We have formulas to find areas of shapes like rectangles, triangles, and circles (pi, anyone?).
What if we want to find the area of a less-reasonable shape? Think of sea monkeys. Sure, we all want them. We want to race them and bring them along on space adventures—but how many can we fit in our two-dimensional space ship? We have to use integrals to figure out the area of each sea monkey before heading into orbit.
Let R be the region between the graph y = f ( x ) = x^{2} + 1 and the x-axis on the interval [0,4]:
R is a weird shape, and we don't have any formula that says how to find its area (yet!).
In this section we're going to look at ways to approximate the areas of shapes that are formed, like R, by graphing non-negative functions on specified intervals.
A non-negative function is as it sounds: a function that never outputs a negative y-value.
Sample Problem
The following function is non-negative (hitting zero is allowed):
Sample Problem
The following function is not non-negative:
Back to R—we can approximate the area of R by drawing rectangles that more-or-less cover R, and calculating the total area covered by those rectangles.
There are several different procedures for drawing these rectangles. The most important ones to know are the Left-Hand Sum, the Right-Hand Sum, and the Midpoint Sum.
These are examples of Riemann Sums. There's also a procedure called the Trapezoid Sum, which draws trapezoids instead of rectangles.
The first step in any of these procedures is to chop up the original interval into sub-intervals, usually all of the same size.
On each sub-interval we draw a rectangle whose base is that sub-interval. The height of each rectangle depends on which procedure we're using.
With a Left-Hand Sum (LHS) the height of the rectangle on a sub-interval is the value of the function at the left endpoint of that sub-interval.
We can find the values of the function we need using formulas, tables, or graphs.
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. One way to find these function values is to calculate them using a formula for the function.
Left-Hand Sums with Formulas
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. One way to find these function values is to calculate them using a formula for the function.
Left-Hand Sums with Tables
In order to find a left-hand sum we need to know the value of the function at the left endpoint of each sub-interval. We can take a left-hand sum if we have a table that contains the appropriate function values.
Sample Problem
Some values of the decreasing function f ( x ) are given by the following table:
- Use a Left-Hand Sum with 2 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [0,4].
Answer. We don't know what the function f looks like, but we know these points are part of it:
Dividing the interval [0,4] into 2 equally-sized sub-intervals gives us sub-intervals of length 2.
The height of the rectangle on [0,2] is f ( 0 ) = 20, so the area of this rectangle is
height ⋅ width = 20(2) = 40.
The height of the rectangle on [2,4] is f ( 2 ) = 17, so the area of this rectangle is
height ⋅ width = 17(2) = 34.
Adding the areas of these rectangles, we estimate the area between the graph of f and the x-axis on [0,4] to be
40 + 34 = 74.
- Use a Left-Hand Sum with 4 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [0,4].
Answer. Dividing the interval [0,4] into 4 evenly-sized sub-intervals produces sub-intervals of length 1:
Sub-interval [0,1]: This rectangle has height f ( 0 ) = 20 and width 1, so its area is 20.
Sub-interval [1,2]: This rectangle has height f (1) = 18 and width 1, so its area is 18.
Sub-interval [2,3]: This rectangle has height f (2) = 17 and width 1, so its area is 17.
Sub-interval [3,4]:This rectangle has height f (3) = 11 and width 1, so its area is 11.
Adding the areas of these rectangles, we estimate the area between the graph of f and the x-axis on [0,4] to be
20 + 18 + 17 + 11 = 66.
- Are the estimates in parts (b) and (c) over- or under-estimates for the area between the function f and the x-axis on the interval [0,4]?
Answer. We don't know what the function f looks like exactly, but we know it's a decreasing function that passes through these dots:
That means it must look something like this:
and so our estimates in (b) and (c) were both overestimates, because the rectangles covered extra area:
- Could we use the left-hand sum with more than 4 sub-intervals to estimate the area between the function f and the x-axis on the interval [0,4]?
Answer. No. The table doesn't contain enough data for us to divide the interval [0,4] into more than 4 sub-intervals. If we tried to use 8 sub-intervals, for example, we would need to know f (0.5), and that value isn't in the table.
Left-Hand Sum with Graphs
When finding a left-hand sum, we need to know the value of the function at the left endpoint of each sub-interval. We can find these values by looking at a graph of the function.
Left-Hand Sum Calculator Shortcuts
When working through left-hand sums, we need to multiply every function value we use by the width of the rectangle. We can use these observations to work more efficiently and make better use of the calculator. Check out the examples for more ideas.
Left-Hand Sum with Math Notation
It's possible to write the process for taking a LHS as a nice tidy formula. Assume that we're using sub-intervals all of the same length and we want to estimate the area between the graph of f ( x ) and the x-axis on the interval [a,b].
An interval of the form [a,b] has length ( b – a ). This is true whether or not the numbers a and b are positive or negative. For example, the interval [-1,10] has length 11.
If we wish to divide the interval [a,b] into n equal sub-intervals, each sub-interval will have length
This quantity is often called Δ x:
We split the interval [a,b] into n sub-intervals, each of length . We need to know the value of f at each endpoint except at x = b (the right-most endpoint of the original interval).
We need to know what those endpoints are. The first one is x = a. The next endpoint is a + Δx. The next is (a + Δx) + Δx = (a + 2Δx). Then (a + 2Δx) + Δx = (a + 3Δx).
When do we stop? We know that
This means if we start at a and take n steps of size Δx, we'll get to b, which is the end of the original interval. The last endpoint we need for a left-hand sum is the one just before b, which is
a + ( n – 1 )Δx
These endpoints are often labeled like this, so we don't need to write down as much:
To take a LHS we find the value of f at every endpoint but the last, add these values, and multiply by the width of a sub-interval. In this case, the width of a sub-interval is Δx.
[f (x_{0}) + f (x_{1}) + f (x_{2}) + ... + f (x_{ n – 1 })](width) = [f (x_{0}) + f (x_{1}) + f (x_{2}) + ... + f (x_{ n – 1 })]Δx
Using a left-hand sum with n sub-intervals, we estimate the area between the graph of f and the x-axis on the interval [a,b] is
LHS(n) = [f (x_{0}) + f (x_{1}) + f (x_{2}) + ... + f (x_{ n – 1 })]Δx.
If we wanted to be extra fancy, we could use summation notation. Using i to keep track of which endpoint we're on, we can write the left-hand sum as
.
This formula is the same thing as the calculator shortcut. It's a short, tidy way to write down the process for taking a left-hand sum. There are two important things to remember.
- For a left-hand sum, the last endpoint you use is x_{ n – 1 }:
- After adding up the values of f at the appropriate endpoints, multiply by the width of a sub-interval:
Left-Hand Sum with Sub-Intervals of Different Lengths
All the left-hand sums we've found so far had sub-intervals that were all the same length. It doesn't need to be this way. There are some situations where we want to use sub-intervals of different lengths.
Sample Problem
Values of the function f are shown in the table below. Use a left-hand sum with the sub-intervals indicated by the data in the table to estimate the area between the graph of f and the x-axis on the interval [1,8].
Answer. The sub-intervals given in this table aren't all the same. Most of them are 2, but one is 1.
On sub-interval [1,3] the height of the rectangle is f (1) = 2 and the width is 2, so the area is
2(2) = 4.
On sub-interval [3,4] the height of the rectangle is f (3) = 5 and the width is 1, so the area is
5(1) = 5.
On sub-interval [4,6] the height of the rectangle is f (4) = 3 and the width is 2, so the area is
3(2) = 6.
On sub-interval [6,8] the height of the rectangle is f (6) = 5 and the width is 2, so the area is
5(2) = 10.
Adding the areas of the rectangles, we estimate the area between f and the x-axis on [1,8] to be
4 + 5 + 6 + 10 = 25.
Practice:
Let R be the region between the graph y = f ( x ) = x^{2} + 1 and the x-axis on the interval [0,4]: Use a left-hand sum with two sub-intervals to approximate the area of R. | |
To take a left-hand sum we first divide the interval in question into sub-intervals of equal size. Since we're looking at the interval [0,4], each sub-interval has size 2. On the first sub-interval, [0,2], we do the following: - Go to the left endpoint of the sub-interval (0).
- Go straight up until you hit the function.
Figure out the y-value of the function where you hit it (f ( 0 ) = (0)^{2} + 1 = 1). - Make a rectangle whose base is the subinterval and whose height is the y-value you just found:
Finally, calculate the area of this rectangle: (height) ⋅ (width) = 1 ⋅ 2 = 2 Now we do the same stuff on the second sub-interval, [2,4]. - Go to the left endpoint of the sub-interval (2).
- Go straight up until you hit the function.
Figure out the y-value of the function where you hit it (f ( 2 ) = (2)^{2} + 1 = 5). - Make a rectangle whose base is the sub-interval and whose height is the y-value you just found:
Finally, calculate the area of this rectangle: (height) ⋅ (width) = 5 ⋅ 2 = 10 Adding the areas of the rectangles together, we see that the rectangles cover an area of size 12: This is less than the area we're trying to find, because the two rectangles don't cover all of R. That's ok, because we're only estimating. In order to find a better estimate, we can use more rectangles. We start by dividing the original interval into more sub-intervals (each sub-interval will now be smaller). On each sub-interval we do the same stuff we did before. | |
Let R be the region between the graph y = f ( x ) = x^{2} + 1 and the x-axis on the interval [0,4]. Use a Left-Hand Sum with 4 sub-intervals to estimate the area of R.
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Since we're looking at the interval [0,4], each sub-interval has length 1. First sub-interval ([0,1]): - Go to the left endpoint of the sub-interval (0).
- Go straight up until you hit the function and figure out the y-value of the function where you hit it (f ( 0 ) = (0)^{2} + 1 = 1).
- Make a rectangle whose base is the subinterval and whose height is the y-value you just found:
Calculate the area of the rectangle: (height) ⋅ (width) = 1 ⋅ 1 = 1 Second sub-interval ([1,2]): - Go to the left endpoint of the sub-interval (1).
- Go straight up until you hit the function and figure out the y-value of the function where you hit it (f (1) = (1)^{2} + 1 = 2).
- Make a rectangle whose base is the subinterval and whose height is the y-value you just found:
Calculate the area of the rectangle: (height) ⋅ (width) = 2 ⋅ 1 = 2 Third sub-interval ([2,3]): - Go to the left endpoint of the sub-interval (2).
- Go straight up until you hit the function and figure out the y-value of the function where you hit it (f ( 2 ) = (2)^{2} + 1 = 5).
- Make a rectangle whose base is the subinterval and whose height is the y-value you just found:
Calculate the area of the rectangle: (height) ⋅ (width) = 5 ⋅ 1 = 5 Fourth sub-interval ([3,4]): - Go to the left endpoint of the sub-interval (3).
- Go straight up until you hit the function and figure out the y-value of the function where you hit it (f (3) = (3)^{2} + 1 = 10).
- Make a rectangle whose base is the subinterval and whose height is the y-value you just found:
Calculate the area of the rectangle: (height) ⋅ (width) = 10 ⋅ 1 = 10 The area covered by these four rectangles is 1 + 2 + 5 + 10 = 18. We're still not quite covering the whole area we want, but we're closer. | |
Let S be the region between the graph of g and the x-axis on the interval [0,4]. Use a left-hand sum with 2 sub-intervals to estimate the area of S. Is this an under-estimate or an over-estimate? | |
Dividing the interval [0,4] into 2 sub-intervals gives us sub-intervals of length 2. On the sub-interval [0,2] the height of the rectangle is g (0), which we can see from the graph is 0. The area of this rectangle is 0. On the sub-interval [2,4] the height of the rectangle is g (2), which we can see from the graph is 4. The area of this rectangle is height ⋅ width = 4(2) = 8. Adding the areas of the rectangles together, we estimate the area of S is 0 + 8 = 8. To determine if this is an over-estimate or under-estimate, we need to compare the area that is part of S that we didn't cover (area we missed) to the area that isn't part of S but got covered anyway (extra area!). The missed area is shaded in black, the extra area in pink: There are more black shaded area than pink shaded area in that picture. This means the area we missed is bigger than the extra area we covered, so overall we have an underestimate. | |
- Let W be the region between the graph of f and the x-axis on the interval [-20,20].
Use a left-hand sum with 4 sub-intervals to estimate the area of W. - Let Z be the region between the graph of g and the x-axis on the interval [-4,0].
LHS graphs 4 - without pink - Use a left-hand sum with 2 sub-intervals to estimate the area of Z.
- Use a left-hand sum with 4 sub-intervals to estimate the area of Z.
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We divide the interval [-4,0] into two sub-intervals of size 2, so each rectangle has width 2.
On sub-interval [-4,-2] the rectangle has height g (0) = 0, so the area of this rectangle is also 0.
On sub-interval [-2,0] the rectangle has height g (-2) = 4, so the area of this rectangle is
4(2) = 8. We estimate that the area of Z is
0 + 8 = 8. - We divide the interval [-4,0] into four sub-intervals of size 1, so each rectangle has width 1.
On sub-interval [-4,-3] the height of the rectangle is g (-4) = 0 so the area of the rectangle is 0. On sub-interval [-3,-2] the height of the rectangle is g (-3) = 2 so the area of the rectangle is 2(1) = 2.
On sub-interval [-2,-1] the height of the rectangle is g (-2) = 4 so the area of the rectangle is 4. On sub-interval [-1,0] the height of the rectangle is g (-1) = 6 so the area of the rectangle is 6. Adding the areas of these rectangles, we estimate that the area of Z is 0 + 2 + 4 + 6 = 12. | |
Let f ( x ) = x^{2} + 2 and let R be the region between the graph of f and the x-axis on the interval [0,8]. Use a left-hand sum with 4 sub-intervals to estimate the area of R.
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We're going to go through the same process we've been using. We cut the original interval into 4 sub-intervals of length 2. On the sub-interval [0,2] the height of the rectangle is f ( 0 ) = 0^{2} + 2 = 2 so the area of the rectangle is f ( 0 )(2) = 2(2). (Yes, we know two times two is four, but we don't want to do that just yet. Trust us.) On the sub-interval [2,4] the height of the rectangle is f ( 2 ) = 2^{2} + 2 = 6 so the area of the rectangle is f ( 2 )(2) = 6(2). On the sub-interval [4,6] the height of the rectangle is f (4) = 4^{2} + 2 = 18 so the area of the rectangle is f (4)(2) = 18(2). On the sub-interval [6,8] the height of the rectangle is f (6) = 6^{2} + 2 = 38 so the area of the rectangle is f (6)(2) = 38(2). Adding the areas of these rectangles, we find f ( 0 )(2) + f ( 2 )(2) + f (4)(2) + f (6)(2) = 2(2) + 6(2) + 18(2) + 38(2). We can use the distributive property to pull out the common factor 2 and rewrite this as [f (0) + f (2) + f (4) + f (6)](2) = [2 + 6 + 18 + 38](2). In other words, we take the value of the function at the left endpoint of each sub-interval and add these values. Then we multiply by the width of each sub-interval or rectangle. We never use the value of the function at the right-most endpoint of the original interval. | |
Let f ( x ) = 4x and let R be the region between the graph of f and the x-axis on the interval [1,2]. Use a left-hand sum with 4 sub-intervals to estimate the area of R.
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Dividing the interval [1,2] into 4 equally-sized sub-intervals gets us sub-intervals of length 0.25. We'll be using the values of f at x = 1, 1.25, 1.5, and 1.75. Here are the values of f we need: We add the values of f and multiply by the width of a sub-interval: Remember, we can do also take a left-hand sum when we're given a table or graph for the function instead of a formula. | |
Let f ( x ) = 2x on [2,10]. Find LHS(5). That is, use a left-hand sum with 5 sub-intervals to estimate the area between the graph of f and the x-axis on [2,10].
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We have a = 2, b = 10, and we want 5 sub-intervals. The width of a sub-interval is Then our endpoints are We need the value of f at every endpoint but the last: Using the formula,
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Use a left-hand sum to estimate the area between the graph of g and the x-axis on the interval [0,10].
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We can see from the graph what the value of g is at x = 0, 2, 7, and 8. We have four sub-intervals of lengths 2, 5, 1, and 2: On the sub-interval [0,2] we have a rectangle with height g (0) = 1 and width 2, so its area is 2. On the sub-interval [2,7] we have a rectangle with height g (2) = 3 and width 5, so its area is 15. On the sub-interval [7,8] we have a rectangle with height g (7) = 0, so its area is 0. On the sub-interval [8,10] we have a rectangle with height g (8) = 5 and width 2, so its area is 10. Adding the areas of the rectangles together, we estimate that the area between the graph of g and the x-axis on [0,10] is 2 + 15 + 0 + 10 = 27. | |
Let R be the area between the graph of f ( x ) = x^{2} + 1 and the x-axis on the interval [0,4].
- Draw R and the 8 rectangles that result from using a Left-Hand Sum with 8 sub-intervals to approximate the area of R.
- Use the Left-Hand sum with 8 sub-intervals to approximate the area of R (you might want a calculator).
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,.5]:The left endpoint of this sub-interval is 0.The height of this rectangle is
f ( 0 ) = 0^{2} + 1 = 1
so the area is
(height) ⋅ (width) = 1 ⋅ (0.5) = .5
Sub-interval [.5,1]:
The left endpoint of this sub-interval is 0.5. The height of this rectangle is
f (.5) = .5^{2} + 1 = 1.25
so the area is
(height) ⋅ (width) = 1.25 ⋅ (.5) = .625
Sub-interval [1,1.5]. The left endpoint of this interval is 1. The height of this rectangle is
f (1) = 1^{2} + 1 = 2
so the area is
(height) ⋅ (width) = 2 ⋅ (.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 ⋅ (.5) = 1.625
Sub-interval [2,2.5]:
The height of this rectangle is
f ( 2 ) = 2^{2} + 1 = 5
so the area is
(height) ⋅ (width) = 5 ⋅ (.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) = 1 ⋅ (.5) = 3.625.
Sub-interval [3,3.5]. The height of this rectangle is
f (3) = 3^{2} + 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) = 1 ⋅ (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.
- Let S be the area between the graph of y = f ( x ) = 2^{x} and the x-axis on the interval [1,6].
- Use a Left-Hand Sum with 2 subintervals to approximate the area of S. Draw S and the rectangles used in this Left-Hand Sum on the same graph.
- Use a Left-Hand Sum with 5 subintervals to approximate the area of S. Draw S 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 S?
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:
- 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:
- The table below shows some values of the increasing function f ( x ).
- Use a left-hand sum with one sub-interval to estimate the area between the graph of f and the x-axis on the interval [2,8].
- Use a left-hand sum with three sub-intervals to estimate the area between the graph of f and the x-axis on the interval [2,8].
- Are your answers in (a) and (b) over- or under-estimates of the actual area between the graph of f and the x-axis on the interval [2,8]?
Answer
- If we use only one sub-interval, that means we're not breaking up the original interval up at all. We will 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 in (a) and (b) 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:
- Some values of the decreasing function g are given in the table below:
- Use a left-hand sum with 3 sub-intervals to estimate the area between the graph of g and the x-axis on the interval [-1,2].
- Use a left-hand sum with 2 sub-intervals to estimate the area between the graph of g and the x-axis on the interval [-1,2].
- Are your answers in (a) and (b) over- or under-estimates for the actual area between the graph of g and the x-axis on the interval [-1,2]?
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,.5] the rectangle has height g (-1) = 34 and width 1.5, so its area is
34(1.5) = 51.
On [.5,2] the rectangle has height g (.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 in (a) and (b) were both overestimates. The function g ( x ) and the rectangles from the estimate in (a) 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.
- Let f ( x ) = x^{2} + 6x + 9. Use a left-hand sum with 6 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [-6,-3].
Answer
Dividing the interval [-6,-3] into 6 sub-intervals produces sub-intervals of length .5. We need the following values of f:
We add them, and multiply by the width of a sub-interval:
- Let f ( x ) = -x^{2} + 2x + 8. Use a left-hand sum with 8 sub-intervals to estimate the area between the graph of f and the x-axis on the interval [0,4].
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
Let g be a function with values given by the table below.
Use a left-hand sum with 3 sub-intervals to estimate the area between the graph of g and the x-axis on the interval
[0,12].
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:
Let h be a function with values given by the table below. Use a left-hand sum with 9 sub-intervals to estimate the area between the graph of h and the x-axis on the interval [-9,9].
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
The function f ( x ) on the interval [0,30] is graphed below. Use a left-hand sum with 3 sub-intervals to estimate the area between the graph of f and the x-axis on this interval.
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:
Use a left-hand sum with the sub-intervals indicated by the data in the table to estimate the area between the graph of f and the x-axis on the interval [-10,1].
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
Use a left-hand sum with the sub-intervals indicated by the data in the table to estimate the area between the graph of g and the x-axis on the interval [0,5π].
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
Use a left-hand sum to estimate the area between the graph of h and the x-axis on the interval [2,7].
Answer
From the graph we can tell what h is at x = 0, 1, 2, 3, 5, and 7. Since the problem says to look at the interval [2,7], we don't care about the values of h at 0, 1, or 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