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Use rectangles to estimate the area under the parabola y 2x2 from 0 to 1 (the parabolic region S EXAMPLE 1 illustrated to the left) (1, 2) SOLUTION We first notice that the area must be somewhere between 0 and 2 because S is contained in a rectangle of side lengths 1 and 2, but we can certainly do better than that. Suppose we divide the region into four strips by X,and x = - as in the Figure (a) below. 4 drawing vertical lines x = 4 y 2 x2 у У (1, 2) (1, 2) y 2x2 1 Video Example S4 S3 х х 1 1 3 1 1 4 2 4 4 2 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 words, the heights of these rectangles are the values of the function f(x) = 2x2 at the right endpoints of the subintervals 0, and and the heights are 2 4 and 2(1)2. If we let R4 be the sum of Each rectangle has width the areas of these approximating rectangles, we get 2 .2 2 . 21) 1 R4 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 2(0)2 1 . 4 4 We see that the area is larger than L4, So we have lower and upper estimates for A: X < A < X

Use rectangles to estimate the area under the parabola y 2x2 from 0 to 1 (the parabolic region S EXAMPLE 1 illustrated to the left) (1, 2) SOLUTION We first notice that the area must be somewhere between 0 and 2 because S is contained in a rectangle of side lengths 1 and 2, but we can certainly do better than that. Suppose we divide the region into four strips by X,and x = - as in the Figure (a) below. 4 drawing vertical lines x = 4 y 2 x2 у У (1, 2) (1, 2) y 2x2 1 Video Example S4 S3 х х 1 1 3 1 1 4 2 4 4 2 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 words, the heights of these rectangles are the values of the function f(x) = 2x2 at the right endpoints of the subintervals 0, and and the heights are 2 4 and 2(1)2. If we let R4 be the sum of Each rectangle has width the areas of these approximating rectangles, we get 2 .2 2 . 21) 1 R4 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 2(0)2 1 . 4 4 We see that the area is larger than L4, So we have lower and upper estimates for A: X < A < X

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section11.3: Hyperbolas
Problem 35E
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Use rectangles to estimate the area under the parabola y 2x2 from 0 to 1 (the parabolic region S EXAMPLE 1 illustrated to the left) (1, 2) SOLUTION We first notice that the area must be somewhere between 0 and 2 because S is contained in a rectangle of side lengths 1 and 2, but we can certainly do better than that. Suppose we divide the region into four strips by X,and x = - as in the Figure (a) below. 4 drawing vertical lines x = 4 y 2 x2 у У (1, 2) (1, 2) y 2x2 1 Video Example S4 S3 х х 1 1 3 1 1 4 2 4 4 2 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 words, the heights of these rectangles are the values of the function f(x) = 2x2 at the right endpoints of the subintervals 0, and and the heights are 2 4 and 2(1)2. If we let R4 be the sum of Each rectangle has width the areas of these approximating rectangles, we get 2 .2 2 . 21) 1 R4 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 2(0)2 1 . 4 4 We see that the area is larger than L4, So we have lower and upper estimates for A: X < A < X
Transcribed Image Text:Use rectangles to estimate the area under the parabola y 2x2 from 0 to 1 (the parabolic region S EXAMPLE 1 illustrated to the left) (1, 2) SOLUTION We first notice that the area must be somewhere between 0 and 2 because S is contained in a rectangle of side lengths 1 and 2, but we can certainly do better than that. Suppose we divide the region into four strips by X,and x = - as in the Figure (a) below. 4 drawing vertical lines x = 4 y 2 x2 у У (1, 2) (1, 2) y 2x2 1 Video Example S4 S3 х х 1 1 3 1 1 4 2 4 4 2 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 words, the heights of these rectangles are the values of the function f(x) = 2x2 at the right endpoints of the subintervals 0, and and the heights are 2 4 and 2(1)2. If we let R4 be the sum of Each rectangle has width the areas of these approximating rectangles, we get 2 .2 2 . 21) 1 R4 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 2(0)2 1 . 4 4 We see that the area is larger than L4, So we have lower and upper estimates for A: X < A < X
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