- Let
*W* be the area between the graph of and the x-axis on the interval [1,4].

- Use a Left-Hand Sum with 3 subintervals to approximate the area of
*W*. Draw *W* and the rectangles used in this Left-Hand Sum on the same graph.

- Use a Left-Hand Sum with 6 subintervals to approximate the area of
*W*. Draw *W* and the rectangles used in this Left-Hand Sum on the same graph.

- Are your approximations in parts (b) and (c) bigger or smaller than the actual area of
*W*?

Answer

- Dividing the interval [1,4] into 3 sub-intervals of equal length gives us sub-intervals of length 1.Sub-interval [1,2]:The height of this rectangle is

so the area of the rectangle is

height ⋅ width = 1(1) = 1.

Sub-interval [2,3]. The height of this rectangle is

so the area of the rectangle is

Sub-interval [3,4]. The height of this rectangle is

so the area of the rectangle is

Adding the areas of these rectangles, we estimate that the area of *W* is

- Dividing the interval [1,4] into 6 sub-intervals of equal length gives us sub-intervals of length .

Sub-interval [1,1.5]. The height of this rectangle is

so the area is

Sub-interval [1.5,2]. The height of this rectangle is

so the area is

Sub-interval [2,2.5]. The height of this rectangle is

so the area is

Sub-interval [2.5,3]. The height of this rectangle is

so the area is

Sub-interval [3,3.5]. The height of this rectangle is

so the area is

Sub-interval [3.5,4]. The height of this rectangle is

so the area is

Adding the areas of all these rectangles, we estimate that the area of *W* is

- The approximation in (b) is an overestimate, because when we add the areas of the rectangles we find the area of
*W* plus some extra:

Similarly, the approximation in (c) is an overestimate, because again the rectangles cover more area than *W*: