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* (1) = 2^{1} = 2

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(2.5) = 5.

On the second sub-interval the height of the rectangle is *f* (3.5) = 2^{3.5}

and the width is 2.5, so the area of this second rectangle is 2^{3.5}(2.5) ≅ 28.28.

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

28.28 + 5 = 33.28.

- 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* (1) = 2^{1} = 2

so the area of the rectangle is

height ⋅ width = 2(1) = 2.

Sub-interval [2,3]: 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 [3,4]: 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 [4,5]: 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 [5,6]: The height of the rectangle is

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

so the area of the rectangle is

height ⋅ width = 32(1) = 32.

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

2 + 4 + 8 + 16 + 32 = 62.

- The approximation found in (b) is an underestimate. The rectangles don't cover all of
*S*, so the area covered by the rectangles is less than the area of *S*.

The approximation found in (c) is also an underestimate, since again the rectangles don't cover all of *S*: