The Fundamental Theorem of Calculus

Establishing Connections Between Approaches

(Part 3)

Notes to Student:

Now that you have explored the Riemann Rectangle Approximation Method (module 1) and the Accumulation Function (module 2) approaches to answering our basic area question a foundational basis has been laid for the unifying Fundamental Theorem of Calculus . All three labs are to be used as a supplement by you to better internalize the "big picture" and should not replace the formal approach of the proofs and theory used to establish the processes in general which are being rigorously presented in your text and in class. Refer back to all three modules as you read the text and listen to your teacher to help solidify the notational and conceptual aspects of this very important section of calculus. As you work through this module you will need to "scroll" down the pages using the control bar on the far right of the screen.

Introduction:

In the first two modules you saw that the area question can be approached through the summing of rectangular areas based on various selection schemes or by defining an area accumulation function (the antiderivative). It is the goal of this third module to connect the two approaches in order to obtain a unified system.

Objectives Defined:

The goal of this third module is to connect the two approaches in determining the area beneath a curve and to establish the proper notation for generalized results. We shall achieve this goal by:

1. Revisiting the Riemann Rectangle Approximation Method.

2. Revisiting the Accumulation Function Method.

3. Examining their connections to each other.

The Rectangle Approximation Method Revisited:

We first revisit the Riemann Rectangle Approximation Method from module 1. It seemed reasonable from that investigation that the approximation to the area under the curve improved as we increased the number of rectangles regardless of the scheme used. This supposition can be written symbolically as:

The notation is the height of the ith rectangle and is the width of each rectangle. The product is simply the area of the ith rectangle and sigma notation is used to designate their sum for a given number, n, of them. This summation is a simplified version of a more generalized Riemann Sum named after Bernhard Riemann, but will serve us adequately for now. If we let the number of rectangles increase without bound we have the following limit:

This equation says that the exact area under a curve is the infinite sum of approximating rectangles provided that our function is non-negative over a specified interval. If our function takes on both positive and negative values, then we have a net change/signed area value. We utilize Leibniz's notation for this limit and write the following definition.

Provided this limit exists, we call this infinite sum the definite integral from a to b . In short, we have

The area under the curve on [a,b] = if f is continuous and non-negative on [a,b].

Key Items for You to Note about the Definite Integral.

1. The definite integral is a number which may be interpreted as the area beneath a curve.

2. The value of the definite integral always represents a net change.

3. "a" is the starting interval value and is called the "lower limit of integration".

4. "b" is the ending interval value and is called the "upper limit of integration".

5. "f(x)" is the curve we seek the net change on and is called the "integrand".

6. " " is a notational device to signify an infinite sum.

7. "dx" is a notational device indicating the independent variable.

Example:

Suppose we compute the area for f(x) = on [0,4] by using a right-hand Riemann approximation with 20 rectangles. Place the cursor in the red restart Maple command and press "Enter".

> restart:
with(student):
with(plots):
setoptions(labels=["",""],font=[SYMBOL,20]):
f:=x->x^2: /* function.
Digits:=6:
a:=0: /* left endpoint.
b:= 4: /* right endpoint.
n:=20: /* number of rectangles.
dx:=(b-a)/n:
`Approximate Area`:=Sum(f(1+k*dx)*dx,k=1..n)=value(rightsum(f(x),x=0..4.0,n))*`sq.units`;
display(seq(rightbox(f(x),x=a..b,NumRects),NumRects =5..20),insequence=true,title="Right-Hand Approximating\nRectangles",titlefont=[TIMES,BOLD,14]);

Warning, the name changecoords has been redefined

Click on the picture and use the animation toolbar which will appear at the top to view the animation.

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The true area can be found by computing the infinite limit we defined above which we wrote as:

Maple is powerful enough to handle such a limit! If the cursor is not currently in the red restart Maple command press "Enter" until it is and then press Enter again to execute the code.

> restart:with(plots):dx:=(4-0)/n:
an1:=plot(x^2,x=0..4,color=red,thickness=3):
an2:=plot(x^2,x=0..4,y=0..20,filled=true,color=aquamarine,view=[0..4,0..20]):Area:=eval(limit(Sum((0+k*dx)^2*dx,k=0..n-1),n=infinity)):
an3:=display(textplot([3.5,5,convert(Area,string)],color=blue)):
Limit(Sum((0+k*dx)^2*dx,k=1..n),n=infinity)=Int(x^2,x=0..4);
display(an1,an2,an3,textplot({[3,5.1,"Area ="]},color=blue,font=[TIMES,BOLD,14]));

Warning, the name changecoords has been redefined

> /* Press Enter to Continue

The exact area can be determined if the limit from above can be evaluated. This is not usually easy to do by hand but a powerful software package like Maple can give you excellent results. Furthermore, what if the function takes on both positive and negative values? When this occurs then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis and the sum of the areas of the rectangles that lie below the x-axis. Thus, the definite integral can be interpreted as a net area . When a definite integral is computed one must take care to know whether the function takes on positive and negative values so that if a true area value is desired, the appropriate signed area can be dealt with. Consider the following example:

Let

- 4x

Calculate the net area and the total area between f(x) and the x-axis on [0, 3]. Place the cursor in the red restart command and press "Enter".

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> restart:
with(plots):
f:=x->x^3-4*x:
Area1:=Int(f(x),x=0..2):
g:=t->t^3-4*t:
k:=int(f(x),x=0..2)+int(g(x),x=2..3):
Area2:=Int(g(x),x=2..3):
`Net Area`:=Area1+Area2=k*` sq.units`;
l:=abs(int(f(x),x=0..2))+int(g(x),x=2..3):
`Total Area`:=abs(Area1)+Area2=l*`sq.units`;
A1:=plot(f(x),x=0..2,filled=true,color=aquamarine,title="Area 1 & Area 2",titlefont=[TIMES,BOLD,12]):
A2:=plot(g(x),x=2..3,filled=true,color=yellow):
display(A1,A2,textplot({[1.1,-2,"Area 1"],[2.6,2,"Area 2"]},color=blue,font=[TIMES,BOLD,14]));

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Conclusion: The exact net area can be found by computing the definite integral which is the limit of an infinite sum of rectangular areas which we call the Riemann Sum. However, this value can only be determined if we can actually compute the infinite limit . Otherwise, we are stuck to choosing a finite number of rectangles and using them to obtain an approximation to the definite integral's value.This is in practice very cumbersome to do by hand for non-trivial problems.

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The Accumulation Function Revisited - FTC 1:

Let's look at f(x) = again. The definite integral computes the signed area under this function over an interval [a,b]. Place the cursor in the red restart Maple command to see the symbolic representation of this.

> restart:
Area:=Int(x^2,x=a..b);

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In Module 2 we were able to determine the area under this curve by defining a function which we called The Accumulation Function, A(x), which we also learned is called an antiderivative of f(x). It is customary to actually use F(x) to represent the antiderivative of f(x) .Thus, instead of using A(x) we will now use F(x) in its place. Since the Accumulation function determines the signed area up to a point, x, in an interval we can write this fact as:

The Fundamental Theorem of Calculus Part 1

If a function, f, is continuous on a closed interval [a,b] then we can define a function on [a,b] as

.

which is continuous on [a,b] and also differentiable on the open interval (a,b) with F'(x) = f(x).

F(x) yields the net accumulation of the integral from an initial starting point, a, up to a variable ending point x. It is very important to note that F(x) is a continuous function and not a number unless x is given a specific value. For example, if the interval is [0,2] and we specify x to be 1, then F(1) yields the net accumulation from 0 to 1. The two diagrams below show the distinction:

x unspecified x = 1

F(x) = F(1) = =

From Module 2 we observed that which we have now formally stated: F'(x) = f(x) . If we take the derivative of the integral for F(x) we see the following:

F(x) = F'(x) = = = f(x)

This says that if you take the derivative of the antiderivative, F(x), you get back the original function f(x) .

The implication is that differentiation and antidifferentiation are inverse processes . It also implies that every continuous function f(x) has an antiderivative. Finding the antiderivative as module 2 showed may not be an easy matter.

Key Items for You to Note about the Fundamental Theorem of Calculus Part 1 (FTC 1).

1. The FTC 1 defines the accumulator function as an integral.

2. The integral in the FTC 1 equation has its upper limit of integration as a variable thus making the integral's value dependent on x.

3. "a" is a scalar starting interval value.

4. "x" is a variable ending interval value.

5. "f(t)" is the curve we seek the net change on and is called the "integrand".

6. " " is a notational device to signify an infinite sum.

7. "dt" is a notational device indicating the independent variable.

8. The FTC 1 implies that the derivative of the antiderivative yields the integrand.

9. The FTC 1 yields the important relationship that differentiation and antidifferentiation are inverse processes.

Problems You Are Responsible For

Q1. Write the FTC 1 equation.

Q2. State the conditions necessary for the FTC 1 to hold.

Q3. Draw an example picture of the Accumulation Function up to a point x.

Q4. Decide whether the FTC 1 represents a value or a function.

Q5. Summarize in your own words what the FTC 1 equation says.

Q6. Explain in your own words with proper symbolism the phrase " differentiation is the inverse process of integration".

Q7. Evaluate: .

Q8. Find the values of A(0), A(1), A(3), A(5), and A(6) from the graph shown below if A(x) represents the accumulation function.

Q9. Create a scenario where L'Hospital's Rule may be applied to the FTC 1.

Q10. Suppose the upper limit of integration in the FTC 1 is a function like sinx. How does that affect the result of the derivative of the integral shown below?

The Establishment of the FTC 2:

In module 2 we explored how the net accumulation beneath a curve seemed to be related to the value of the Accumulation Function, A(x), evaluated at the interval endpoints. We explored this as:

Does A(b) - A(a) = The net area?

To see how this might be so, consider the interval [a,b]. Using the accumulation function A(x) and letting x = a and then b we get:

A(a) = and A(b) =

Subtracting yields: A(b) - A (a) = -

But, = 0 because there is no accumulation beneath an individual point.

Hence,

A(b) - A (a) = which is precisely the limit established by the Riemann Sum Method.

The Fundamental Theorem of Calculus Part 2

If f is continuous on [a,b] and F(x) is any antiderivative of f , then

The Net Accumulation = = F(b) - F(a), where F'(x) = f(x).

This means that if we cannot evaluate the infinite limit of the Riemann Sum to get the exact accumulation, we can still find it if we can find the antiderivative of the given curve. This is perhaps one of the greatest achievements in the field of mathematics! This provides us with another means to evaluate definite integrals without resorting to messy Riemann Rectangle Approximation Methods.

= F(b) - F(a)

The equation above has tremendeous implications. In words it says this:

The definite integral over a closed interval can be determined by simply evaluating the antiderivative at the endpoints. This is exactly what you intuitively explored in Module 2!

Of course, we still have the difficulty of actually finding the antiderivative but now we know that if we can find it, we can compute the definite integral which can in turn be interpreted as a net accumulation.

Key Items for You to Note about the Fundamental Theorem of Calculus Part 2 (FTC 2).

1. The FTC 2 returns the net accumulation over an interval [a,b].

2. The definite integral in the FTC 2 may be interpreted as area if the integrand function is non-negative over the interval.

3. The FTC 2 holds if the integrand function does not have any infinite discontinuities on the interval [a,b].

3. "a" is a scalar starting interval value.

4. "b" is a scalar ending interval value.

5. "f(x)" is the curve we seek the net change on and is called the "integrand".

6. " " is a notational device to signify an infinite sum.

7. "dx" is a notational device indicating the independent variable.

8. The FTC 2 states that the value of a definite integral is simply the difference in the antiderivative evaluated at the interval endpoints

9. The FTC 2 offers a tremendeous advantage over the cumbersome Riemann Sum Approximation Method provided that we can find the antiderivative.

Problems You Are Responsible For

Q1. Write the FTC 2 equation.

Q2. State the conditions necessary for the FTC 2 to hold.

Q3. Draw an example picture of the FTC 2 on an interval [a,b].

Q4. Decide whether the FTC 2 represents a value or a function.

Q5. Summarize in your own words what the FTC 2 equation says.

Q6. Give an example where the FTC 2 fails to be applicable.

Q7. Evaluate: .

Q8. Separate the figure into 4 distinct regions and create a definite integral for each which would represent the accumulation.

Q9. Create 3 different integrals which all have a net accumulation of 4.

Q10. Since the derivative of an object's position equation is its velocity, the integral of an object's velocity graph should be its net accumulated distance. If an object has a velocity equation given by v(t) = - ft/sec, what is the net displacement over an interval [0,5] seconds?

Using the Fundamental Theorem of Calculus: Applications:

There are a wealth of applications for integration (antidifferentiation).

Things to keep in mind:

1. The definite integral returns a value that represents the net accumulation.

2. If you evaluate the definite integral of any rate graph you will get the total change.

Examples of Integration using Maple:

Run the following Maple code examples by always starting with the cursor in the red restart command line.

Example 1

Evaluate the definite integral

 .

> restart:
with(plots):
k:=evalf(Int(x*sin(x),x=0..7)):
Int(x*sin(x),x=0..7)=k;
a1:=plot(x*sin(x),x=0..7,y=-6..6,color=red,thickness=3):
a2:=plot(x*sin(x),x=0..7,y=-6..6,color=green, filled = true,view=[0..7,-6..6], title="Graph of xsinx",titlefont=[TIMES,BOLD,14]):
display(a1,a2);

Warning, the name changecoords has been redefined

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To find the total area bounded by the curve above, we need to identify when the curve is above and below the horizontal axis on the interval [0,7] .

> restart:with(plots):
`Total Area`:= Int(x*sin(x),x=0..Pi)+abs(Int(x*sin(x),x=Pi..2*Pi))+Int(x*sin(x),x=2*Pi..7);`Total Area`:=evalf(%)*`sq.units`;
a3:=plot(x*sin(x),x=0..7,y=-6..6,color=red,thickness=3):
a4:=plot(x*sin(x),x=0..7,y=-6..6,color=green, filled = true,view=[0..7,-6..6], title="Graph of xsinx",titlefont=[TIMES,BOLD,14]):
display(a3,a4);

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Note: We need to consider the integral as signed area.

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Example 2:

As stated earlier, it is importatnt for you to understand that if you integrate a rate function you get a net accumulation quantity. For example, if we have a rate graph of water leaking out for 15 minutes, then the area beneath the rate graph yields the net amount of water that leaked out over the 15 minute interval. Follow the example below closely to make this clearer.

One of the earliest pollution problems brought to the attention of the Environmental Protection Agency (EPA) was the case of the Sioux Lake in South Dakota. For years a small paper plant located nearby had been discharging waste containing carbon tetrachloride into the waters of the lake. At the time the EPA learned of the situation, the chemical was entering the lake at a rate of 16 cubic yards/year . The agency immediately ordered the installation of filters designed to slow (and eventually stop) the flow of carbon tetrachoride from the mill. It took three years to actually implement the program of intervention and the filters during which time the rate of inflow was a steady 16 cubic yards per year. Once the filters were installed, the flow declined. From the time that the filters were installed until the time the flow stopped, the rate of flow was well modeled by the quadratic

Rate (in cubic yards/year) = - 14 t + 49

where t is time measured in years since the EPA learned of the situation.

A. Plot the curve which models the flow of carbon tetrachloride into the lake beginning at the time the EPA first learned of the situation. Press "Enter" to get the cursor in the red restart command and then press "Enter" agian to obtain the graph.

> restart:
with(plots):
f:=t->t^2-14*t+49:
amount:=t->piecewise(t<3,16,t>3,t^2-14*t+49):
a1:=plot(amount(t),t=0..10,title="Flow Rate Model in Cubic Yards per Year",labels=[years,rate],titlefont=[TIMES,BOLD,14],labelfont=[TIMES,BOLD,12],thickness=3,color=blue,view=[0..7,0..20]):
a2:=plot(amount(t),t=0..7,color=plum,filled=true):
display(a1,a2,textplot([2,7,"Total Pollutants"],font=[TIMES,BOLD,14],color=blue));

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How many years elapsed between the time the EPA learned of the situation and the time the pollution flow stopped entirely? Can you write two separate integrals to compute the total amount of pollutants which came into the lake? The code below displays the solution. Study it carefully. Press "Enter" to continue.

> k:=int(16,t=0..3)+int(f(t),t=3..7):
`Total Pollutants`:=Int(16,t=0..3)+Int(f(t),t=3..7)= k*`cubic yards`;

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Example 3:

This third example combines both parts of the FTC. As stated earlier, sometimes we do not have a nice formula for our function and, thus, the antiderivative cannot be found. But, if we know that we are dealing with a rate graph we can still approximate the definite integral by creating a grid beneath the graph and calculating the area by counting the squares. Work through the four parts of this example.

The graph below displays the velocity of a 6 hour jog by a student.

> restart:with(plots):
a1:=plot((2.1*sin(x)-2.2*cos(x/3+2))+1.1,x=0..6,thickness=3,view=[0..6,0..5],color=blue,tickmarks=[8,6],title="Jogger's Velocity Graph",titlefont=[TIMES,BOLD,14],labels=[hours,`mile/hr`],labelfont=[TIMES,BOLD,12]):
a2:=plot({1,2,3,4,5},x=0..6,view=[0..6,0..5.1],color=red):
a3:=implicitplot({x=1,x=2,x=3,x=4,x=5,x=5.99},x=0..6,y=0..5,color=red,numpoints=1):
display(a1,a2,a3,tickmarks=[6,6]);

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Answer the following questions:

Q1. What is the approximate number of miles jogged?

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Q2. When was the jogger accelerating?

Q3. Assuming the jogger's unknown velocity function is written as v(t), write a definite integral to represent the jogger's total distance traveled.

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Q4. Try to sketch the displacement graph for this jogger.

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Conclusions:

You have now completed all three Labs for Integration. From these labs you should have seen the following:

1.The area under a continuous curve can be determined by a limiting process of approximating rectangles (The Riemann Sum). There are many ways in which to define the rectangles but they converge to the same value as the number of rectangles increases without bound. The process is usually quite cumbersome to do by hand and is not a preferred method.

2. The area under a continuous curve can be determined by evaluating the antiderivative of a function at the endpoints over a given interval (Fundamental Theorem of Calculus). This is by far the preferred method but its weakness lies in the fact that one must be able to find the antiderivative function.

3. The Fundamental Theorem of Calculus is composed of two parts which when taken together display the inverse processes of differentiation and antidifferentiation (integration). We call the FTC 1 and FTC 2 but we refer to both them taken together as The Fundamental Theorem of Calculus.

4. Definite integrals return net accumulations, thus, if a function takes on both positive and negative values one needs to account for the sign in the negative cases.

5. If the curve being integrated is that of a rate function, then the value of the definite integral can be interpreted as a total amount of change (net quantity). This has many applications which will be investigated in later classes.