<Text-field style="_cstyle433" layout="Heading 1"><Font bold="true">Summing It up with Riemann, Definite Integrals, and the Fundamental Theorem of Calculus</Font></Text-field>

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<Text-field style="Heading 1" layout="Heading 1">Introduction</Text-field>

OBJECTIVE: Visualize the relationship among Riemann sums, the definite integral, and the Fundamental Theorem of the integral calculus. Use each of the concepts to compute or approximate the signed area under the graph of a function representing various applications.

In practical applications, the signed area between the graph of a function and the x or t-axis can represent a wide variety of different things. For example, if a function gives the rate of water flowing into a storage tank as a function of time, the total signed area between the graph of the flow-rate function and the t-axis represents the total amount of water that accumulates or drains from the tank during the time interval. To see that this is the case, you should recall something that you have probably known for a long time, that is, change in a quantity is equal to its rate of change times time. Some familiar examples of this are "distance is equal to speed times time" and "amount is equal to rate times time." If water flows into a tank at a constant rate of 30 liters per second for 100 seconds, then it easy to determine that the total water added to the tank during this time is 300\303\227100 = 30000 liters. These simple rules are based, of course, on the assumption that the rate of change is constant during the time interval. Now that we are more sophisticated, we know that rates of change do not have to be constant and can, in fact, vary continuously over a time interval. How do we calculate the total change over a time interval when the rate varies from one instant to the next? Calculus provides the answer, and we begin by looking at a technique for estimating change from rates of change using a summation process called Riemann sums.

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<Text-field style="Heading 1" layout="Heading 1">Part I: An Example</Text-field>

The idea of Riemann sums builds upon what you already know, that is change is equal to rate of change times time. Suppose that f(t) represents the rate of change of some quantity Q, that is, LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2J0YrLUYjNiYtRiw2JFEiZkYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RMCZBcHBseUZ1bmN0aW9uO0YnL0Y3USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQS8lKXN0cmV0Y2h5R0ZBLyUqc3ltbWV0cmljR0ZBLyUobGFyZ2VvcEdGQS8lLm1vdmFibGVsaW1pdHNHRkEvJSdhY2NlbnRHRkEvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZQLUkobWZlbmNlZEdGJDYlLUYjNiQtRiw2JFEidEYnRjZGNi8lJ2l0YWxpY0dGQUY9RjYtRjo2LlEiPUYnRmVuRj1GP0ZCRkRGRkZIRkpGTC9GT1EsMC4yNzc3Nzc4ZW1GJy9GUkZbby1JJm1mcmFjR0YkNigtRiM2JC1GLDYkUSNkUUYnRjZGNi1GIzYkLUYsNiRRI2R0RidGNkY2LyUubGluZXRoaWNrbmVzc0dRIjFGJy8lK2Rlbm9tYWxpZ25HUSdjZW50ZXJGJy8lKW51bWFsaWduR0ZfcC8lKWJldmVsbGVkR0ZBRjZGK0Y2. The quantity Q might represent the amount of water in a reservoir, the distance traveled by a moving object, or the amount of money in an investment portfolio. In these types of practical applications, it is often times easier to consider the forces that influence change and thereby determine f(t) than it is to determine Q directly. But if you know f(t), there is a way to estimate the change in Q over an interval of time, even though the rate of change varies from one instant to the next. Here it is. Let's consider a specific example. Suppose that 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 for 0 <= t <= 15, and that we wish to calculate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= over this time interval. First, let's graph f(t).

restart; with(student):

f := t -> 1/4 - cos(Pi*t/12); plot( f(t), t=0..15, labels=["t", "y = f(t)"], thickness = 2, color=green );

We know from the definition of differentials that over a short interval of time LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= is approximately equal to 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. This suggests a method of attack. If we partition the 15-hour period into very short time intervals, called subintervals, and then approximate the change in Q over each of these subintervals and add them up, we should get an approximation of the total LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= as t goes from 0 to 15. We let dt be the length or duration of each of the subintervals and evaluate f(t) at some instant in each subinterval. Where should we pick t in each subinterval? If the subintervals are short enough, it shouldn't make too much difference. (You might guess, however, that it would probably be better to pick t somewhere near the middle of each subinterval since that would most likely give a value of f(t) that is near its average value over each subinterval.) We can say all of this mathematically as follows. First we divide the total 15-hour period into n subintervals each of length dt and number each subinterval from 1 to n. Then, 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 which is approximately equal to 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, where h = dt and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRImlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== is any value of t taken from the LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1cEdGJDYlLUkjbWlHRiQ2JVEiaUYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GLzYlUSN0aEYnRjJGNS8lMXN1cGVyc2NyaXB0c2hpZnRHUSIwRicvRjZRJ25vcm1hbEYn subinterval. The sum that we form in this way to approximate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= is called a Riemann sum.

There is a geometric view that can be attached to the Riemann sum, 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. The product, 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, can be thought of as the area of a rectangle that is h units wide and 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 units high. In each subinterval, we can draw this rectangle that is h units wide and 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 units high. The base of each rectangle is on the x or t axis, and the opposite side is on the graph of f(t). The Riemann sum is the sum of the areas of these rectangles. If we admit that 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 could be negative, then the area of a rectangle can be negative; hence we add signed areas in a Riemann sum, which itself can be positive, negative, or 0. The product can be negative when 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 is positive and h is negative. This would occur if we were to sum back through the time history of f(t). It makes sense that if Q increases as time progresses, then it would decrease if were to go backwards through the same time interval, and vice versa. To illustrate the geometry of Riemann sums, we use the Maple leftbox( ) and rightbox( ) functions from the student package. The arguments of leftbox( ) and rightbox( ) are the function f(t), the independent variable t, and a plot range.

Let's use leftbox( ) on the function we have been considering and use 15 rectangles.

leftbox( f(t), t=0..15, 15);

Note that the rectangles under the t axis have negative areas, whereas the ones above have positive areas. Try changing the number of rectangles in the leftbox( ) command to see that the total signed area of the rectangles gets closer to the signed area under the graph of f(t). You can also change the function and the bounds on the interval to explore other possibilities if you wish.

<Text-field style="Heading 1" layout="Heading 1">You Try It: Part I</Text-field>

In addition to leftbox( ), Maple has written rightbox( ) and middlebox( ) commands so that you can see the geometry of a Riemann sum when we pick points other than the left endpoint in each subinterval. The arguments are the same as for leftbox( ). Use these commands to visualize the Riemann sums for the function considered in the preceding section and for some other functions that you choose. Here is an example. This may take a few minutes to execute. First, try it on the function we have selected, then try it with one of your own by changing the appropriate entries in the commands that follow. Be careful to use correct notation when inserting a function.

f := t -> -3*t^3; a := -2; b := 2; n := 20; leftbox( f(t), t=a..b, n, title="Left Reimann Sum" , labels=["t", "y"]); middlebox( f(t), t=a..b, n, title="Middle Reimann Sum",labels=["t", "y"] ); rightbox( f(t), t=a..b, n, title="Right Reimann Sum",labels=["t", "y"]);

Geometrically, which looks as if it would give the best results? Try it with another function.

<Text-field style="Heading 1" layout="Heading 1">Part II: Summing It Up with Riemann</Text-field>

We need to calculate the total signed area of the rectangles. First we calculate h and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRImlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ== for each of the subintervals, and then we add up the signed areas of the rectangles. We use the left endpoint in each interval for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUklbXN1YkdGJDYlLUkjbWlHRiQ2JVEiY0YnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1GIzYkLUYvNiVRImlGJ0YyRjUvRjZRJ25vcm1hbEYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGPQ==.

f := t -> 1/4 - cos(Pi*t/12): a := 0: b := 15: n := 15: area := leftsum( f(t), t=a..b, n ): printf( `The total signed area of the rectangles is %a`, evalf( area ) );

To see what happens as we increase the number of rectangles, we combine the leftbox( ) command together with the commands in the preceding section and put them all in a single section, the one that follows. With this we can easily change f(t), n, a, and/or b.

f := t -> 1/4 - cos(Pi*t/12): a := 0: b := 15: n := 30: area := leftsum( f(t), t=a..b, n ): printf( `The total signed area of the rectangles is %a`, evalf( area ) ); leftbox( f(t), t=a..b, n );

If we let the number of subintervals and corresponding rectangles approach infinity, we might expect the Riemann sum to approach the actual signed area under the graph of f(t), provided f is continuous on [a, b], and that this value will be exactly equal to LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= over the interval.

<Text-field style="Heading 1" layout="Heading 1">You Try It: Part II</Text-field>

Try out your own function by changing the definition of f. Be careful to use the correct notation. Replace any of the appropriate terms with functions and domains of your own choosing.

f := t -> exp(-t^2/2)/sqrt(2*Pi): a := -2: b := 2: n := 30: area := leftsum( f(t), t=a..b, n ): printf( `The total signed area of the rectangles is %a`, evalf( area ) ); leftbox( f(t), t=a..b, n , view=[-3..3,0..0.5],labels=["t","y"]);

<Text-field style="Heading 1" layout="Heading 1">Part III: Take it To the Limit - The Definite Integral</Text-field>

To evaluate the limit of the Riemann sum, we form the sum without assigning a specific value to n and then take the limit as n goes to infinity. In the first cell that follows, we use Maple to find an expression for the Riemann sum in terms of n, and in the next cell we take the limit.

unassign('n'): f := t -> 1/4 - cos(Pi*t/12): a := 0: b := 15: area := leftsum( f(t), t=0..15, n ): area := simplify( value( area ), 'trig' );

Now take the limit.

limit( area, n = infinity );

This is the exact value for the signed area under the graph of f(t), and hence it is also the exact value for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc=. The limit of a Riemann sum is called a definite integral and is specified by the following notation, LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2KEYrLUYjNiotSShtc3Vic3VwR0YkNictSSNtb0dGJDYwUSYmaW50O0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lK2ZvcmVncm91bmRHUS5bMTQ0LDE0NCwxNDRdRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnL0krbXNlbWFudGljc0dGJFEmaW5lcnRGJy8lJmZlbmNlR0Y8LyUqc2VwYXJhdG9yR0Y8LyUpc3RyZXRjaHlHRjwvJSpzeW1tZXRyaWNHRjwvJShsYXJnZW9wR0Y8LyUubW92YWJsZWxpbWl0c0dGPC8lJ2FjY2VudEdGPC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRlYtRiw2JFEiYUYnL0ZBUSdpdGFsaWNGJy1GLDYkUSJiRidGZm4vJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMkYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGKy1GIzYmLUYsNiRRImZGJ0Zmbi1GNzYtUTAmQXBwbHlGdW5jdGlvbjtGJ0ZARkZGSEZKRkxGTkZQRlJGVEZXLUkobWZlbmNlZEdGJDYlLUYjNiQtRiw2JFEidEYnRmZuRmZuRjpGQEZmbkYrLUknbXNwYWNlR0YkNiYvJSdoZWlnaHRHUSYwLjBleEYnLyUmd2lkdGhHUSYwLjNlbUYnLyUmZGVwdGhHRmZwLyUqbGluZWJyZWFrR1ElYXV0b0YnLUY3NjBRMCZEaWZmZXJlbnRpYWxEO0YnRjpGPUZARkNGRkZIRkpGTEZORlBGUkZURldGXnBGZm4tRjc2LlEiPUYnRjpGQEZGRkhGSkZMRk5GUEZSL0ZVUSwwLjI3Nzc3NzhlbUYnL0ZYRmZxLUYjNictSSdtdW5kZXJHRiQ2JS1GNzYuUSRsaW1GJ0Y6RkBGRkZIRkpGTEZOL0ZRUSV0cnVlRidGUkZUL0ZYUSwwLjE2NjY2NjdlbUYnLUYjNiYtRiw2JFEibkYnRmZuLUY3Ni5RJyZyYXJyO0YnRjpGQEZGRkhGSkZMRk5GUEZSRlRGVy1GLDYkUSgmaW5maW47RidGZm5GZm4vJSxhY2NlbnR1bmRlckdGPEYrLUZicDYmRmRwL0ZocFEmMC40ZW1GJ0ZqcEZccS1Gam82JS1GIzYoLUkrbXVuZGVyb3ZlckdGJDYnLUY3Ni5RJiZTdW07RidGOkZARkZGSC9GS0ZhckZML0ZPRmFyRmByRlJGVEZici1GIzYmLUYsNiRRImlGJ0ZmbkZicS1JI21uR0YkNiVRIjFGJ0Y6RkBGZm5GZnJGUkZfc0YrLUZicDYmRmRwL0ZocFEkNS4wRidGanBGXHEtRiM2J0YrLUYjNiZGY29GZm8tRmpvNiUtRiM2JC1JJW1zdWJHRiQ2JS1GLDYkUSJjRidGZm4tRiM2JEZjdEZmbkZeb0ZmbkY6RkBGZm4tRjc2LVExJkludmlzaWJsZVRpbWVzO0YnRkBGRkZIRkpGTEZORlBGUkZURlctRiw2JFEiaEYnRmZuRmZuRitGZm5GOkZARmZuRitGZm5GK0Zmbg== where 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 . We can use Maple to evaluate definite integrals as in the following cell.

deltaQ := int( f(t), t=0..15 );

Comparing this result with the value of the Riemann sum calculated above, we see that they are the same.

It is important to remember that a definite integral is defined as the limit of a Riemann sum. We are sure that the definite integral of a function will exist (i.e., the limit of the Riemann sum will be a finite number), provided that f(t) is continuous on the closed interval over which we perform the summation.

<Text-field style="Heading 1" layout="Heading 1">You Try It: Part III</Text-field>

Leftbox Approximations

For each of the following functions, use the group of commands copied into the section below to estimate the definite integral using a Riemann sum. First use the leftbox( ) command to show the graph and some rectangles (use fewer than 50 rectangles for a clear image). Then use the commands in the next cell to see if you can make an estimate of the definite integral that is accurate to 3 digits by varying n. Make note of how many subintervals you need to use to do this. After you make an estimate, use the definite integral command to see Maple's decimal estimate of the exact value, and compare it with yours. Here are some functions to try. We show (c) as an example and find by trial and error that 5836 rectangles are needed to get three-digit accuracy. a) 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 for 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

b) 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 for 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

c) 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 for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JKG1mZW5jZWRHRiQ2JC1GIzYmLUkjbW5HRiQ2JFEkMC41RicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVElJmxlO0YnRjovJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRkMvJSlzdHJldGNoeUdGQy8lKnN5bW1ldHJpY0dGQy8lKGxhcmdlb3BHRkMvJS5tb3ZhYmxlbGltaXRzR0ZDLyUnYWNjZW50R0ZDLyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGUi1GLDYlUSJ0RicvJSdpdGFsaWNHUSV0cnVlRicvRjtRJ2l0YWxpY0YnRjpGOkY9LUY3NiRRIjNGJ0Y6RjpGK0Y6

d) 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 for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JKG1mZW5jZWRHRiQ2JC1GIzYmLUkjbW5HRiQ2JFEiMUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RJSZsZTtGJ0Y6LyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZDLyUpc3RyZXRjaHlHRkMvJSpzeW1tZXRyaWNHRkMvJShsYXJnZW9wR0ZDLyUubW92YWJsZWxpbWl0c0dGQy8lJ2FjY2VudEdGQy8lJ2xzcGFjZUdRLDAuMjc3Nzc3OGVtRicvJSdyc3BhY2VHRlItRiw2JVEidEYnLyUnaXRhbGljR1EldHJ1ZUYnL0Y7USdpdGFsaWNGJ0Y6RjpGPS1GNzYkUSI0RidGOkY6RitGOg==

e) 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 for LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JKG1mZW5jZWRHRiQ2JC1GIzYmLUkjbW5HRiQ2JFEiMUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RJSZsZTtGJ0Y6LyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZDLyUpc3RyZXRjaHlHRkMvJSpzeW1tZXRyaWNHRkMvJShsYXJnZW9wR0ZDLyUubW92YWJsZWxpbWl0c0dGQy8lJ2FjY2VudEdGQy8lJ2xzcGFjZUdRLDAuMjc3Nzc3OGVtRicvJSdyc3BhY2VHRlItRiw2JVEidEYnLyUnaXRhbGljR1EldHJ1ZUYnL0Y7USdpdGFsaWNGJ0Y6RjpGPS1GNzYkUSIyRidGOkY6RitGOg==

You can change the appropriate terms below to explore the functions listed above.

f := t -> 2/t^2: a := 0.5: b := 3: n := 15: leftbox( f(t), t=a..b, n , view=[-1..3,0..10]);

n := 5836; area := leftsum( f(t), t=a..b, n ); printf( `The total signed area of the rectangles is %a`, evalf( area ) );

evalf(int(f(t), t=a..b));

Middlebox Approximations

Write a series of commands to calculate the Riemann sum using the midpoint of each subinterval. Use your series of commands to repeat the problems in exercise 1. Compare the number of rectangles that are required to get three-digit accuracy using the left Riemann sum and the midpoint Riemann sum, and discuss what you find. Use the command middlebox( ) to visualize the midpoint rectangles if you wish.

<Text-field style="Heading 1" layout="Heading 1">Part IV: Sometimes There is an Easier Way - The Fundamental Theorem of Calculus</Text-field>

Remember that our motivation for the area problem is to calculate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= over an interval of its domain LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1JKG1mZW5jZWRHRiQ2JC1GIzYmLUYsNiVRImFGJy8lJ2l0YWxpY0dRJXRydWVGJy8lLG1hdGh2YXJpYW50R1EnaXRhbGljRictSSNtb0dGJDYtUSUmbGU7RicvRj1RJ25vcm1hbEYnLyUmZmVuY2VHUSZmYWxzZUYnLyUqc2VwYXJhdG9yR0ZHLyUpc3RyZXRjaHlHRkcvJSpzeW1tZXRyaWNHRkcvJShsYXJnZW9wR0ZHLyUubW92YWJsZWxpbWl0c0dGRy8lJ2FjY2VudEdGRy8lJ2xzcGFjZUdRLDAuMjc3Nzc3OGVtRicvJSdyc3BhY2VHRlYtRiw2JVEidEYnRjlGPEZDRkNGPy1GLDYlUSJiRidGOUY8RkNGK0ZD, given that we know its derivative function, 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. Our approach has been:

1. to approximate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= by dividing the interval of the domain into a sequence of subintervals

2. to then approximate the 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's over each of these with 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 (i.e., over a short interval, change is approximately equal to rate times time)

3. to finally adding them up in a Riemann sum and take the limit as the number of subintervals goes to infinity.

As the number of subintervals increases, the approximation improves, approaching the exact LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc=. That is, as n goes to infinity, the limit of the Riemann sum gives the exact value of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc=, and this we call the definite integral of f(t) over the interval [a, b]. You are probably wondering, why all the fuss? Wouldn't it be easier simply to find Q = F(t), an antiderivative of f(t), and calculate 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 = 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 ? Yes, in fact we can do exactly that. This can make the whole business of evaluating a definite integral a lot easier, provided there is a simple formula for the antiderivative, which isn't always the case. Let's do it for the integral 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, where 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.

f := t -> 1/4 - cos(Pi*t/12);

First, we find an antiderivative of f(t) and then calculate LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2Ji1GLDYlUSgmRGVsdGE7RicvJSdpdGFsaWNHUSZmYWxzZUYnLyUsbWF0aHZhcmlhbnRHUSdub3JtYWxGJy1JI21vR0YkNi1RMSZJbnZpc2libGVUaW1lcztGJ0Y3LyUmZmVuY2VHRjYvJSpzZXBhcmF0b3JHRjYvJSlzdHJldGNoeUdGNi8lKnN5bW1ldHJpY0dGNi8lKGxhcmdlb3BHRjYvJS5tb3ZhYmxlbGltaXRzR0Y2LyUnYWNjZW50R0Y2LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdGTi1GLDYlUSJRRicvRjVRJXRydWVGJy9GOFEnaXRhbGljRidGN0YrRjc= .

F:=unapply(int(f(t), t)+C,t);

deltaQ := (a,b)->F(b)-F(a);

deltaQ( 0, 15 );

This is the same value we obtained by taking the limit of a Riemann sum! This result is fundamental. It tells us that we can evaluate the definite integral of a function, f(t), either by 1) taking the limit of a Riemann sum or 2) finding an antiderivative of f(t), let's call it F(t), and taking the difference of this function over the interval of summation. We can summarize this as follows: LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2KEYrLUYjNiotSShtc3Vic3VwR0YkNictSSNtb0dGJDYwUSYmaW50O0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lK2ZvcmVncm91bmRHUS5bMTQ0LDE0NCwxNDRdRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnL0krbXNlbWFudGljc0dGJFEmaW5lcnRGJy8lJmZlbmNlR0Y8LyUqc2VwYXJhdG9yR0Y8LyUpc3RyZXRjaHlHRjwvJSpzeW1tZXRyaWNHRjwvJShsYXJnZW9wR0Y8LyUubW92YWJsZWxpbWl0c0dGPC8lJ2FjY2VudEdGPC8lJ2xzcGFjZUdRJjAuMGVtRicvJSdyc3BhY2VHRlYtRiw2JFEiYUYnL0ZBUSdpdGFsaWNGJy1GLDYkUSJiRidGZm4vJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMkYnLyUvc3Vic2NyaXB0c2hpZnRHUSIwRidGKy1GIzYmLUYsNiRRImZGJ0Zmbi1GNzYtUTAmQXBwbHlGdW5jdGlvbjtGJ0ZARkZGSEZKRkxGTkZQRlJGVEZXLUkobWZlbmNlZEdGJDYlLUYjNiQtRiw2JFEidEYnRmZuRmZuRjpGQEZmbkYrLUknbXNwYWNlR0YkNiYvJSdoZWlnaHRHUSYwLjBleEYnLyUmd2lkdGhHUSYwLjNlbUYnLyUmZGVwdGhHRmZwLyUqbGluZWJyZWFrR1ElYXV0b0YnLUY3NjBRMCZEaWZmZXJlbnRpYWxEO0YnRjpGPUZARkNGRkZIRkpGTEZORlBGUkZURldGXnBGZm4tRjc2LlEiPUYnRjpGQEZGRkhGSkZMRk5GUEZSL0ZVUSwwLjI3Nzc3NzhlbUYnL0ZYRmZxLUYjNictSSdtdW5kZXJHRiQ2JS1GNzYuUSRsaW1GJ0Y6RkBGRkZIRkpGTEZOL0ZRUSV0cnVlRidGUkZUL0ZYUSwwLjE2NjY2NjdlbUYnLUYjNiYtRiw2JFEibkYnRmZuLUY3Ni5RJyZyYXJyO0YnRjpGQEZGRkhGSkZMRk5GUEZSRlRGVy1GLDYkUSgmaW5maW47RidGZm5GZm4vJSxhY2NlbnR1bmRlckdGPEYrLUZicDYmRmRwL0ZocFEmMC40ZW1GJ0ZqcEZccS1Gam82JS1GIzYoLUkrbXVuZGVyb3ZlckdGJDYnLUY3Ni5RJiZTdW07RidGOkZARkZGSC9GS0ZhckZML0ZPRmFyRmByRlJGVEZici1GIzYmLUYsNiRRImlGJ0ZmbkZicS1JI21uR0YkNiVRIjFGJ0Y6RkBGZm5GZnJGUkZfc0YrLUZicDYmRmRwL0ZocFEkNS4wRidGanBGXHEtRiM2J0YrLUYjNiZGY29GZm8tRmpvNiUtRiM2JC1JJW1zdWJHRiQ2JS1GLDYkUSJjRidGZm4tRiM2JEZjdEZmbkZeb0ZmbkY6RkBGZm4tRjc2LVExJkludmlzaWJsZVRpbWVzO0YnRkBGRkZIRkpGTEZORlBGUkZURlctRiw2JFEiaEYnRmZuRmZuRitGZm5GOkZARmZuRitGZm5GK0Zmbg== = 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 or, in terms of Q and LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVEjZFFGJy8lJ2l0YWxpY0dRJXRydWVGJy8lLG1hdGh2YXJpYW50R1EnaXRhbGljRicvRjhRJ25vcm1hbEYnLUYjNiQtRjE2JVEjZHRGJ0Y0RjdGOi8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRi8lKWJldmVsbGVkR1EmZmFsc2VGJ0Y6, 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. This all makes good sense if you think of it as a generalization of our old friend, change is equal to rate times time. This result is so important that it is called the Fundamental Theorem of Calculus, Part 2. The theorem assures us that we can evaluate a definite integral in this simpler way whenever f(t) is continuous over the interval of summation and provided a formula exists for the antiderivative of f(t). The Fundamental Theorem, Part 2, helps us in two ways. First, as we have discovered, it tells us that instead of taking the limit of a Riemann sum, we can evaluate the definite integral of a function f(t) by taking the difference of values of an antiderivative F(t) over the interval of integration, provided there is a formula for F(t). But what if, as is often the case, there is no algebraic formula for F(t)? Well, now we can turn the problem around. Even when we don't have a formula for F(t), we can evaluate a change in the value of F(t) over an interval [a, b] by taking the limit of a Riemann sum. Specifically,

F(b) - F(a) = 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 = 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. This is the second way that the Fundamental Theorem, Part 2, helps us out. But in this last case, how do we know that the value we calculate is actually a difference of an antiderivative of f(t)? That is, if there is no algebraic formula for F(t), how do we know that the value we calculate by taking the limit of a Riemann sum with f(t) is actually F(b) - F(a), where F(t) is an antiderivative of f(t)? The Fundamental Theorem of Calculus, Part 1, provides the answer. It assures us that if f(t) is continuous on [a, b], then f(t) has an antiderivative F(t) that is defined for all t in [a, b] and we can use the definite integral, that is, the limit of a Riemann sum, to form F(t). Specifically, 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. As a result, we can say for sure that 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, where F(t) is an antiderivative of f(t).

<Text-field style="Heading 1" layout="Heading 1">You Try It: Part IV</Text-field>

Now you try it with the functions in the list that follows. First, calculate the limit of a Riemann sum, and then find an antiderivative and calculate the difference over the interval of summation. Change the limits on the interval (i.e., a and b) as well as the function definition for each exercise. We show (d) as an example. (Note that because Maple has difficulty evaluating some limits symbolically, we use the evalf( ) and limit( ) commands, which gives a numeric estimate of the limit that is precise to the number of digits available in your computer's numeric processor.

a) LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2KEYrLUYjNiYtRiw2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RMCZBcHBseUZ1bmN0aW9uO0YnL0Y6USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGRC8lKXN0cmV0Y2h5R0ZELyUqc3ltbWV0cmljR0ZELyUobGFyZ2VvcEdGRC8lLm1vdmFibGVsaW1pdHNHRkQvJSdhY2NlbnRHRkQvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZTLUkobWZlbmNlZEdGJDYkLUYjNiQtRiw2JVEidEYnRjZGOUZARkBGQC1GPTYtUSI9RidGQEZCRkVGR0ZJRktGTUZPL0ZSUSwwLjI3Nzc3NzhlbUYnL0ZVRlxvLUYjNidGKy1GIzYnLUkjbW5HRiQ2JFEiM0YnRkAtRj02LVExJkludmlzaWJsZVRpbWVzO0YnRkBGQkZFRkdGSUZLRk1GT0ZRRlQtRiM2JC1JJW1zdXBHRiQ2JUZlbi1GY282JFEiMkYnRkAvJTFzdXBlcnNjcmlwdHNoaWZ0R1EiMEYnRkBGK0ZALUY9Ni1RKCZtaW51cztGJ0ZARkJGRUZHRklGS0ZNRk8vRlJRLDAuMjIyMjIyMmVtRicvRlVGaHAtRmNvNiRRIjZGJ0ZARkBGK0ZARitGQA== for 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

Change the functions and ranges below to explore the various behaviors.

f := t -> sqrt(t): a := 1: b := 4: area_approx := leftsum( f(t), t=a..b, N ): area:=evalf(limit( area_approx, N = infinity ));

Now we use the Fundamental Theorem.

F :=unapply(int(f(t), t),t): deltaQ := evalf( F(b) - F(a) );

<Text-field style="Heading 1" layout="Heading 1">Part V: Calculus Gives Birth To New Functions</Text-field>

The Natural Log Function As we indicated before, many functions have antiderivatives that cannot be expressed with algebraic formulas. One example is the function 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. It is important to realize that while the antiderivatives of this function cannot be represented by an algebraic formula, an antiderivative does, nonetheless, exist. We can use the Fundamental Theorem, Part 1, to define its antiderivatives, thus giving birth to a new function that can only be defined using calculus. Specifically, an antiderivative of LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2J0YrLUYjNiYtRiw2JFEiZkYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RMCZBcHBseUZ1bmN0aW9uO0YnL0Y3USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGQS8lKXN0cmV0Y2h5R0ZBLyUqc3ltbWV0cmljR0ZBLyUobGFyZ2VvcEdGQS8lLm1vdmFibGVsaW1pdHNHRkEvJSdhY2NlbnRHRkEvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZQLUkobWZlbmNlZEdGJDYlLUYjNiQtRiw2JFEieEYnRjZGNi8lJ2l0YWxpY0dGQUY9RjYtRjo2LlEiPUYnRmVuRj1GP0ZCRkRGRkZIRkpGTC9GT1EsMC4yNzc3Nzc4ZW1GJy9GUkZbby1JJm1mcmFjR0YkNigtRiM2JC1JI21uR0YkNiVRIjFGJ0ZlbkY9RjZGVi8lLmxpbmV0aGlja25lc3NHRmVvLyUrZGVub21hbGlnbkdRJ2NlbnRlckYnLyUpbnVtYWxpZ25HRmpvLyUpYmV2ZWxsZWRHRkFGNkYrRjY= is defined as 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 for x > 0. Since this F(x) function is so important it is given a special name. It is called the natural log of x and is denoted by ln x. There is a very good reason for calling it a logarithm, that is, it exhibits all of the characteristics of a log function. All of the properties of logarithms hold for ln x. Now the question is how do we evaluate this function? While the Fundamental Theorem, Part 2, still holds true for 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 and its antiderivative ln x, that is, 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 it is not much help in evaluating 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 because there is no algebraic formula for ln x. Consequently, we are forced to resort to its definition to find values for ln x. Specifically, ln x is the limit of a Riemann sum. We know that 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 exists because 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 is continuous on any interval from 1 to x, provided x > 0. But how do we represent the values of ln x as decimals? Usually we settle for an approximation that is accurate to some specified precision. One approach is to approximate 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 with a Riemann sum using enough rectangles to give the desired precision. Let's do it. We define our own approximate ln x function in the following section and call it approxln(x). The variable n is the number of rectangles used in the Riemann sum that approximates ln x.

approxln := proc(b, n) local a, h, c, i; a := 1; h := (b - a)/n; c := i -> a + i*h; evalf( sum( (1/c(i))*h, i=0..n - 1 ) ); end:

As an exercise, vary the value of n in the next group of cells to determine how many rectangles are needed to approximate a accurate to three decimal places for 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. We start with n = 12 and calculate the error, that is, the difference between our approximate value and that given by ln(x) which is Maple's command for calculating ln x.

n := 12: approxln(10, n) - ln(10.);

approxln(1/10, n) - ln(0.1);

Let's compare our approximating function approxln( ) with Maple's ln(x) by graphing the two together. First use n = 12, and then replace it with the value that you found above that ensures three decimal places of precision for 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.

n:=12: plot( [ln(x), approxln(x,n)], x=0.1..10, labels=["x", "y"], color=[red,blue], thickness=[1,1]); print(`The Maple ln(x) is plotted in red and our Riemann sum approximation is plotted in blue`);

What do you observe?

<Text-field style="Heading 1" layout="Heading 1">You Try It: The Standard Normal Distribution and the Error Function</Text-field>

When you did the You Try It: Part II, you may have thought that the function 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 seemed a bit unusual, but the bell-shaped graph that resulted probably looked very familiar. This function is a classic example of one that does not have an algebraic form for its antiderivative, and it is called the standard-normal-distribution function, which is used in probability and statistics. Let's examine it in more detail.

f := t -> exp(-t^2/2)/sqrt(2*Pi);

plot( f(t), t=-5..5, labels=["x", "y"], thickness=2, color=green, labels=[x,y] );

Now we consider the definite integral, F(x) = 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 , which gives the area under the bell-shaped curve between -x and x. Because f(x) is an even function, we can rewrite the integral as 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. A function that is closely related to F(x) is the error function, which is denoted by erf x and is defined as follows: 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. We leave it as an exercise for you to show that 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. We will use F(x) in the remainder of this module. Since there is no algebraic formula for the antiderivative of f(x), we will approximate values of F(x) using a Riemann sum with a sufficient number of rectangles to ensure the desired precision.

Using the method that we used for ln x in Part V, write Maple commands to approximate values of F(x) to three decimal places of accuracy on the interval 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. First use a left-hand Riemann sum, and then use a midpoint Riemann sum. To test your commands, plot the graph of 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 and the graph of your own F(x) to see if the functions match. (In Maple, the error function is erf(x).) Comment on what you find. We illustrate the Riemann sum with the middlebox( ) command.

middlebox( f(t), t=0..-4, 15 );

Note that the signed areas are negative because we are summing from right to left, and therefore h < 0 and so is 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.

If you need help with this exercise, the following section, contains our solution. Try doing it on your own before you check out what we did.

<Text-field style="Heading 1" layout="Heading 1">Solution</Text-field>

approxf := proc(b, n) local a, h, c, i; a := 0; h := (b - a)/n; c := i -> a + (i + 1/2)*h; evalf( sqrt(2/Pi) * sum( exp(-c(i)^2/2)*h, i=0..n - 1 ) ); end:

approxf( -0.1, 4 );

evalf( erf(-0.1/sqrt(2)) );

plot( [approxf(x,n), erf(x/sqrt(2))-.03], x=-2..2, labels=["x", "y"], color=[red,blue], thickness=[2,2]);

NOTE: The "-.03" is just to allow the overlapping functions to show.

There is one final note. Though the use of a Riemann sum to estimate a definite integral is a very direct approach, it is not very efficient. More efficient methods of estimating definite integrals have been developed, and some of these are studied in the module on numerical integration that is entitled "Riemann, Trapezoids, and Simpson."