Discovery project

A DISCOVERY PROJECT AREA FUNCTIONS In this section, we will discuss four main problems; the fourth problem will ask us to make a conclusion based on the results of the first three problems.

Study each problem, its solution and explanations carefully. The results will be the subject of discussion in the next section. Once again, that this section is aimed at providing a “sneak-peek� into the idea of the fundamental theorem.

Here we go:

PROBLEM 1 a)

Draw the line y = 2t + 1 and use geometry to find the area under this line, above the t-axis, and between the vertical lines t = 1 and t = 3.

b)

If x > 1, let A(x) be the area under the region that lies under the line y = 2t + 1 between t = 1 and t = x. Sketch this region and use geometry to find an expression for A(x).

c)

Differentiate the area function A(x). What do you notice?

Solution a).

To start with, we graph the function f(t) = 2t + 1 on the interval [1, 3]:

The area Z under the line above the t–axis and between the vertical lines t = 1 and t = 3 takes the shape of a trapezium.

So, using geometry, the required area is found by computing the area of the trapezium, that is, Area of Z = Area of trapezium.

Which means

Area of Z

=

½[F(ta) + F(tb)]I

=

½ × 2[3 + 7]

=

½ [10]

=

10

Thus, the area under the line y = 2t + 1 from t = 1 to t = 3 is 10.

b).

In this second part, the problem here is to find a general expression for the area A(x), given that x > 1 and that the area under y = 2t + 1 from t = 1 to t = x. In part (a) above, we found the area under the line y = 2t + 1 on the interval [1, 3] to be the area of a trapezium.

If we use the trapezium formula,

½ [F(ta) + F(tb)]I Then,

F(ta) F(tb) I

and

= 3 = 2t + 1 =

t–1

So, the formula becomes

A(x)

=

½ [F(ta) + F(tb)] I

A(x)

=

½ [3 + (2t + 1)][t – 1]

A(x)

=

½ [3 + 2t + 1][t – 1]

A(x)

=

½ [4 + 2t][t – 1]

A(x)

=

½ [4 + 2t][t – 1]

A(x)

=

½ [4 + 2t][t – 1]

A(x)

=

½ [2t2 + 2t - 4]

A(x)

=

½ [2t2] + ½[2t] - ½[4]

A(x)

=

t2 + t – 2

Hence, an expression that represents the area A(x) under the line y = 2t + 1 from t = 1 to t = 3 is

A(x)

=

Such region would look like this:

t2 + t – 2

c).

Now we know that the area function representing the area under y = 2t + 1 on the interval [1, x] is

A(x)

=

t2 + t – 2

So, if we differentiate A(x) , we have

dx/dt = d/dt (t2 + t - 2) dx/dt = [d/dt (t2)] + [d/dt (t)] – [d/dt (2)] dx/dt = 2t + 1 – 0 dx/dt = 2t + 1 Notice that when we differentiated the area function

A(x)

=

t2 + t – 2

We actually got back the original equation of the line:

y

=

2t + 1

At this point, it seems that one process has undone the other and we are back to where we started. The significance of this result will be discussed in the following section.

PROBLEM 2 a)

If x ≥ -1, let

A(x)

=

∫

x

(1 + t 2) dt

-1

A(x) represents the area of a region. Sketch that region. b)

Use the fact that

∫

b a

(x) dx

=

(b 2 – a2)/2

to find an expression for A(x). c)

Find A’(x).What do you notice?

d)

If x

≥ -1, and h is a small positive number, then

A(x + h ) – A(x) represents the area of a region. Describe

and sketch that region. e)

Draw a rectangle that approximates the region in part (d). By comparing the areas of these two regions, show that

A(x + h ) – A(x) h f)

≈

1 + x2

Use part (e) to give an intuitive explanation for the result in part (c).

Solution

a)

The graph above is derived from the fact that, given the equation

A(x)

∫

=

and x ≥ -1, so,

x

(1 + t 2) dt

-1

f(t) = 1 + t2,, and the required interval is [-1, x]. Hence, the graph above represents the

area function A(x):

b)

From this area function:

A(x)

=

∫

x

(1 + t 2) dt

-1

We want to find an expression for A(x). In other words, we are simply asked to evaluate the integral on the given interval. Using property 4 of the definite integral, we have

A(x)

=

∫

x -1

+

(1) dt

∫

x -1

(t 2) dt

So, we have

A(x)

= 1[x – (-1)]

A(x)

= 1[x + 1] + [(x3 + 1)/3]

A(x)

=

A(x)

= 1 + (1/3) + x + (x3/3)

A(x)

=

+ [(x3 + 1)/3]

x + 1 + (x3/3) + (1/3)

4/3 + x + (x3/3)

Hence, the area of the region can be expressed as

A(x)

= 4/3 + x + (x3/3)

c) In (b) above, we found that the area under the curve y = 1 + t2 from t = -1 to t = x is given by the equation

A(x)

=

∫

x -1

(1 + t 2) dt

Which is then evaluated to produce

A(x)

=

4/3 + x + (x3/3)

So now, we want to find A’(x):

A’(x)

=

[d/dx(4/3)]

A’(x)

=

0 + 1 + x2

A’(x)

=

1 + x2

+

[d/dx(x)] +

[d/dx(x3/3)]

Once again, we obtain a result similar to that in problem 1(c); we obtain an expression representing an area, differentiate it, and in the end, we get back the original function. But in this case, there is slight difference. Recall that the integrand here is

y = 1 + t2 The value of its integral (with respect to x, on the interval [-1, x]), is

A(x)

=

4/3 + x + (x3/3)

which, when differentiated, gives

A’(x) =

1 + x2

You’ll find that that A’(x) is somewhat similar to the integrand. There is a connection, which will be explained in the next section.

d) Here, we are given that x ≥ -1, and h is a small positive number. We are given that A(x + h ) – A(x) represents the area of a region. So, we have this: a graph of the function y = 1+t 2, showing the area on the interval [-1, (x+h)]; where x is a number such that x > -1, and h is a small positive number greater than x).

Let’s consider the area difference:

A(x + h )

–

A(x)

At this point, it is clear that we are dealing with integrals over adjacent intervals.

From the graph above, the area A(x + h) is represented by the area on the interval [-1, (x+h)], that is, the blue region plus the orange region (the whole area). On the other hand, the area A(x) is represented by the area on the interval [-1, x], that is, the blue region only. Let’s assume that f(t) = 1 + t2 Using integral notation, we have

A(x)

=

∫

x -1

f(t) dt

and

A(x+h)

∫

=

x+h -1

f(t) dt

So, using the subtraction rule (property 6) of integrals, we have

A(x+h) – A(x)

=

∫

A(x+h) – A(x)

=

∫

x+h -1

f(t) dt

–

∫

x -1

f(t) dt

This is equal to x+h x

f(t) dt

From the graph, you can see that the integral

∫

x+h x

f(t) dt

is represented by the orange region. Therefore, the region representing A(x + h) – A(x) is the orange region (which we will call R).

e)

In (d) above, we found that the region representing A(x + h) – A( x) is the orange region. We can also use a rectangle to approximate this region, which would look like this:

From the graph, we can say that the orange rectangle provides an approximation of R. We also find that the area of R

is approximately equal to the area of the rectangle with height f(x) and width h, that is,

Area of rectangle

=

hf(x)

(Where f(x) = 1 + x2). Thus, we say that, for h > 0,

A(x+h) – A(x)

=

hf(x)

Dividing both sides by h gives

A(x + h ) – A(x) h

≈

hf(x) h

≈

f(x)

≈

f(x)

resulting in

A(x + h ) – A(x) h Thus,

A(x + h ) – A(x) h

f) From parts (c) and (e) above, you’ll find that we’ve arrived at the same result. In the first instance, we obtained the result by differentiating the area function and we found that the derivative of the integral A(x) with respect to the upper limit x is actually equal to the integrand (y = 1 + t2) evaluated at x (which is equal to y = 1 + x2) On the other hand, we obtained the result in (e) by finding the difference of areas, and we found that, for a small value of h, the difference of areas is approximately equal to y = 1 + x2.

Eventually, you’ll find that, in (c) and (e) above, we have simply found the derivative of the integral in two different ways: in the first case, we differentiate the area function (which is an algebraic representation of the given integral), and in the second case, we applied the definition of a derivative. We’ll discuss these observations in detail in the next section.

PROBLEM 3 a)

Draw the graph of the function f(x) = cos (x2) in the viewing rectangle [0, 2] by [-1.25, 1.25].

b)

If we define a new function g by

g(x)

=

∫

x 0

(cos t 2) dt

then g(x) is the area under the graph of f from 0 to x [until f(x) becomes negative, at which point g(x) becomes a difference of areas]. Use part (a) to determine the value of x at which g(x) starts to decrease.

[Unlike the integral in problem 2, it is impossible to evaluate the integral defining g to obtain an explicit expression for g(x)]. c)

Use the integration command on your calculator or computer to estimate g(0.2), g(0.4), g(0.6), g(0.8), ................ g(2). Then use these values to sketch a graph of g.

d)

Use the graph from part (c) to sketch the graph of g’ using the interpretation of g’(x) as the slope of a tangent line. How does the graph of g’(x) compare with the graph of f?

Solution a) The graph of f(x) = cos (x2) is shown below:

[Note the given graphing rectangle:[0, 2] by [-1.25, 1.25]]. Below is a table of values generated from g:

x

g(x)

x

g(x)

0.1

0.9999900000

1.1

0.9495038699

0.2

0.1999680024

1.2

0.9739445541

0.3

0.2997570911

1.3

0.9746890778

0.4

0.3989772129

1.4

0.949778977

0.5

0.4968840292

1.5

0.8991848529

0.6

0.5922705167

1.6

0.8255167459

0.7

0.6833787905

1.7

0.7345512315

0.8

0.7678475376

1.8

0.6353654311

0.9

0.8427176986

1.9

0.5398550484

1.0

0.9045242379

2.0

0.4614614624

Table 1

b) In the table above, we approximate the values of g(x) to 10 decimal places for accuracy. If you look at the table carefully, you’ll find that the value of g(x) begins to decrease when x = 1.4.

c) Below is a table of values of g(x) {for 0.2 ≤ x ≤ 2.0}:

x

g(x)

0.2

0.1999680024

0.4

0.3989772129

0.6

0.5922705167

0.8

0.7678475376

1.0

0.9045242379

1.2

0.9739445541

1.4

0.949778977

1.6

0.8255167459

1.8

0.6353654311

2.0

0.4614614624

Table 2 Using the values in table 2 above, we have a graph of g:

d) In this part, we have a small problem: we cannot derive an explicit expression for the area function g. So, we’ll have to use the table of values to approximate g’ and use the estimated values of g’ to draw its graph. Recall that the derivative f’ is defined as the slope of a tangent line, that is,

f(x + h ) – f(x) h

≈

f '(x)

≈

g '(x)

In this case, we replace f with g:

g(x + h ) – g(x) h Let’s start by computing g ’(0.2).

If we let h = 0.2, we have

g’(0.2) = [g(0.2 + 0.2) – g(0.2)]/0.2 g’(0.2) = [g(0.4) – g(0.2)]/0.2 g’(0.2) = 0.9950460525 Next, we’ll compute g ’(0.4).

If we let h = 0.2, we have

g’(0.4) = [g(0.4 + 0.2) – g(0.4)]/0.2 g’(0.4) = [g(0.6) – g(0.4)]/0.2 g’(0.4) = 0.9664665190 If we let h = – 0.2, we have

g’(0.4) =

[g(0.4 – 0.2) – g(0.4)]/ –0.2

g’(0.4) =

[g(0.2) – g(0.4)]/ –0.2

g’(0.4) =

0.9950460525

We take the average of both values to obtain a more accurate estimate of g’(0.4):

g’(0.4) =

½[0.9950460525 + 0.9664665190]

g’(0.4) =

0.9807562858

Next, we’ll compute g ’(0.6).

If we let h = 0.2, we have

g’(0.6) =

[g(0.6 + 0.2) – g(0.6)]/0.2

g’(0.6) =

[g(0.8) – g(0.6)]/0.2

g’(0.6) =

0.8778851045

If we let h = – 0.2, we have

g’(0.6) =

[g(0.6 – 0.2) – g(0.6)]/ –0.2

g’(0.6) =

[g(0.4) – g(0.6)]/ –0.2

g’(0.6) =

0.9664655190

We take the average of both values to obtain a more accurate estimate of g’(0.6):

g’(0.6) =

½[0.8778851045 + 0.9664665190]

g’(0.6) =

0.9221753118

Next, we’ll compute g ’(0.8).

If we let h = 0.2, we have

g’(0.8) =

[g(0.8 + 0.2) – g(0.8)]/0.2

g’(0.8) =

[g(1.0) – g(0.8)]/ 0.2

g’(0.8) =

0.6833835015

If we let h = – 0.2, we have

g’(0.8) =

[g(0.8 – 0.2) – g(0.8)]/ –0.2

g’(0.8) =

[g(0.6) – g(0.8)]/ –0.2

g’(0.8) =

0.8778851045

We take the average of both values to obtain a more accurate estimate of g’(0.8):

g’(0.8) =

½ [0.8778851045 + 0.6833835015]

g’(0.8) =

0.780634303

Next, we’ll compute g ’(1.0).

If we let h = 0.2, we have

g’(1.0) =

[g(1.0 + 0.2) – g(1.0)]/0.2

g’(1.0) =

[g(1.2 ) – g(1.0)]/ 0.2

g’(1.0) =

0.347101581

If we let h = – 0.2, we have

g’(1.0) =

[g(1.0– 0.2) – g(1.0)]/ –0.2

g’(1.0) =

[g (0.8) – g(1.0)]/ –0.2

g’(1.0) =

0.6833835015

We take the average of both values to obtain a more accurate estimate of g’(1.0):

g’(1.0) =

½ [0.347101581 + 0.6833835015]

g’(1.0) =

0.5152425413

Next, we’ll compute g ’(1.2).

If we let h = 0.2, we have

g’(1.2) =

[g(1.2 + 0.2) – g(1.2)]/0.2

g’(1.2) =

[g(1.4 ) – g(1.2)]/ 0.2

g’(1.2) =

–0.1208278855

If we let h = – 0.2, we have

g’(1.2) =

[g(1.2– 0.2) – g(1.2)]/ –0.2

g’(1.2) =

[g(1.0) – g(1.2)]/ –0.2

g’(1.2) =

0.347101581

We take the average of both values to obtain a more accurate estimate of g’(1.2):

g’(1.2) =

½ [–0.1208278855

g’(1.2) =

0.1131368478

+

0.347101581]

Next, we’ll compute g ’(1.4).

If we let h = 0.2, we have

g’(1.4) =

[g(1.4 + 0.2) – g(1.4)]/0.2

g’(1.4) =

[g(1.6) – g(1.4)]/ 0.2

g’(1.4) =

–0.6213111555

If we let h = – 0.2, we have

g’(1.4) =

[g(1.4– 0.2) – g(1.4)]/ –0.2

g’(1.4) =

[g(1.2) – g(1.4)]/ –0.2

g’(1.4) =

–0.1208278855

We take the average of both values to obtain a more accurate estimate of g’(1.4):

g’(1.4)

=

½ [–0.1208278855 – 0.6213111555]

g’(1.4)

=

– 0.3710695205

Next, we’ll compute g ’(1.6).

If we let h = 0.2, we have

g’(1.6) =

[g(1.6 + 0.2) – g(1.6)]/0.2

g’(1.6) =

[g(1.8 ) – g(1.6)]/ 0.2

g’(1.6) =

–0.9507565740

If we let h = – 0.2, we have

g’(1.6) =

[g(1.6 – 0.2) – g(1.6)]/ –0.2

g’(1.6) =

[g(1.4) – g(1.6)]/ –0.2

g’(1.6) =

–0.6213111555

We take the average of both values to obtain a more accurate estimate of g’(1.6):

g’(1.6) =

½ [–0.9507565740 – 0.6213111555]

g’(1.6) =

– 0.7860338648

Next, we’ll compute g ’(1.8).

If we let h = 0.2, we have

g’(1.8) =

[g(1.8 + 0.2) – g(1.8)]/0.2

g’(1.8) =

[g(2.0) – g(1.8)]/ 0.2

g’(1.8) =

–0.8695198435

If we let h = – 0.2, we have

g’(1.8) =

[g(1.8 – 0.2) – g(1.8)]/ –0.2

g’(1.8) =

[g(1.6) – g(1.8)]/ –0.2

g’(1.8) =

–0.9507565740

We take the average of both values to obtain a more accurate estimate of g’(1.8):

g’(1.8) =

½ [–0.9507565740 – 0.8695198435]

g’(1.8)

–0.9101382088

=

Finally, we’ll compute g’(2.0).

If we let h = – 0.2, we have

g’(2.0) =

[g(2.0– 0.2) – g(2.0)]/ –0.2

g’(2.0) =

[g(1.8) – g(2.0)]/ –0.2

g’(2.0) =

– 0.8695198435

Note that we cannot have h = 0.2 because g’(2.0) is the endpoint in this context. Likewise, we cannot also have h = 0.2 because g’(2.0) is the other endpoint . So now, let’s compile the derivatives we’ve obtained:

x 0.2

g'(x) 0.9950460525

0.4

0.9807562858

0.6

0.9221753118

0.8

0.7806343030

1.0

0.5152425413

1.2

0.1131368478

1.4

– 0.3710695205

1.6

– 0.7860338648

1.8

– 0.9101382088

2.0

– 0.8695198435

Table 3

Now, using table 3 which is a table of derivatives, we can then graph g’ on the interval 0.2 ≤ x ≤ 2:

Now, carefully compare the graph above with that of the original function y = f (x) :

So, what do you notice?

BOTH GRAPHS ARE THE SAME!!!!!!

Here’s a summary of what we’ve done to solve problem 3: We graphed the function f(x) = cos (x2) and used it to determine the value of x at which the integral function g decreases.

We evaluated g at the given values of x and used the results sketch its graph.

We used the graph of g to derive g’ using a table of values. Note that in problems 1 and 2, we were able to express the integrals as explicit expressions. This is impossible for problem 3. Therefore, the only way to derive g’ was to use a table of values defining the integral g.

After obtaining the graph of g’ and comparing with the graph of f, we found that both graphs were the same.

We’ll look into these results a bit further in the following section.

PROBLEM 4 Suppose f is a continuous function on the interval [a, b] and we define a new function g by the equation

g(x)

∫

=

b a

f(t) dt

Based on your results in problems 1 – 3, conjecture an expression for g’(x).

Solution To answer this question, we need to analyze the solutions to the problems 1 through 3.

Let’s look at problem 1.

We find that the integrand is f(t) = 2t + 1, the lower and upper limits respectively, are t = 1 and t = x. So then, we have the equation

A(x)

=

∫

x 1

(2t + 1) dt

Which when evaluated, using the properties of the integral, gives

A(x)

= x2 + x – 2

Differentiating A gives

A' (x) = 2x + 1 Thus, if

f (t)

=

A(x)

=

A' (x)

=

and

2t + 1

∫

x 1

(2t + 1) dt

2x + 1

It follows that

A'(x)

=

d/dx

∫

x

(2t + 1) dt

1

=

Let’s look at problem 2.

We are given the integral

A(x)

=

∫

x -1

(1 + t 2) dt

Which when evaluated, gives

A(x) =

4/3 + x + (x3/3)

Differentiating A gives

A’(x) = Thus, if

1 + x2

2x + 1

=

f(x)

f(t) =

and

1 + t2

A(x)

=

∫

A’(x)

=

1 + x2

x

(1 + t 2) dt

-1

It follows that

A'(x)

=

d/dx

∫

x -1

(1 + t 2) dt

=

1 + x2

=

f(x)

And finally, problem 3: The way problem 3 was evaluated was quite different because it was impossible to derive an expression for the given integral. Hence, we had to “improvise”, using a table of values. We found that, if

f(x) = cos (x2) g(x)

and

=

∫

x

(cos t 2) dt

0

AND the graphs of g’and f are the same, it implies that

g’(x) = That is,

g'(x)

=

d/dx

∫

x 0

f (x)

(cos t 2) dt

=

f (x)

In the end, we find that we arrive at the same results for problems 1, 2 and 3: For problem 1, we found that A’(x)

=

f (x)

For problem 2, we found that A’(x)

=

f (x)

For problem 3, we found that g’(x)

=

f (x)

Thus, at this point, we can conclusively say that, for a function

(which is continuous on [a, b])

y = f(x) AND a new function

g(x)

=

∫

b a

f(t) dt

Then the derivative of the area function g is given by

g'(x)

=

d/dx

∫

b a

f(t) dt

=

f(x)

In the following section, we discuss one of the most important theorems in calculus:

THE FUNDAMENTAL

THEOREM OF CALCULUS. The discussion will be based on the results of this discovery project.

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