4.1  THE DEFINITE INTEGRAL

 

We shall begin our study of the integral calculus in the same way in which we began with the differential calculus - by asking a question about curves in the plane.

 

Suppose f is a real function continuous on an interval I and consider the curve  y=f(x). Let a<b where a,b are two points in I, and let the curve be above the x-axis for x between a and b; that is, f(x)≥ 0. We then ask: What is meant by the area of the region bounded by the curve y= f(x), the x-axis, and the lines x=a and x=b? That is, what is meant by the area of the shaded region in Figure 4.1.1? We call this region the region under the curve y=f(x) between a and b.

 

 

 

 

 

 

 

Figure 4.1.1  The Region under a Curve

 

The simplest possible case is where f is a constant function; that is, the curve is a horizontal line f(x)=k, where k is a constant and k≥ 0, shown in Figure 4.1.2. In this case the region under the curve is just a rectangle with height k and width b- a, so the area is defined as

                           Area = k. (b-a).

 

The areas of certain other simple regions, such as triangles, trapezoids and semicircles, are given by formulas from plane geometry.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.2

 

The area under any continuous curve y=f(x) will be given by the definite integral, which is written

                     

 

Before plunging into the detailed definition of the integral, we outline the main ideas.

 

First, the region under the curve is divided into infinitely many vertical strips of infinitesimal width dx. Next, each vertical strip is replaced by a vertical rectangle of height f(x), base dx, and area f(x) dx. The next step is to form the sum of the areas of all these rectangles, called the infinite Riemann sum (look ahead to Figures 4.1.3 and 4.1.11). Finally, the integral ∫ ba f(x)dx is defined as the standard part of the infinite Riemann sum.

 

The infinite Riemann sum, being a sum of rectangles, has an infinitesimal error. This error is removed by taking the standard part to form the integral.

 

It is often difficult to compute an infinite Riemann sum, since it is a sum of infinitely many infinitesimal rectangles. We shall first study finite Riemann sums, which can easily be computed on a hand calculator.

 

Suppose we slice the region under the curve between a and b into thin vertical strips of equal width. If there are n slices, each slice will have width Δx=(b -a)/n. The interval [ a, b] will be partitioned into n subintervals

where  

The points  x0,x1,……xn are called partition points. On each subinterval [ xk-1,  xk], we form the rectangle of height f( xk-1). The kth rectangle will have area

                              

                                   f( xk-1)·Δx

 

From Figure 4.1.3, we can see that the sum of the areas of all these rectangles will be fairly close to the area under the curve. This sum is called a Riemann sum and is equal to

                                

                                   f( x0) Δx +  f( x1)Δx +… f( xn-1)Δx.

It is the area of the shaded region in the picture. A convenient way of writing Riemann sums is the “∑-notation” { ∑ is the capital Greek letter sigma},

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.3  the Riemann Sum

 

The a and b indicate that the first subinterval begins at a and the last subinterval ends at b.

 

We can carry out the same process even when the subinterval length Δx does not divide evenly into the interval length b-a. But then, as Figure 4.1.4 shows, there will be a remainder left over at the end of the interval [a,b], and the Riemann sum will have an extra rectangle whose width is this remainder. We let n be the largest integer such that

                                    a + n Δx ≤ b,

and we consider the subintervals

                   [x0, x1],…，[xn-1, xn], [xn, b],

 

where the partition points are

                   x0 =a,   x1 =a+Δx,  x2 =a+2Δx,……，xn =a+n Δx, b.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.4

 

xn  will be less than or equal to b but xn +Δx will be greater than b. Then we define the Riemann sum to be the sum

 

____________________________

 

Thus given the function f, the interval [a,b], and the real number Δx> 0, we have defined the Riemann sum _____ f(x) Δx. We repeat the definition more concisely.

 

DEFINITION

 

Let  a<b and let Δx be positive real number. Then the Riemann sum ___ is defined as the sum

 

Where n is the largest integer such that a + n Δx ≤ b, and

 

are the partition points.

If xn=b, the last term f(xn) (b -xn) is zero. The Riemann sum___f(x)Δx is a real function of three variables a, b, and Δx,

 

The symbol x which appears in the expression is called a dummy variable (or bound variable), because the value of ______ f(x) Δx does not depend on x. The dummy variable allows us to use more compact notation,writing f(x)Δx just once instead of writing f(x0) Δx , f(x1) Δx, f(x2) Δx, and so on.

 

From Figure 4.1.5 it is plausible that by making Δx smaller we can get the Riemann sum as close to the area as we wish.

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.5

EXAMPLE  1   Let f(x)=___. In Figure 4.1.6, the region under the curve from x = 0

      to x =2 is a triangle with base 2 and height 1, so its area should be

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.6

 

      Let us compare this value for the area with some Riemann sums. In figure 4.1.7, we take Δx =_____. The interval [0, 2] divides into four subintervals ________, ___ and ____. We make a table of values of f(x) at the lower endpoints.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.7

     The Riemann sum is then

 

In Figure 4.1.8, we take  Δx =_____. The table of values is as follows.

 

The Riemann sum is

 

We see that the value is getting closer to one.

Finally, let us take a value of Δx that does not divide evenly into the interval length 2. Let Δx . We see in Figure 4.1.9 that the interval then divides into three subintervals of length 0.6 and one of length 0.2, namely [0, 0.6], [0.6, 1.2], [1.2, 1.8], [1.8, 2.0],

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.8

The Riemann sum is

 

Example 2  Let f(x) =_______, defined on the closed interval I=[-1,1]. The region under the curve is a semicircle of radius 1. We know from plane geometry that the area is π/2, or approximately 3.14/2=1.57. Let us compute the values of some Riemann sums for this function to see how close they are to 1.57. First take Δx=____as in Figure 4.1.10(a). We make a table of values.

 

The Riemann sum is then

 

Next we take Δx=____. Then the interval [-1, 1] is divided into ten subintervals as in Figure 4.1.10 (b). Our table of values is as follows.

 

 

Figure 4.1.10

 

The Riemann sum is

 

 

Thus we are getting closer to the actual area π/2 ~ 1.57.

 

By taking  Δx small we can get the Riemann sum to be as close to the area as we wish.

 

Our next step is to take Δx to be infinitely small and have an infinite Riemann sum. How can we do this? We observe that if the real numbers a and b are held fixed, then the Riemann sum

 

is a real function of the single variable Δx. (The symbol x which appears in the expression is a dummy variable, and the value of

 

depends only on Δx and not on x.)  Furthermore, the term

 

is defined for all real  Δx >0. Therefore by the Transfer Principle,

 

is defined for all hyperreal dx>0. When dx>0 is infinitesimal, there are infinitely many subintervals of length dx, and we call

an infinite Riemann sum(Figure 4.1.11).

 

 

 

 

 

 

 

 

 

 

Figure 4.1.11 Infinite Riemann Sum

 

We may think intuitively of the Riemann sum

 

as the infinite sum

 

where H is the greatest hyperinteger such that a +H dx≤ b, (Hyperintegers are discussed in Section 3.8.) H is positive infinite, and there are H +2 partition points x0, x1,…，xH, b. A typical term in this sum is the infinitely small quantity f(xK) dx where K is hyperinteger, 0 ≤ K<H , and xK = a + K dx.

 

The infinite Riemann sum is a hyperreal number. We would next like to take the standard part of it. But first we must show that it is a finite hyperreal number and thus has a standard part.

 

THEOREM 1

       Let f be a continuous function on an interval I, let a<b be two points in I, and let dx be a positive infinitesimal. Then the infinite Riemann sum

is a finite hyperreal number.

 

PROOF  Let B be a real number greater than the maximum value of f on [a,b].

         Consider first a real number Δx >0. We can see from figure 4.1.12 that the

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.12

 

      Finite Riemann sum is less than the rectangular area B.(b-a);

              

Therefore by the Transfer Principle,

In a similar way we let C be less than the minimum of f on [a, b] and show that

Thus the Riemann sum ________f(x)dx is finite.

 

We are now ready to define the central concept of this chapter, the definite integral. Recall that the derivative was defined as the standard part of the quotient Δy/Δx and was written dy/dx. The “definite integral” will be defined as the standard part of the infinite Riemann sum

 

and is written ___ f(x) dx. Thus the Δx is changed to dx in analogy with our differential notation. The ___ is changed to the long thin S, i.e., ∫, to remind us that the integral is obtained from an infinite sum. We now state the definition carefully.

 

DEFINITION

        Let f be a continuous function on an interval I and let a <b be two points in I.

        Let dx be a positive infinitesimal. Then the definite integral of from a to b with

        respect to dx is defined to be the standard part of the infinite Riemann sum with

        respect to dx, in symbols

 

       

 

 

        We also define  

 

 

By this definition, for each positive infinitesimal dx the definite integral

is a real function of two variables defined for all pairs(u,w) of elements of I. The symbol x is a dummy variable since the value of

does not depend on x.

 

In the notation ___ f(x) dx for the Riemann sum and ___ f(x) dx for the integral, we always use matching symbols for the infinitesimal dx and the dummy variable x. Thus when there are two or more variables we can tell which one is the dummy variable in an integral. For example, x²t can be integrated from 0 to 1 with respect to either x or t. With respect to x,

 

(where dx =1/H), and we shall see later that

 

With respect to t, however,

 

and we shall see later that

 

The next two examples evaluate the simplest definite integrals. These examples do it the hard way. A much better method will be developed in Section 4.2.

 

EXAMPLE 3  Given a constant C >0, evaluate the integral ___ c dx.

         Figure 4.1.13 shows that for every positive real number Δx, the finite Riemann

         sum is

 

        

         By the Transfer Principle, the infinite Riemann sum in Figure 4.1.14 has the same value,

 

 

        

         Taking standard parts,

                      

 

 

            This is the familiar formula for the area of a rectangle.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.13                                  figure 4.1.14

 

EXAMPLE 4  Given b> 0, evaluate the integral____x dx.

        The area under the line y=x is divided into vertical strips of width dx.

        Study Figure 4.1.15. The area of the lower region A is the infinite Riemann

        Sum

(1)                            area of A = ____  x dx.

        By symmetry, the upper region B has the same area as A;

(2)                             area of A = area of B.                        

   Call the remaining region C, formed by the infinitesimal squares along the diagonal. Thus

(3)are of  A + area of  B + area of  C= b².

       Each square in C has height dx except the last one, which may be smaller, and the widths add up to b, so       

(4)    0 ≤ area of C ≤ b dx.

      Putting (1)-(4) together,

 

 

   Since b dx is infinitesimal,

   

 

 

   Taking standard parts, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Figure 4.1.15

 

PROBLEMS FOR SECTION 4.1

 

Compute the following finite Riemann sums. If a hand calculator is available, the Riemann sums can also be computed with ____.

1  __(3x + 1)__x,   _x= __         2 ____(3x+1)__x,     _x = ___

3  __(3x + 1)__x,   _x= __         4 _____________,     _x = ___

5  ___________,   _x= __          6 _____________,     _x = 1

7  __(2x - 1)__x,   _x= 2          8  ____(x²-1)__x,     _x = ___

9  __(x² - 1)__x,   _x= __         10 ____(x²-1)__x,     _x = ___

11__(5x² - 12)__x,   _x= 2         12 ____(5x²-12)__x,     _x =1

13__(1 + 1/x)__x,   _x= __         14   ____10-2x__x,     _x = ___

15__ __x,   _x= __               16    ____2x3Δx,     Δx = ___

17____Δx,   Δx= __              18 _____x-4Δx,        Δx = 2

19__sinx Δx,   Δx=__/4         20 ____sin²xΔx,     Δx = /4

21__exΔx,   Δx=1/5            22 ____xexΔx,     Δx = 1/5

23__inx Δx,   Δx= 1           24 _ ___,          Δx =1

□25   let b be a positive real number and n a positive integer. Prove that if Δx=b/n,

 

                               

      Using the formula 1 +2 …+ (n-1) = _________, prove that

 

 

 

□26 let H be a positive infinite hyperinteger and dx = b/H. Using the Transfer Principle and Problem 25, prove that ____ dx = b²/2

 

□27 let b be a positive real number, n a positive integer, and Δx = b/n. Using the formula

 

Prove that

                    

 

 

□28  use problem 27 show that ___dx = b3/3

 

 

 

4.2  FUNDAMENTAL  THEOREM  OF  CALCULUS

In this section we shall state five basic theorems about the integral, culminating in the Fundamental Theorem of Calculus. Right now we can only approximate a definite integral by the laborious computation of a finite Riemann sum. At the end of this section we will be in a position easily to compute exact values for many definite integrals. The key to the method is the Fundamental Theorem. Our first theorem shows that we are free to choose any positive infinitesimal we wish for dx in the definite integral.

 

THEOREM 1

Given a continuous function f on [a, b] and two positive infinitesimals dx and du, the definite integrals with respect to dx and du are the same,