over the interval [2,4]. Place the cursor in the red
restart
Maple command and press "Enter".
The Accumulation Function
Fundamental Theorem of Calculus
An Exploratory Introduction-Part 2
Notes to Student:
Now that we have explored the Riemann Rectangle Approximation Method we are ready to investigate the area question further. Pay particular attention to the notation used in the lab and the stated objectives.
Introduction:
Once again, we will focus on seeking the answer to the area question posed below.
"Given a function f which is continuous and non-negative over a closed interval [a,b], how do we find the area enclosed beneath the curve and above the horizontal axis over the designated interval ?"
In this second module we will discover an alternative method to the Riemann Rectangle Approximation Method that we examined in our first module. Proceed to the next section.
Objectives Defined:
Time: 50 minutes
The purpose of this module is to explore and define a new method of determining the area beneath a curve and further extensions other than the Riemann Rectangle Approximation Method. This goal will be achieved by:
1. Defining a continuous function called the Accumulation Function, A(x).
2. Exploring animated graphs of a rectangle, triangle, trapezoid, and parabola with their respective Accumulation Functions.
3. Formulating generalizations drawn upon observations and computations.
Proceed to the next section.
Focus:
Suppose we wish to
find the area beneath
the curve
over the interval [2,4]. Place the cursor in the red
restart
Maple command and press "Enter".
>
restart:
with(plots):
f(x):=3*x^2:
a1:=plot(3*x^2,x=2..4,title="What is the Area\n Beneath the Curve?",filled=true,view=[0..4,0..60],font=[TIMES,BOLD,14],titlefont=[TIMES,BOLD,14],color=aquamarine):
a2:=plot(3*x^2,x=2..4,view=[0..4,0..60],thickness=3, color=red):
display(a1,a2);
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent2.gif)
Up to this point the only approach to this problem has been the Riemann Rectangle Approximation Method. This method as you recall was very labor intinsive and notationally cumbersome to do by hand. It is our goal to devise another approach to determining the area beneath a curve by constructing a new function which we will call " The Accumulation Function" . Proceed to the next section.
>
Defining the Accumulation Function:
We will define a function which "accumulates" area over an interval. We'll call this function the Accumulation Function and designate it as A(x) . Let's see how this function might behave for a rectangle, a triangle, a trapezoid, and a parabola. Work through each figure and the questions which accompany them. Examine the graphs carefully taking time to view and analyze the animations. Proceed to the next section.
The Rectangle:(Animated)
Consider a very simple rectangle lying in the first quadrant over the interval [0,2]. Place the cursor in the red restart Maple command and press "Enter".
>
restart:
with(plots):
a:=0:
b:=2:
c:=0:
d:=10:
n:=20:
f:=x->1:
plot(f,x=a..b,filled=true,title="Simple Rectangle\nArea = base x height",tickmarks=[2,2], titlefont=[TIMES,BOLD,14],color=aquamarine);
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent3.gif)
> /*Press Enter to Continue
We can see that the area is 2 . Our Accumulation Function , A(x) , needs to return a value of 2 when our inputs for x vary across the same interval [0,2]. The code below will graph the Accumulation Function, A(x), on the same graph as the rectangle. The graph animates the "sweeping out" of the area beneath the rectangle as x varies from 0 to 2 along with the plot of A(x). In other words, it shows the connection between the swept out area and the Accumulation Function, A(x) . Press "Enter" to continue.
>
a:=0:
b:=2:
c:=0:
d:=3:
n:=20:
Digits:=4:
f:x->1:g:=value(Int(f(t),t=a..x)):
an1:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):
for i from 1 to n do
k:=a+(b-a)*i/n:
h:=evalf(value(Int(f(x),x=a..k))):
an2[i]:=plot(f(x),x=a..k,filled=true,view =[a..b,c..d],color=AQUAMARINE):
an3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue,thickness=3):
an4[i]:=display(textplot([1,2,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):
end do:
p:=plots[display]([seq(an2[i],i=1..n)],insequence=true):q:=plots[display]([seq(an3[i],i=1..n)],insequence=true):
r:=plots[display]([seq(an4[i],i=1..n)],insequence=true):
display(r,an1,p,q,title="Rectangle and A(x)",titlefont=[TIMES,BOLD,14],tickmarks=[2,4],TEXT([0.6,2],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));
![[Maple Plot]](Accumulationimages/AccumulationStudent4.gif)
1. "Click" on the figure above and use the Animation Toolbar which will appear at the top of the page . Run the animation several times.When done, place the cursor back at the prompt > and press "Enter" to continue.
>
2. The blue function is the Accumulation Function, A(x), while the red function defines the rectangle and is the constant function f(x) = 1 over [0,2]. As x increases, the area increases, and hence, the height of A(x) increases.
Q1. As the rectangle's area filled, in what manner did the area function A(x) grow? Explain your answer (You may view the animation as often as you like by clicking on it. Place the cursor at the prompt > and press "Enter" when finished ).
>
Q2. From the graph determine A(x) for the inputs x = 0, 0.2, 0.4, 0.6, 0.8, 1, and 2. Create a t-table for your work.
x-input | A(x) value
|
|
|
...
>
Q3. If our interval is designated as [a,b], then what are the values of A(a) & A(b)?
>
Q4. Write out the equation A(b) - A(a) = . How does this compare to the area?
>
Q5. Using the graph and your t-table from Q2, write an equation for A(x).
>
Q6. Compute A'(x) and compare it to the rectangle's defining function, f(x) . What do you notice?
>
The Triangle:(Animated)
Our second familiar shape is the triangle. Execute the following Maple commands by placing the cursor in the red restart command and pressing "Enter". The code displays a triangle over the interval [0,2]
>
restart;
with(plots):
a:=0:
b:=2:
c:=0:
d:=10:
n:=20:
f:=x->2*x:
plot(f(x),x=a..b,filled=true,title="Simple Triangle\nArea = (base x height)/2",tickmarks=[2,2], titlefont=[TIMES,BOLD,14],color=aquamarine);;
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent5.gif)
We see that the area is 4 . Our Accumulation Function , A(x) , needs to return a value of 4 when our inputs for x vary across the same interval [0,2]. The graph below animates the "sweeping out" of the area beneath the triangle as x varies from 0 to 2 along with the plot of A(x). In other words, it shows the connection between the swept out area and the Accumulation Function, A(x) . Press "Enter" to continue.
>
a:=0:
b:=2:
c:=0:
d:=5:
n:=20:
Digits:=4:
f:x->2*x:
g:=value(Int(f(t),t=a..x)):
an1:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):
for i from 1 to n do
k:=a+(b-a)*i/n:
h:=evalf(value(Int(f(x),x=a..k))):
an2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],color=AQUAMARINE):
an3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue, thickness=3):
an4[i]:=display(textplot([1.1,3.6,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):
end do:
p:=plots[display]([seq(an2[i],i=1..n)],insequence=true):q:=plots[display]([seq(an3[i],i=1..n)],insequence=true):
r:=plots[display]([seq(an4[i],i=1..n)],insequence=true):
display(r,an1,p,q,title="Triangle and A(x)",titlefont=[TIMES,BOLD,14],tickmarks=[2,2],TEXT([0.7,3.6],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));
![[Maple Plot]](Accumulationimages/AccumulationStudent6.gif)
> /*Press Enter to Continue
1. "Click" on the figure above and use the Animation Toolbar which will appear at the top of the page . Run the animation several times.When done, place the cursor back at the prompt > and press "Enter" to continue.
>
2. The blue function is the Accumulation Function, A(x), while the red function defines the triangle and is the function f(x) = 2x over [0,2]. As x increases, the area increases, and hence, the height of A(x) increases.
Q1. As the triangle's area filled, in what manner did the area function A(x) grow? Explain your answer (You may view the animation as often as you like by clicking on it. Place the cursor at the prompt > and press "Enter" when finished ).
>
Q2. From the graph determine A(x) for the inputs x = 0, 0.5, 1, 1.5, and 2. Create a t-table for your work.
x-input | A(x) value
|
|
|
...
>
Q3. If our interval is designated as [a,b], then what are the values of A(a) & A(b)?
>
Q4. Write out the equation A(b) - A(a) = . How does this compare to the area?
>
Q5. Using the graph and your t-table from Q2, write an equation for A(x).
>
Q6. Compute A'(x) and compare it to the triangle's defining function, f(x) . What do you notice?
>
The Trapezoid:(Animated)
We now investigate another familiar shape: the trapezoid. Execute the following Maple commands by placing the cursor in the red restart command and pressing "Enter". The display is a simple trapezoid over the interval [1,3].
>
restart;
with(plots):
a:=1:
b:=3:
c:=0:
d:=20:
n:=20:
f:=x->2*x+2:
plot(f(x),x=a..b,filled=true,title="Simple Trapezoid\nArea = [(base1+base2) x height]/2",tickmarks=[3,8], titlefont=[TIMES,BOLD,14],view=[0..3,0..8],color=aquamarine);
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent7.gif)
Confirm that the area is 12 . Our Accumulation Function , A(x) , needs to return a value of 12 when our inputs for x vary across the same interval [1,3]. The code below will graph the Accumulation Function, A(x), on the same graph as the trapezoid. The graph animates the "sweeping out" of the area beneath the trapezoid as x varies from 1 to 3 along with the plot of A(x). In other words, it shows the connection between the swept out area and the Accumulation Function, A(x) . Press "Enter" to continue.
>
a:=1:
b:=3:
c:=0:
d:=16:
n:=20:
Digits:=4:
f:x->2*x+2:
g:=value(Int(f(t),t=a..x)):
an1:=plot(f(x),x=a..b,view=[(a-1)..b,c..d],color=red,thickness=3):
for i from 1 to n do
k:=a+(b-a)*i/n:
h:=evalf(value(Int(f(x),x=a..k))):
an2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],color=AQUAMARINE):
an3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue,thickness=3):
an4[i]:=display(textplot([1.8,11,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):
end do:
p:=plots[display]([seq(an2[i],i=1..n)],insequence=true):
q:=plots[display]([seq(an3[i],i=1..n)],insequence=true):
r:=plots[display]([seq(an4[i],i=1..n)],insequence=true):
display(r,an1,p,q,title="Trapezoid and A(x)",titlefont=[TIMES,BOLD,14],tickmarks=[3,8],TEXT([1.3,11],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));
![[Maple Plot]](Accumulationimages/AccumulationStudent8.gif)
> /*Press Enter to Continue
1. "Click" on the figure above and use the Animation Toolbar which will appear at the top of the page . Run the animation several times.When done, place the cursor back at the prompt > and press "Enter" to continue.
>
2. The blue function is the Accumulation Function, A(x), while the red function defines the trapezoid and is the function f(x) = 2x + 2 over [1,3]. As x increases, the area increases, and hence, the height of A(x) increases.
Q1. As the trapezoid's area filled, in what manner did the area function A(x) grow? Explain your answer. (You may view the animation as often as you like by clicking on it. Place the cursor at the prompt > and press "Enter" when finished ).
>
Q2. From the graph determine A(x) for the inputs x = 1, 1.5, 2, 2.5, and 3. Create a t-table for your work.
x-input | A(x) value
|
|
|
...
>
Q3. If our interval is designated as [a,b], then what are the values of A(a) & A(b)?
>
Q4. Write out the equation A(b) - A(a) = ...How does this compare to the area?
>
Q5. Using the graph and your t-table from Q2, write an equation for A(x).
>
Q6. Compute A'(x) and compare it to the trapezoid's defining function, f(x) . What do you notice?
>
A More Difficult Figure:(Animated)
Now we turn our attention to a more challenging area. Consider the function
f(x) =
over the interval [0,4]. Execute the following Maple commands by placing the cursor in the red
restart
command and pressing "Enter".
>
restart;
with(plots):
a:=0:
b:=4:
c:=0:
d:=20:
n:=20:
f:=x->x^2:
plot(f(x),x=a..b,filled=true,title="Simple Parabola\nArea = ?",tickmarks=[4,8], titlefont=[TIMES,BOLD,14],view=[0..4,0..16],color=aquamarine);
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent10.gif)
> /*Press Enter to Continue
We do not have an elementray formula to compute the area of the shaded figure! Our Accumulation Function , A(x) , needs to return the correct area value when our inputs for x vary across the same interval [0,4]. But how can we come up with the correct function? The code below will graph the Accumulation Function, A(x), on the same graph as the parabola and animate the "sweeping out" of the area beneath the parabola as x varies from 0 to 4. It once again shows the connection between the swept out area and the Accumulation Function, A(x) and should help us find the defining A(x). Press "Enter" to continue.
>
a:=0:
b:=4:
c:=0:
d:=25:
n:=20:
Digits:=4:
f:x->x^2:
g:=value(Int(f(t),t=a..x)):
an1:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):
for i from 1 to n do k:=a+(b-a)*i/n:
an2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],color=AQUAMARINE):
h:=evalf(value(Int(f(x),x=a..k))):
an4[i]:=display(textplot([1.9,15,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):
an3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue,thickness=3):
end do:
p:=plots[display]([seq(an2[i],i=1..n)],insequence=true):q:=plots[display]([seq(an3[i],i=1..n)],insequence=true):
r:=plots[display]([seq(an4[i],i=1..n)],insequence=true):
display(r,an1,p,q,title="Parabola and A(x)",titlefont=[TIMES,BOLD,14],TEXT([1.2,15],'`A(x) = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));
![[Maple Plot]](Accumulationimages/AccumulationStudent11.gif)
1. "Click" on the figure above and use the Animation Toolbar which will appear at the top of the page . Run the animation several times.When done, place the cursor back at the prompt > and press "Enter" to continue.
>
2. The
blue
function is the
Accumulation Function, A(x),
while the
red
function
defines the parabola
and is the function
f(x) =
over [0,4]. As x increases, the area increases, and hence, the height of A(x) increases.
Q1. As the parabola fills, in what manner did the area function A(x) grow? Explain your answer. (You may view the animation as often as you like by clicking on it. Place the cursor at the prompt > and press "Enter" when finished ).
>
Q2. From the graph approximate A(x) for the inputs x = 1, 2, 3, and 4? Create a t-table for your work
x-input | A(x) value
|
|
|
...
>
Q3.Using your data in Q2 and graph, find a suitable A(x).
>
Q4. How does A'(x) compare to the parabola's defining function, f(x) ? Is this consistent with the results of the rectangle, triangle, and trapezoid?
>
Q5. Determining a function whose derivative is equal to a given function is called antidifferentiation . With that in mind, does the Accumulation Function, A(x), which you wrote in Q3 have its derivative equal to the function, f(x)? If it does, then we say that A(x) is an antiderivative of f(x) .
>
Q6. If our interval is designated as [a,b], then what are the values of A(a) & A(b)?
Q7. What is the area beneath the parabola? Explain your answer.
>
Generalizing Our Results:
After the previous examples you should be able to complete the following statements.
1. A(x) is called the _____________________.
2. A(x) defines a continuous __________________.
3. A'(x) = __________________.
4. The process of determining a function whose derivative equals a given function is called _____________________.
5. A(b) - A(a) = _________________________.
Finding the Antiderivative:(Animated)
The Accumulation Functions (antiderivatives) we previously found were fairly easy to see from the graphs. In general, the antiderivative is easily determined for many elementary functions but there are examples where finding the antiderivative is either very difficult or impossible . Furthermore, what if the given function, f(x), dips below the axis somewhere within the given interval? How does that affect the Accumulation Function? This next example is one such instance where the function, f(x), takes on both positive and negative values. Consider the function f(x) = xsin(x) . Execute the Maple commands below by placing the cursor in the red restart command and pressing "Enter".
>
restart;
with(plots):
a:=-1:
b:=2*Pi:
c:=-5:
d:=5:
n:=20:
f:=x->x*sin(x):
an1:=plot(f(x),x=a..b,view=[a..b,c..d],color=aquamarine,filled=true,title="The Graph of\n f(x) = xsin(x)",titlefont=[TIMES,BOLD,14]):
an2:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3):
display(an1,an2);
Warning, the name changecoords has been redefined
![[Maple Plot]](Accumulationimages/AccumulationStudent13.gif)
Obviously this function's antiderivative is much more challenging! Not only is the antiderivative going to be difficult to find but how do we interpret A(x)'s values now? Is it still area if it goes below the horizontal axis? Before executing the next set of commands conjecture what you think the Accumulation Function's graph might look like.
>
a:=-1:
b:=2*Pi:
c:=-8:
Digits:=4:
d:=5:
n:=20:
f:x->x*sin(x):
g:=value(Int(f(t),t=a..x)):
an1:=plot(f(x),x=a..b,view=[a..b,c..d],color=red,thickness=3,axesfont=[TIMES,BOLD,14]):
for i from 1 to n do
k:=a+(b-a)*i/n:
an2[i]:=plot(f(x),x=a..k,filled=true,view=[a..b,c..d],color=GREEN):
an3[i]:=plot(g(x),x=a..k,view=[a..b,c..d],color=blue,thickness=3):
h:=evalf(value(Int(f(x),x=a..k)));
an4[i]:=display(textplot([3.9,4.2,convert(h,string)],font=[TIMES,BOLD,14],color=blue)):
end do:
p:=plots[display]([seq(an2[i],i=1..n)],insequence=true):
q:=plots[display]([seq(an3[i],i=1..n)],insequence=true):
r:=plots[display]([seq(an4[i],i=1..n)],insequence=true):
display(r,an1,p,q,title="f(x)=xsin(x)",titlefont=[TIMES,BOLD,16],TEXT([2,4.2],'`Accumulator = `',FONT(TIMES,BOLD,14),COLOR(RGB,0,0,1)));
![[Maple Plot]](Accumulationimages/AccumulationStudent14.gif)
1. "Click" on the figure above and use the Animation Toolbar which will appear at the top of the page . Run the animation several times.When done, place the cursor back at the prompt > and press "Enter" to continue
>
Q1. Are the results consistent with what you predicted? Explain why A(x) looks as it does.
>
Q2. Can we still interpret the result for A(x) as an "area accumulator" or is there a better way to define it. Please discuss your thoughts.
The Accumulator Function (antiderivative) is given by executing the code below. Press "Enter" to see the antiderivative.
> A(x):=int(f(x),x);
Q3. Now compute A'(x). Does it equal f(x) as it did in the other examples?
>
Using the A(x) found in Q2, Compute A(b)-A(a) on the interval [-1, 2pi]. How does it compare with the accumulatuion value seen on the graph?
Even though we cannot interpret the ending value as area, we can refer to it as a net change or as signed area . A(x) takes on both positive and negative values for differing values of x.
> /*Open the next section with the cursor.
Summarizing our Results:
After completing this module it is hoped that you understand the following items.
1. We can define a continuous function, A(x), which accumulates signed area over a given interval.
2. The net accumulation of A(x) over a closed interval [a,b] is simply determined by A(b) - A(a).
3. If we can find the Accumulation Function, A(x), for a given function, f(x), and f(x) is non-negative over a closed interval [a,b], then A(b) - A(a) represents the area beneath f(x) and the horizontal axis over the prescribed interval. This is far easier to perform than the Riemann Rectangle Approximation Method.
4. There are instances when finding the Accumulation Function is either very difficult or impossible. This is the major drawback to this method.
5. The Accumulation Function, A(x), is often called an antiderivative of f(x).
6. The Accumulation Function, A(x), has the unique property that its derivative A'(x) = f(x).