Area Problem and the Riemann Sum Part III

I will discuss here an important topic which I pulled out from the Area Problem and the Riemann Sum series Part I. As I said, I don't want that post to be very long. In case you skip that part, follow the link above to read and understand the first part. When you pass the subtopic about "Options in choosing the point x_i or the height of the Rectangle" click the link to bring you back here.

In Calculus, It is very important to study and understand clearly the Area Problem because it will give us one of the interpretations of a definite integral that will lead us to the definition of the definite integral. So I decided to make a separate post to elaborate the techniques of estimating the area under a curve.

Area of the Region Between the Function and the x-axis

Again, we are going to assume that we’ve got a function f(x) that is positive on some interval [a,b]. What we want to do is determine the area of the region between the function and the x-axis.

For example, determine the area between f(x) = x² + 1 on [0,2]. We are going to determine the area of the shaded region below.

Seems a hard task to do but from what we learned on the first part of the series if we are going to enclose the shaded region with a rectangle we can estimate the area by dividing up the interval into n subintervals each of width,

Now, in each interval we have the width and we have the height which is is given by the function value at a specific point in the interval. We can then find the area of each of these rectangles, add them up and this will be an estimate of the area. For better understanding keep on reading.

Using the value at the Right Endpoint

This is the first technique or option that we are going to use for estimating the area. If we divide up the interval into 4 subintervals and use the function value at the right endpoint of each interval to define the height of the rectangle. This gives,

By simply looking at the figure, we can conclude that by choosing the height as we did each of the rectangles will over estimate the area since each rectangle takes in more area than the graph each time. Let’s continue to estimate the area. The width of each of the rectangles is ½ [(b-a)/n = (2-0)/4]. The height of each rectangle is determined by the function value at the right endpoint and so the height of each rectangle is nothing more that the function value at the right endpoint. Here is the estimated area.

Using the value at the Left Endpoint

This is our second option. We could have taken the rectangle heights to be the function value at the left endpoint. Using the left endpoints as the heights of the rectangles will give the following graph and estimated area.

In this case we can clearly see that our estimation will be an underestimation since each rectangle misses some of the area each time.

Using the value at Midpoint

The last option is the common point for getting the heights of the rectangles that is often more accurate. Instead of using the right or left endpoints of each sub interval we could take the midpoint of each subinterval as the height of each rectangle. Here is the graph for this case.

So, it looks like each rectangle will over and under estimate the area. This means that the approximation this time should be much better than the previous two choices of points. Here is the estimation for this case.

We’ve now got three estimates. For comparison’s sake the exact area is

So, both the right and left endpoint estimation did not do all that great of a job at the estimation. The midpoint estimation however did quite well.

Careful! Do not Draw any Conclusion yet

Be careful to not draw any conclusion about how choosing each of the points will affect our estimation.

• In this case, because we are working with an increasing function choosing the right endpoints will overestimate and choosing left endpoint will underestimate.
• If we were to work with a decreasing function we would get the opposite results. For decreasing functions the right endpoints will underestimate and the left endpoints will overestimate.
• Also, if we had a function that both increased and decreased in the interval we would, in all likelihood, not even be able to determine if we would get an overestimation or underestimation.

Easiest Way to Get Better Approximation

The three methods is quite satisfying because none of the estimations above really did a great of a job at estimating the area. So we want a better approximation. To this, we could try to find a different point to use for the height of each rectangle but that would be useless and there wouldn’t be any guarantee that the estimation would be in fact be better. So what we are looking is the best way for getting better approximations that would work for any function we would chose to work with whether increasing function or decreasing function.

Let's continue to work, suppose we double the number of rectangles that we used and see what happens. Below are the graphs showing the eight rectangles and the estimations for each of the three choices for rectangle heights that we used above.

Here are the area estimations for each of these cases.

Now, after learning all these methods of approximating the area of a curve what can you conclude? I know you already get it. The more rectangles we used the much better approximation we get.

Accordingly, the best way to get better approximation is to take more rectangles or to increase n subintervals.

Back to Part I of the series

At this point, You are ready to continue learning the concept and the relation of Riemann Sum to Definite Integral. Kindly proceed to toward the end of part I series.

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credit: Paul Dawkins, Lamar University©2013 www.FroydWess.com

Area Problem and the Riemann Sum Part II

This is the part II of the series Area Problem and the Riemann Sum. It is important not to miss the part I before you continue on this topic.

Riemann Sum

Riemann Sum is the total sum of n adjacent rectangles sideways on an interval [a,b] with a uniform positive widths and with heights determined by arbitrary values of a function.

When is Riemann Sum have Positive values

• If f take positive values along the interval [a,b], just like in Figure 4, the function f is positive. All the heights of adjacent rectangles take positive values. In this case, we have adjacent rectangles with a positive heights and a positive widths. Therefore, The Riemann sum that approximates the area between the x-axis and f have positive values.
• A positive area represents the fact that the particular rectangle is above the x-axis. 

When is Riemann Sum have Negative values

• However, if f were to take some negative values along the interval [a,b], then some of the heights would be assigned negative values. Therefore, negative height values multiplied by negative width values give negative area values.
• A negative area represents that the rectangle is below the x-axis (since its height is given by a negative value of f ).

What if function f takes both positive and negative values

• If f is continuous and takes positive and negative values along [a,b], then any corresponding Riemann sum approximates a net area where the net area equals the difference between the area of regions above the x-axis but under f and the area of regions below the x-axis but above f.
• In short, the areas of the rectangles above the x-axis minus the areas of the rectangles below the x-axis.

Exact Net Area Between f and the x-axis

Until now, I know after all these things, you might be asking, How to get the exact area then?

Here is the idea, I suppose you have the knowledge about Limit. Using the Riemann sum approximation, we will define the exact net area between f and the x-axis as the limit of the sum of the areas of approximating rectangles as the number of rectangles approach infinity (n → ∞) .

Assume that f is a continuous function along the interval [a,b] . The net area A of the region bounded by f and the x-axis is given by

where n represents the number of approximating rectangles with a uniform width equal to Δx = (b − a)/n and where x_i represents some point in the ith subinterval of [a,b] .

Definite Integral

As long as f is continuous the value of the limit is independent of the sample points x_i used. The definite integral of f from a to b is the number

provided the limit exists.

Sample Riemann Sum Problems

1. Use a Riemann sum to estimate the area between f and the x-axis along the interval [1,3] forf (x) = 0.25x³ .

Answer: Using four rectangles and right end points as the sample point in each subinterval:

2. Find the area between the curve y = x³ and the x-axis over the interval [0,2]. Use a sketches to show how to obtain over and under estimates for the area using Riemann sums.

Let y = f (x) . Note that the function increases. Arbitrarily select a number of rectangles like four. Determine the width of each rectangle.

For an over-estimate, select heights in a manner that too much area is included. A right Riemann sum will do the trick as below.

For an under-estimate, select heights in a manner that too little area is included. A left Riemann sum will suffice as shown.

For your exercise selects heights in the midpoints, make the graph and find the area. Let me know how well you get the answer by providing feedback on the comment section.

The next topic is Finding Areas using Definite Integration. See you there.

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credit: Randell Simpson (Temple College)©2013 www.FroydWess.com

Area Problem and the Riemann Sum Part I

To give you a brief review, remember that Calculus is a branch of mathematics that addressed two problems: finding instantaneous rate of change and the area between a curve and the x-axis. In differential calculus you have learned that the derivative gives the instantaneous rate of change. It answers the question, How fast is "something" changing?. The antiderivative or integral gives the area between a curve and the x- axis. It answers the question, How much is the "total"?.

In my next lectures you are going to learn to compute the area under a curve and area between a curve using definite integration. I know you are excited to solve problems, but before that you must focus to learn this three-part series lectures first.

Area of a Rectangle

For now, I am going to discuss the techniques for estimating the area under a curve. Let us consider a rectangle. How can we compute the area of rectangle with height h and length b - a ? Yes, it is very easy simply the product  h ⋅ (b − a). 

Area Under a Curved

We are going to assume a region that is bounded with a curved side, What we want to do is determine the area.  Consider, for example, the region shown below that is bounded by f, the x-axis, and the lines x = a and x = b .

From the first look, we can't possibly imagine how to compute for the exact area of R. We can say that getting the area of the region between f and the x-axis is really a hard task to do.  This problems made our Mathematicians of long ago exerted a lot of effort to uncover the solutions for this problems.

Enclosed the region R with a Rectangle

Enclosing a rectangle over R as in Figure 3 reveals a simple process for easy computing an estimate for the area.

Figure 3 shows that the product h ⋅ (b − a) gives a gross yet reasonable estimate for the area in R.  If we let A equals the area of R, then clearly we can say that  h ⋅ (b − a) > A since h is greater than or equal to all values of f along the interval [a,b] .

Divide the Rectangle for Much Better Estimate

To obtain a much better estimate,  we divide the rectangle with height h into n rectangles of some uniform width equal to (b − a) /n with heights determined in some manner by f. 

In Figure 4 below, each rectangle intersects f at the top left corner so that the height of each rectangle is determined by f(x_i ) where x_i corresponds to the left end point of the subintervals of [a,b] that comprise the widths of the rectangles. The sum of the areas of the rectangles provides a much improved estimate of the area of R.

The number of rectangles used to generate an estimate is arbitrary, but we can conclude that the greater the number of rectangles used the more accurate the estimate will be. Similarly, the choice to determine the height of each rectangle using f(x_i ) where x_i  corresponds to the left end point of the subintervals of [a,b] is also arbitrary. The points x_i  could refer to the right end points of the subintervals of [a,b] , the midpoints of the subintervals of [a,b] , or even just any random point within each subinterval of [a,b] .

Options in choosing the point x_i or the height of the Rectangle

I just want to highlight or give emphasis to this crucial part of estimating the area because it is important for you to know what would be the difference if you choose the right end points, the left end points or the midpoints. To answer the question, I made a separate post because I don't want this post to be very long. Just follow the link Area Problem and the Riemann Sum Part III and at the end of the post you will find a link to bring you back here.

Welcome back

I suppose you already read and learned the different methods of estimating the area. Let's continue from Figure 4, we are going to write an approximation formula for A, the area of R. Since the number of rectangles used for the estimate is arbitrary, let's represent the number of rectangles with n. Recall that the rectangles span [a,b], and each rectangle has a width given by (b − a) /n . We will let x₁, x₂, . . ., x_n be sample points taken from the first, second, . . ., and nth subintervals. Accordingly, we have the following.

The notation above is easy to evaluate if n let's say five, but if n is large number, the notation becomes somewhat very long and complex. In that case, we are going to use sigma notation. The Greek letter sigma, Σ, indicates the summation of terms. Letting x_i  be some arbitrary point within each closed subinterval of [a,b], and letting Δx represent the width of each rectangle, (b − a) /n , we

obtain the following.

The letter i is called the index of summation and stands for the ith (that is, the first, second, . . .,nth) subinterval. The sum is called a Riemann sum.

I will let you absorb all these things first before we continue with the Riemann Sum. If you have any clarification just leave it on the comment section below. Now if you are ready you can proceed to Part II of the series.

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credit: Randell Simpson (Temple College)©2013 www.FroydWess.com

Area Under a Curve and Between a Curve - Set 1 Answers key

It's nice seeing you here. These are the answers for the problems involving Area Under a Curve and Between Two Curves. You can always check back the problems by following the link above.

 1. 4 units²

2. 9 units²

3. 32/3 units²

4. 6 units²

5. 76/3 units²

6. 1/3 units²

7. 64/3 units²

8. 142/3 units²

9. units²

10. 18 units²

If you skip this particular topic try to visit Finding Areas using Definite Integration.

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Area Under a Curve and Between a Curve - Set 1 Problem

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I think you are ready and excited to solve some area problems. But not too fast, before you continue I would like to remind you for a few pointers. If you would like to review first then visit Finding Areas using Definite Integration

• Each problems required a graph. It is much easier if you make graph for each solutions. 
• In most cases the bounding region(enclosed) , which will give the limits of integration, is difficult to determine without a graph. 
• It is hard to determine which of the functions is the upper function and which is the lower function without a graph. 
• Finally,  unlike the area under a curve, the area between two curves will always be positive.  If you get a negative number or zero, you’ve made a mistake somewhere and you will need to go back and find it.

Begin. Good luck!

1. Find the area of the region bounded by y = 2x, y = 0, x = 0 and x = 2.

2. Find the area of the region enclosed between the curves y = x ² - 2x + 2 and -x ² + 6 .

3. Find the area of the region enclosed by y = (x-1) ² + 3 and y = 7.

4. Find the area of the region bounded by x = 0 on the left, x = 2 on the right, y = x ³ above and y = -1 below.

5. Find the area under the curve y = x² + 1 between x = 0 and x = 4 and the x-axis.

6. Determine the area of the region enclosed by y = x ² and y = x ^1/2.

7. Determine the area of the region bounded by y = 2x ² + 10 and y = 4x + 16.

8. Determine the area of the region bounded by by y = 2x ² + 10, y = 4x + 16, x = -2, and       x = 5.
9. Determine the area of the region enclosed by y = sinx, y = cosx, x = , and the y axis.
10. Determine the area of the region enclosed by , y = x-1.

You can check your works and see if you got it right by following this link Answers to Area Under a Curve and Between a Curve - Set 1 Problem.

More Area problems

11. Determine the area of the region bounded by x = -y ² + 10 and x = (y-2) ².

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Definite Integrals - Set 1 Answers key

If you are confident enough for your answers in Definite Integral problems Set 1. Check the results. I suppose you get all of them right.

1. 176

2. 8

3. ln(5)

4. ln(4)

5. 2

6.

7. 103

8. 38

9.

10.

Solve applied problems involving Definite Integrals

1. The change in cost is 9.9 thousand pesos.

2. 96,000 pesos

How did you find Definite Integral? Let me know, How well you learned the topic. Share on our comment section. Thanks.

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Definite Integral - Set 1 Problems

Time to challenge yourself by evaluating the definite integrals of the following problems below.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Solve applied problems involving Definite Integrals

1. The marginal cost of a certain electronics firm is given by the equation

where C′ is in thousands of pesos and the quantity x produced is in hundreds of units per day. If the number of units produced in a given day changes from two hundred to five hundred units, what is the change in cost?

2. The marginal cost function for producing x units of ACU is 3x² − 200x + 1500 pesos. Find the increase in cost if production is increased from 90 to 100 units.

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