Definite Integrals Exercises

Answer

To find the length of each sub-interval, we take the length of the original interval (4) and divide by the number of sub-intervals we want to chop it into (8).

On each sub-interval, we go to the left endpoint of that sub-interval and go up until we hit the function. The y-value of the function there is the height of the rectangle on that sub-interval.

We need to find the area of each rectangle. Sub-interval [0, 0.5]:

The left endpoint of this sub-interval is 0.

The height of this rectangle is

f (0) = 0² + 1 = 1

so the area is

(height) ⋅ (width) = 1 ⋅ (0.5) = 0.5

Sub-interval [0.5, 1]:

The left endpoint of this sub-interval is 0.5. The height of this rectangle is

f (0.5) = (0.5)² + 1 = 1.25

so the area is

(height) ⋅ (width) = 1.25 ⋅ (0.5) = 0.625

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

f (1) = 1² + 1 = 2

so the area is

(height) ⋅ (width) = 2 ⋅ (0.5) = 1

Sub-interval [1.5, 2]:

The height of this rectangle is

f (1.5) = (1.5)² + 1 = 3.25

so the area is

(height) ⋅ (width) = 3.25 ⋅ (0.5) = 1.625

Sub-interval [2, 2.5]:

The height of this rectangle is

f (2) = 2² + 1 = 5

so the area is

(height) ⋅ (width) = 5 ⋅ (0.5) = 2.5

Sub-interval [2.5, 3]:

The height of this rectangle is

f (2.5) = (2.5)² + 1 = 7.25

so the area is

(height) ⋅ (width) = 7.25 ⋅ (0.5) = 3.625.

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

f (3) = 3² + 1 = 10

so the area is

(height) ⋅ (width) = 10 ⋅ (0.5) = 5.

Sub-interval [3.5, 4]:

The left endpoint of this sub-interval is 3.5.

The height of this rectangle is

f (3.5) = (3.5)² + 1 = 13.25

so the area is

(height) ⋅ (width) = 13.25 ⋅ (0.5) = 6.625

Adding up the areas of all 8 rectangles, we get

0.5 + 0.625 + 1 + 1.625 + 2.5 + 3.625 + 5 + 6.625 = 21.5.

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) = 2¹ = 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) = 2¹ = 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) = 2² = 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) = 2³ = 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) = 2⁴ = 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) = 2⁵ = 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.

Both approximations are underestimates, since the rectangles don't cover all of S.

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 first approximation is an overestimate, because when we add the areas of the rectangles we find the area of W plus some extra:

Similarly, the second approximation is an overestimate, because again, the rectangles cover more area than W:

Answer

• If we use only one sub-interval, that means we're not breaking up the original interval up at all. We'll have one rectangle of width 6 and height f(2) = 3.

The area of this rectangle is

height ⋅ width = 3(6) = 18.

Since this is the only rectangle we're using, we estimate that the area between the graph of f and the x-axis is 18.

To use three sub-intervals we need to break the original interval [2, 8] into sub-intervals of length 2.

On sub-interval [2, 4] the height of the rectangle is

f (2) = 3

so the area of this rectangle is

height ⋅ width = 3(2) = 6.

On sub-interval [4, 6] the height of the rectangle is f(4) = 5 so the area of this rectangle is

height ⋅ width = 5(2) = 10.

On sub-interval [6, 8] the height of the rectangle is f(6) = 12 so the area of this rectangle is

height ⋅ width = 12(2) = 24.

Adding the areas of all the rectangles, we estimate that the area between the graph of f and the x-axis is

6 + 10 + 24 = 40.

The estimates are under-estimates. The function f is increasing, so there is area between f and the x-axis on the interval [2, 8] that isn't covered by the rectangles:

Answer

• With 3 sub-intervals we split the original interval [-1, 2] into sub-intervals of length 1, so each rectangle will have width 1.

On [-1, 0] the rectangle has height g(-1) = 34 and area 34(1) = 34.

On [0, 1] the rectangle has height g(0) = 20 and area 20(1) = 20.

On [1, 2] the rectangle has height g(1) = 18 and area 18(1) = 18.

Adding the areas of these rectangles, we estimate that the area between the graph of g(x) and the x-axis on [-1, 2] is

34 + 20 + 18 = 72.

• With 2 sub-intervals we split the original interval [-1, 2] into sub-intervals of length 1.5.

On [-1, 0.5] the rectangle has height g(-1) = 34 and width 1.5, so its area is

34(1.5) = 51.

On [0.5, 2] the rectangle has height g(0.5) = 19 and width 1.5, so its area is

19(1.5) = 28.5.

Adding the areas of these rectangles, we estimate that the area between the graph of g(x) and the x-axis on [-1, 2] is

51 + 28.5 = 79.5

• The estimates were both overestimates. The function g(x) and the rectangles from the first estimate would look something like this:

The rectangles cover more area than we want, so we find an overestimate of the actual area between the graph of g and the x-axis.

Answer

Dividing the interval [-6, -3] into 6 sub-intervals produces sub-intervals of length 0.5. We need the following values of f:

We add them, and multiply by the width of a sub-interval:

Answer

Dividing the interval [0, 4] into 8 sub-intervals produces sub-intervals of length 0.5. We need the following values of f:

We add these values of f and multiply by the width of a sub-interval:

[f (0) + f (.5) + f (1) + f (1.5) + f ( 2 ) + f (2.5) + f (3) + f (3.5)](width)

= [8 + 8.75 + 9 + 8.75 + 8 + 6.75 + 5 + 2.75](.5)

= 28.5

Answer

Breaking the interval [0, 12] into 3 sub-intervals gives us 3 sub-intervals of size 4. We need the following values of g, which we find by looking them up in the table:

We add these values and multiply by the width of a sub-interval:

Answer

Breaking the interval [-9, 9] into 9 sub-intervals gives us sub-intervals of length 2. We can look in the table to find the values of h at

x = -9, -7, -5, -3, -1, 1, 3, 5, 7.

We add these values and multiply by the width of a sub-interval:

[h(-9) + h(-7) + h(-5) + h(-3) + h(-1) + h(1) + h(3) + h(5) + h(7)](width)

= [1 + 2 + 3 + 4 + 3 + 2 + 1 + 0 + 2](2)

= 36

Answer

Breaking the interval [0, 30] into 3 sub-intervals gives us sub-intervals of length 10. From looking at the graph we can see that

We add these and multiply by the length of a sub-interval:

Answer

The table gives us the points on the function. From these points, we can tell what the sub-intervals need to be. We add the areas of the rectangles: 6(3) + 10(1) + 7(4) + 2(3) = 62.

Answer

The table doesn't tell us g(5π), but since we're taking a left-hand sum we don't need to know what g is at the right-most endpoint of the interval. The table provides enough points that we can tell what the sub-intervals need to be. Adding the areas of the rectangles, we get .

Answer

From the graph we can tell what h is at x = 2, 3,  5, and 7. Since the problem says to look at the interval [2, 7], we don't care about the value of h at 7. Knowing h at 2, 3, and 5 means our sub-intervals have lengths 1, 2, and 2.

Adding the areas of the rectangles gives us

4(1) + 6(2) + 8(2) = 32