y 2x Video Example 1 2 EXAMPLE 1 Use rectangles to estimate the area under the parabola y 2x from 0 to 1 the parabolic region 5 illustrated to the left SOLUTION We first notice that the area must be somewhere between 0 and 2 because 5 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 drawing vertical lines x as in the Figure a below and x y X No y 2x 2 Sy 4 Each rectangle has width and the heights are areas of these approximating rectangles we get 1 1 2 y 22 1 2 A 3 4 a 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 and function f x 2x at the right endpoints of the subintervals 0 BI and 2 1 If we let R be the sum of the R 2 2 4 4 201 CO 1 b 15 8 x We see that the area A is less than R so A 15 8 x Instead of using the rectangles above we could use the smaller rectangles whose heights are the values off 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 1 2 4 1 2 0 1 2 1 2 7 x 4 We see that the area is larger than La so we have lower and upper estimates for A 718 X A 15 8 X 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 7 8 1 1 x 11 2 a Using left endpoints b using right endpoiets By computing the sum of these areas of the smaller rectangles 4 and the sum of the areas of the larger rectangles Ra we obtain better lower and upper estimates for A