S₁ y=2x² 1 S3 3 SA (1, 2) y 1 2 (1, 2) 1 1 1 4 2 4 4 (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 13 words, the heights of these rectangles are the values of the function f(x) = 2x² at the right endpoints of the subintervals[0][ and [2,1]. Each rectangle has width 1 - and the heights are 2(¹)², 2(1)², 2(³)², and 2(1)². If we let R4 be the sum of the areas of these approximating rectangles, we get R4 = 1 · 2 (1)² + ¹ · 2(¹)² + 1 · 2(2)² + 1 · 2(1)² 4 4 4 4 We see that the area A is less than R4, so A < 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 intervals. (The leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these approximating rectangles is LÀ G2002 đ 2 2 ) +à 2 ) + + = We see that the area is larger than L4, so we have lower and upper estimates for A:
S₁ y=2x² 1 S3 3 SA (1, 2) y 1 2 (1, 2) 1 1 1 4 2 4 4 (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 13 words, the heights of these rectangles are the values of the function f(x) = 2x² at the right endpoints of the subintervals[0][ and [2,1]. Each rectangle has width 1 - and the heights are 2(¹)², 2(1)², 2(³)², and 2(1)². If we let R4 be the sum of the areas of these approximating rectangles, we get R4 = 1 · 2 (1)² + ¹ · 2(¹)² + 1 · 2(2)² + 1 · 2(1)² 4 4 4 4 We see that the area A is less than R4, so A < 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 intervals. (The leftmost rectangle has collapsed because its height is 0.) The sum of the areas of these approximating rectangles is LÀ G2002 đ 2 2 ) +à 2 ) + + = We see that the area is larger than L4, so we have lower and upper estimates for A:
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question

Transcribed Image Text:n
Ln
Rn
10 0.5700000 | 0.7700000
20 0.6175000 0.7175000
30 0.6337037 0.7003704
50 0.6468000 0.6868000
100 0.6567000 0.6767000
1000 0.6656670 0.6676670

Transcribed Image Text:y
y = 2x²
S3
1
2
3
4
S4
(1, 2)
y
1
1
1
4
4
(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
2x² at the right endpoints of the subintervals
1 3
words, the heights of these rectangles are the values of the function f(x) =
and
'
Each rectangle has width
1
and the heights are 2
2 (4) ²³, 2(4) ²³, 2 (²) ². and 2(1)2. If we let R4 be the sum of the areas of these approximating rectangles, we get
2
R4 = 1 · 2 (¹)² + ¹ · 2 ( ¹ )² + ¹ · 2(2)² + ½ · 2(1) ²
1
1
¹/
4
4
We see that the area A is less than R4, so
A <
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 intervals. (The leftmost rectangle has
collapsed because its height is 0.) The sum of the areas of these approximating rectangles is
L4 = I2(0)2 ²(1) ²
.
+
+
· ²(²)² =
+
4
4
4
4
We see that the area is larger than L4, so we have lower and upper estimates for A:
<A <
We can repeat this procedure with a larger number of strips. The figure shows what happens when we divide the S eight strips of equal width.
y
y
(1, 2)
(1, 2)
y = 2x²
1
1
8
8
(a) Using left endpoints
(b) Using right endpoints
By computing the sum of these areas of the smaller rectangles (L8) and the sum of the areas of the larger rectangles (Rg), we obtain better lower and upper estimates for A:
0.5468750 <A < 0.7968750.
So one possible answer to the question is to say that the true area lies somewhere between 0.5468750 and 0.7968750.
We could obtain better estimates by increasing the number of strips. The table at the left shows the results of similar calculations (with a computer) using n rectangles whose
heights are found with left endpoints (Ln) or right endpoints (Rn). In particular, we see by using 50 strips that the area lies between 0.6468 and 0.6868. With 1000 strips, we
narrow it down even more: A lies between 0.6656670 and 0.6676670. A good estimate is obtained by averaging these numbers: A =
1
2
(1, 2)
3
4
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images

Recommended textbooks for you

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated

Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education

Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY

Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,

