Calculus Online: Lab 2

Welcome to Lab 2 of Math 101, Sections 203, 204, 207 and 209




Instructions

Warning!

If you leave this page or reload it, your work will be lost. Always save your work before leaving or reloading.



Integration and the Fundamental Theorem of Calculus

Our course this term is primarly about integration, which you have seen introduced in class, and the main tool for evaluating integrals, the Fundamental Theorem of Calculus.

Initially, we were led to the concept of integration through the problem of finding the area of a region. Basically, we approximate the area we are interested in by a collection of simple shapes such as rectangles and then add their areas together. In fact, the word integration means "to put together", and it should remind you that this process is simply putting together a large collection of simpler objects to form a more complicated object.

The Fundamental Theorem is important because it gives us a means of evaluating integrals through anti-differentiation. It is used so often that students sometimes think that integration is the same as anti-differentiation. In this lab, we would like to help you see these two important parts of the course more clearly.


Question 1

In this question, we would like to emphasize that an integral is, in some sense, simply a sum. Shown below is the graph of a function. Suppose that we are interested in finding the area under the graph. Our standard trick is to set up a definite integral and evaluate it using the Fundamental Theorem of Calculus. However, this is not always possible: sometimes it is difficult or even impossible to find an antiderivative of the integrand. Then we have to resort to some other method.

Here we have drawn the graph over a grid and we can use this grid to approximate the area. In fact, this is precisely how the process of integration arises.

When you click in a square, it will become illuminated. If you then drag the mouse into a new square, it will also become illuminated. In this way, you can "paint" a region. Shown below the graph, you will see the area of all the illuminated squares. If you click in a square which is illuminated, it will no longer be illuminated.

Part a (2 marks)

Select "Part a" above and shade all the squares which lie entirely below the graph. The area under the graph is guaranteed to be larger than this area.

Part b (2 marks)

Select "Part b" above and shade all the squares for which some part of the square is below the graph. The area under the graph is guaranteed to be smaller than this area.

Part c (2 marks)

Use your results from Parts a and b to estimate the area under the graph. Select "Part c" above and enter your estimate. ( Note: Your result should simply be a number without units.)

Parts a and b give us a range in which we can be sure the area lies. Of course, we can obtain a smaller range, and hence a better estimate, for the area by using a smaller grid. This is the essence of integration.


Question 2

The Fundamental Theorem of Calculus tells us that integration and differentiation are inverse operations; that is, if we first differentiate a function f and then integrate the result, we find the original function f again. In the same way, if we integrate f and then differentiate the result, we find f back again.

In this question, we will show you how this arises in a practical way. Suppose that a particle is moving on the x axis and that its position is measured by $ x(t). $ Remember that the velocity is $ v(t) = x^\prime(t) $ ; that is, the velocity is the rate of change of the position. This means that the average rate of change of the position over a small time interval is approximately the instantaneous velocity. In other words,

\[ v(t_0) \approx \frac{x(t_1) - x(t_0)}{t_1 - t_0} \]

if the interval $ [t_0,t_1] $ is a short time interval or

\[ v(t_0) \Delta t \approx x(t_1) - x(t_0) \]

if $ \Delta t = t_1 - t_0. $

Now if we combine several of these time intervals, and add up the approximate displacements, we find that

\[ \begin{array}{rl} v(t_{n-1})\Delta t + \ldots & + v(t_{2})\Delta t + v(t_1)\Delta t + v(t_0) \Delta t \\ & \approx (x(t_{n}) - x(t_{n-1})) + \ldots + (x(t_3) - x(t_2)) + (x(t_2) - x(t_1)) + (x(t_1) - x(t_0)) \\ & \approx x(t_n) - x(t_0) \end{array} \]

Notice that many terms cancel so we are left with a difference of the starting position and final position. This says that if we add (or "integrate") the displacements over each of the time intervals together, then we have the displacement over the whole time interval. Now if we call $ t_n = b $ and $ t_0 = a $ and let n grow arbitrarily large, we find that

\[ \int_a^b v(t)~dt = x(b) - x(a) \]

Suppose now that we know the velocity of the particle at all times and we would like to know the displacement over some time interval. As we did above, we could use the velocity to form the approximate displacements over some shorter time intervals and then approximate the total displacment. This is illustrated below.

We are considering a particle which starts at x = 0 when t = 0. Suppose we are given the velocity, and we would like to know where the particle is at t = 5. We will break the interval t = 0 to t = 5 into n equal pieces and approximate the displacement over each piece. This is shown below: you will see several bars placed end to end. The width of each bar represents the approximate displacement over that time interval. By moving the red dot, you may vary the number n.

Part a (2 marks)

Estimate the total distance travelled over the time interval t = 0 to t = 5. Select "Part a" above and enter your result. ( Note: Your result should simply be a number. No units please.)

Part b (2 marks)

Estimate the average velocity over this time interval. Select "Part b" above and enter your result. (Again, no units.)

Part c (2 marks)

Estimate the particle's position when it is moving slowest. Select "Part c" above and enter your result. (Just give an approximation based on the diagram above.)

The relationship between the velocity and position studied in this problem is really the Fundamental Theorem of Calculus in a different guise; it says that if we integrate the velocity (which is the derivative of the position), then we obtain the position.


Question 3

Here is the other relationship given by the Fundamental Theorem of Calculus. Let's begin with a function f(x) and form F(x) by integrating:

\[ F(x) = \int_0^x f(t)~dt \]

Then $ F^\prime(x) = f(x). $

Part a (2 marks)

Select "Part a" above. You will see the graph of a function f(x). Move the red dot to a local maximum for the function F(x).

Part b (2 marks)

Select "Part b" above. You will see the graph of the same function f(x). Move the red dot to a local minimum for the function F(x).

Part c (2 marks)

Select "Part c" above and you will see the graph of a function g(x). When you move the dot, you can see how the area under the graph changes. Estimate the value of the function g(3) and enter your estimate in the space provided. (Again, this will just be an estimate.)


Question 4

This question deals with setting up a definite integral. Consider a chain which is hanging vertically. We will find the amount done in lifting the chain up to some level. To do this, we will think of the chain as being broken into many small pieces and find the work done in lifting each small piece. Then we will add together all these small bits of work.

Remember that the work done in lifting a mass m up a height h is mgh where g is a constant. We will suppose that our units are chosen so that g = 1; this means that the work is simply mh.

The density of the chain p measures the ratio of the mass of a section of the chain to its length. This means that if we have a small section of the chain whose length is $ \Delta y, $ then its mass is $ p\Delta y $ and the work required to lift it a distance y is then $ py\Delta y. $

Shown below is the chain broken into a number of pieces, each of which you may lift by clicking on the piece and dragging it up. As you do so, you will see how much work has been done on the graph on the right: the height of the rectangle on the right is py and the width is $ \Delta y $ so that the area of the rectangle represents the amount of work done.

Part a (2 marks)

Determine the length of the chain. Select "Part a" above and enter your result. (No units, please.)

Part b (2 marks)

Determine the density p of the chain. Select "Part b" above and enter your result.

Part c (2 marks)

Of course, the work represented on the graph on the right is an approximation. You can see this on the left: a portion of each piece is lifted higher than it needs to be. Determine the exact amount of work done in lifting the chain. Select "Part c" above and enter your result.

Think about what feature of this graph corresponds to the work required to lift one piece of the chain. Then ask yourself what feature represents the total work to lift all the pieces.



You have now completed the lab and you may save your work. If technical problems prevent your answers from being saved, you will receive a message telling you so. In that case, you will either have to submit your work from the Math labs or try again from some other place. We will NOT accept your answers by e-mail.



How are we doing? Want to complain? Pat us on the back? Please let us know by sending us email.