Use rectangles to estimate the area under the parabola y = y S 3 (3,18) 2x...

Transcribed Image Text:

Use rectangles to estimate the area under the parabola y = y S 3 (3,18) 2x² from 0 to 3 (the parabolic region S illustrated below.) We first notice that the area must be somewhere between 0 and 3 because S is contained in a rectangle of side lengths 3 and 18, but we can 3 3 certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = and x = as in the figure 2 (a) below. -,x= 9 4 4 (3,18) (3,18) We first notice that the area must be somewhere between 0 and 3 because S is contained in a rectangle of side lengths 3 and 18, but we can 3 3 9 as in the figure 4 = certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = x = and x (a) below. 4 2 3/4 3/2 9/4 3 (a) (3,18) 3/4 (212 3/2 9/4 3 (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, 3 3 3 9 [0, ²]· [2, 3] [27], and [2, 3]. (212 (012 (3,18) Q Lumen OHM Assessment ← → C = X C The Velocity Graph Of A Braking X G We can approximate each strip X + ✰ ohm.lumenlearning.com/assessment/showtest.php?action=skip&to=10 L 4] [44] [44] 3 Each rectangle has width and the heights are 2 approximating rectangles, we get 2 3 R₁ = − ³² · · ()*² + ² · · ( ² )* · *· · · ()*+ * · 200² + 4 4 Preview 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 Preview Type here to search 3 - ² + 4·()* L₁ - 1 · 2 (0)² + 1 · 2 ( ³ ) ² + 1 · 2 ( ³ ) ² 2 Preview [ 4 ] +2² (²³) ² ² ( ² ) ² ² ( ² ) ²³ ²² Points possible: 1 This is attempt 1 of 4. Message instructor about this question We see that the area is larger than L4, so we have lower and upper estimates for A: Preview <A < O D Im , and 2(3)². If we let R₁ be the sum of the areas of these Preview Licence 4:53 AM 4/21/2020 x Ę Use rectangles to estimate the area under the parabola y = y S 3 (3,18) 2x² from 0 to 3 (the parabolic region S illustrated below.) We first notice that the area must be somewhere between 0 and 3 because S is contained in a rectangle of side lengths 3 and 18, but we can 3 3 certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = and x = as in the figure 2 (a) below. -,x= 9 4 4 (3,18) (3,18) We first notice that the area must be somewhere between 0 and 3 because S is contained in a rectangle of side lengths 3 and 18, but we can 3 3 9 as in the figure 4 = certainly do better than that. Suppose we divide the region into four strips by drawing vertical lines x = x = and x (a) below. 4 2 3/4 3/2 9/4 3 (a) (3,18) 3/4 (212 3/2 9/4 3 (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, 3 3 3 9 [0, ²]· [2, 3] [27], and [2, 3]. (212 (012 (3,18) Q Lumen OHM Assessment ← → C = X C The Velocity Graph Of A Braking X G We can approximate each strip X + ✰ ohm.lumenlearning.com/assessment/showtest.php?action=skip&to=10 L 4] [44] [44] 3 Each rectangle has width and the heights are 2 approximating rectangles, we get 2 3 R₁ = − ³² · · ()*² + ² · · ( ² )* · *· · · ()*+ * · 200² + 4 4 Preview 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 Preview Type here to search 3 - ² + 4·()* L₁ - 1 · 2 (0)² + 1 · 2 ( ³ ) ² + 1 · 2 ( ³ ) ² 2 Preview [ 4 ] +2² (²³) ² ² ( ² ) ² ² ( ² ) ²³ ²² Points possible: 1 This is attempt 1 of 4. Message instructor about this question We see that the area is larger than L4, so we have lower and upper estimates for A: Preview <A < O D Im , and 2(3)². If we let R₁ be the sum of the areas of these Preview Licence 4:53 AM 4/21/2020 x Ę