Answer

- If we use two sub-intervals, the length of a sub-interval is . The subintervals are then [1,3.5] and [3.5,6]. On the first sub-interval the height of the rectangle is

*f* (3.5) = 2^{3.5}

and the width the of the rectangle is 2.5 (the length of the sub-interval), so the area of the first rectangle is

2^{3.5}(2.5) ≅ 28.28

On the second sub-interval the height of the rectangle is

*f* (6) = 2^{6} = 64

and the width is 2.5, so the area of this second rectangle is

64(2.5) = 160.

Adding the areas of the rectangles together, we estimate that the area of *S* is approximately

28.28 + 160 = 188.28.

- Dividing the interval [1,6] into 5 sub-intervals gives us sub-intervals of length 1, so each rectangle will have width 1. Sub-interval [1,2]: The height of the rectangle is

*f** *( 2 ) = 2^{2} = 4

so the area of the rectangle is

height ⋅ width = 4(1) = 4.

Sub-interval [2,3]: The height of the rectangle is

*f* (3) = 2^{3} = 8

so the area of the rectangle is

height ⋅ width = 8(1) = 8.

Sub-interval [3,4]: The height of the rectangle is

*f* (4) = 2^{4} = 16

so the area of the rectangle is

height ⋅ width = 16(1) = 16.

Sub-interval [4,5]: The height of the rectangle is

*f* (5) = 2^{5} = 32

so the area of the rectangle is

height ⋅ width = 32(1) = 32.

Sub-interval [5,6]: The height of the rectangle is

*f* (6) = 64

so the area of the rectangle is

height ⋅ width = 64(1) = 64

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

4 + 8 + 16 + 32 + 64 = 124.

- The approximations found in (b) and (c) are both overestimates, as we can see the rectangles cover all of
*S* and then some.