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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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: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
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
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
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