.
### Calculation Process
1. **Using Left Endpoints:**
- The sum of the areas of the rectangles using the left endpoints is given by:
\[
L_4 = \frac{1}{4} \times 4(0)^2 + \frac{1}{4} \times 4\left(\frac{1}{4}\right)^2 + \frac{1}{4} \times 4\left(\frac{1}{2}\right)^2 + \frac{1}{4} \times 4\left(\frac{3}{4}\right)^2
\]
- This calculation results in an area of 1.09375.
2. **Using Right Endpoints:**
- Similarly, rectangles using the right endpoints yield a sum calculated as shown, with an area of 1.59375.
### Visual Representation
- **Graph (a):** Displays rectangles using left endpoints with their heights underestimating the curve.
- **Graph (b):** Displays rectangles using right endpoints with heights overestimating the curve.
### Estimation Insights
- We determine the area \( A \) is bounded by:
\[
1.0937500 < A < 1.5937500
\]
- Increasing the number of strips refines this approximation, as demonstrated when dividing \( S \) into eight equal-width strips.
- Using 50 strips, the area is estimated between 1.2936 and 1.3736.
- With 1000 strips, it narrows to between 1.3313340 and 1.3353340. Averaging these numbers gives a good estimate for \( A \).
This educational tool demonstrates numerical integration using approximations, particularly illustrating how increasing the number of subdivisions improves estimation accuracy.
![**Example 1: Estimating the Area Under a Parabola**
**Objective:** Use rectangles to estimate the area under the parabola \( y = 4x^2 \) from 0 to 1, as illustrated in the parabolic region \( S \) to the left.
**Solution:**
We start by noting that the area must be between 0 and 4 because the region \( S \) is within a rectangle of side lengths 1 and 4. We aim to better approximate this area. Consider dividing the region into four strips by drawing vertical lines at \( x = \frac{1}{4}, \, x = \frac{1}{2}, \) and \( x = \frac{3}{4} \), as shown in Figure (a) below.
In Figure (a), the graph shows the curve \( y = 4x^2 \) and strips labeled \( S_1, S_2, S_3, \) and \( S_4 \). Figure (b) illustrates the approximation using shaded rectangles.
**Process:**
To approximate each strip, use a rectangle where the base matches the strip's width, and the height is equivalent to the function's value at the strip's right edge. This approach uses the heights \( f(x) = 4x^2 \) at the right endpoints of the subintervals \( \left[0, \frac{1}{4}\right], \, \left[\frac{1}{4}, \frac{1}{2}\right], \, \left[\frac{1}{2}, \frac{3}{4}\right], \) and \(\left[\frac{3}{4}, 1\right] \).
Each rectangle has a width of \(\frac{1}{4}\) and heights given by the function evaluated at these points:
- \( 4\left(\frac{1}{4}\right)^2 \)
- \( 4\left(\frac{1}{2}\right)^2 \)
- \( 4\left(\frac{3}{4}\right)^2 \)
- \( 4(1)^2 \)
Denote the sum of these rectangle areas as \( R_4 \):
\[
R_4 = \frac{1}{4} \cdot 4\left(\frac{1}{4}\right)^2 + \frac{1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5abad2ef-7d64-4544-b6f7-dd3dfbff74f8%2F5325a096-1030-4433-a96a-a2f7ccedcb42%2Fbe7vl2i_processed.png&w=3840&q=75)
Transcribed Image Text:**Example 1: Estimating the Area Under a Parabola**
**Objective:** Use rectangles to estimate the area under the parabola \( y = 4x^2 \) from 0 to 1, as illustrated in the parabolic region \( S \) to the left.
**Solution:**
We start by noting that the area must be between 0 and 4 because the region \( S \) is within a rectangle of side lengths 1 and 4. We aim to better approximate this area. Consider dividing the region into four strips by drawing vertical lines at \( x = \frac{1}{4}, \, x = \frac{1}{2}, \) and \( x = \frac{3}{4} \), as shown in Figure (a) below.
In Figure (a), the graph shows the curve \( y = 4x^2 \) and strips labeled \( S_1, S_2, S_3, \) and \( S_4 \). Figure (b) illustrates the approximation using shaded rectangles.
**Process:**
To approximate each strip, use a rectangle where the base matches the strip's width, and the height is equivalent to the function's value at the strip's right edge. This approach uses the heights \( f(x) = 4x^2 \) at the right endpoints of the subintervals \( \left[0, \frac{1}{4}\right], \, \left[\frac{1}{4}, \frac{1}{2}\right], \, \left[\frac{1}{2}, \frac{3}{4}\right], \) and \(\left[\frac{3}{4}, 1\right] \).
Each rectangle has a width of \(\frac{1}{4}\) and heights given by the function evaluated at these points:
- \( 4\left(\frac{1}{4}\right)^2 \)
- \( 4\left(\frac{1}{2}\right)^2 \)
- \( 4\left(\frac{3}{4}\right)^2 \)
- \( 4(1)^2 \)
Denote the sum of these rectangle areas as \( R_4 \):
\[
R_4 = \frac{1}{4} \cdot 4\left(\frac{1}{4}\right)^2 + \frac{1
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