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PRIMITIVE, DEFINITE INTEGRAL 

 

 

Def. Primitive of a function. Given a function f(x), the primitive of f(x) is a function F(x) which has f(x) as its derivative i.e. it is a function F(x) such that F'(x) = f(x). If F(x) is a primitive of f(x) then F(x) + c is also a primitive of f(x), where c is any constant. Thus if a function f(x) has one primitive F(x), then along with this one it has an entire family of primitives, infinite in number, of the form F(x) + c. And this family of primitives exhausts the whole set of primitives for f(x) — there are no others.


Syn. Anti-derivative, indefinite integral.


Example. Find the primitive of the function 8x3.

 

Solution. We ask ourselves the question: What function has as its derivative 8x3? To answer the question we think in terms of our memorized differentiation rules and formulas and work backward, trying to figure out what function, when differentiated, would give 8x3. The answer is 2x4 + c where c is any constant.



Definite integral. The definite integral is the most basic, fundamental concept of integral calculus.

ole.gif

Consider the function y = f(x) shown in Fig. 1. We wish to find the area A bounded by the curve y = f(x), the x-axis and the lines x = a and x = b. To find this area we proceed as follows: We divide the interval [a, b] into n sub-intervals, not necessarily of equal length. We denote the length of the first subinterval by Δx1, of the second by Δx2, and so forth up to the final subinterval Δxn. In each subinterval Δxi we arbitrarily choose a point ξi and set up the sum


1)        An = f(ξ1)Δx1 + f(ξ2)Δx2 + ... + f(ξn)Δxn .

 

An is thus equal to the sum of the areas of the rectangles shown in Fig. 1 and represents an approximation to the area A that we wish to compute. We now define the area A to be the following sum:


             ole1.gif


or 

ole2.gif


in which we require that the largest of the sub-intervals, max Δxi, approach zero as the number of sub-intervals n → ∞. We thus make an infinitely fine net on the interval [a, b] in which the largest sub-interval is required to approach zero.

ole3.gif

We have assumed that f(x) ole4.gif 0. If f(x) changes sign, then the limit 2) will give us the algebraic sum of the areas of the segments lying between the curve y = f(x) and the x-axis, where the segments above the x-axis are taken with a plus sign and those below with a minus sign. See Fig. 2.


The need to calculate the limit 2) arises in many problems. For example, suppose that a point is moving along a straight line with variable speed s = s(t). How do we determine the distance d covered by the point in the time from t = a to t = b?


Let us assume that the function s(t) is continuous. Let us divide the interval [a, b] up into n sub-intervals, of length Δt1, Δt2,... , Δtn. To calculate the approximate value for the distance covered in each subinterval Δti, let us suppose that the speed during the period of time is constant i.e. it is equal throughout the subinterval to its actual value at some intermediate point ξ1. The whole distance covered will then be expressed approximately by the sum


             ole5.gif


and the exact value of the distance d covered in the time from a to b, will be the limit of such sums for finer and finer subdivisions; that is, it will be the limit


             ole6.gif


The example just given is only one of many that could be given. Many practical problems lead to the calculation of such a limit.


The limit 2) above is called the definite integral of the function f(x) taken over the interval [a, b].


 

Note. There is a natural question that a reader might ask. Why such a complicated definition? Why not just define the sub-intervals to be equal with the ξ’s placed at the midpoints? I can’t give an answer to that. The definition we have given appears to be the standard, universally used definition. 



Def. Definite (Riemann) integral. The definite integral of a function f(x) taken over the interval [a, b] is denoted by


             ole7.gif


and is defined as


ole8.gif


where f(x)dx is called the integrand and a and b are the limits of integration; a is the lower limit, b is the upper limit.

 

Note the source of the notation for an integral by comparing the left and right members of 3) above: dx corresponds to Δx, f(x) corresponds to f(ξi), and the integral sign ∫ represents the letter S, meaning “sum”.


The intuitive interpretation of a definite integral


             ole9.gif


 is that of the area bounded by the function f(x), the x-axis and the lines x = a and a = b.





The connection between differential and integral calculus. One of the fundamental contributions of Newton and Leibnitz was that they clarified the profound connection that exists between differential and integral calculus, a connection which provides us with a general method of calculating definite integrals for an extremely wide class of functions.


To explain this connection let us give the following example from mechanics: Let us suppose that a point is moving along a straight line with a speed s = s(t), where t is time. We have already shown that the distance Δd covered by our point in the time between t = t1 and t = t2 is given by the definite integral


ole10.gif


Now let us assume that we know the function d = d(t) giving the distance traveled as a function of time as calculated from some initial point A on the straight line. The distance covered in the interval of time [t1, t2] is then given by


5)        Δd = d(t2) - d(t1)


Thus, from 4) and 5), we have


ole11.gif


Now the function d(t) is a primitive of s(t) i.e.


             ole12.gif


Thus 6) provides us with the connection between differential and integral calculus. The relationship 6) tells us that we can compute the definite integral


             ole13.gif


by finding its primitive d(t), its value being given by d(t2) - d(t1).


From this example from mechanics we are led to the well-known formula of Newton and Leibnitz, the so-called Fundamental Theorem of Integral Calculus:


Fundamental Theorem of Integral Calculus. If f(x) is continuous and F(x) is any arbitrary primitive for f(x) i.e. any function such that F'(x) = f(x), then


ole14.gif  


This is often written as


             ole15.gif


where


             ole16.gif


is a shorthand notation for


            F(b) - F(a).



Thus, by this theorem, the problem of calculating the definite integral of a function is reduced to the problem of finding a primitive of the function.





Area function, A(x). The definite integral


             ole17.gif


ole18.gif

represents the area under the curve f(x) between lines x = a and x = b where a and b are regarded as fixed numbers. The integral represents some number, namely the measure of that area. Now let us consider the upper limit b of the integral to be variable. Let us replace b with x. The integral then becomes


ole19.gif

 

and represents a variable area A(x) extending from x = a to a variable point x. In other words,


  ole20.gif  


where A(x) is the area shown in Fig. 3. This area function A(x) is a primitive of the function f(x) i.e. dA/dx = f(x). That this is true is quite obvious. It is quite easy to see that


            dA = f(x)dx

                                                            

and consequently that dA/dx = f(x). A small increase in x of Δx will produce an increase ΔA in

ole21.gif

area of approximately f(x)Δx. See Fig. 4. Thus


ΔA ole22.gif f(x)Δx


which we can rewrite as  


             ole23.gif  

 

and


             ole24.gif



Now let F(x) represent some particular primitive in the family of primitives of f(x). Then, since A(x) is also a primitive,


10)      A(x) = F(x) + c  


where c is a constant. We can evaluate c by substituting in known values: the value of A(x) corresponding to the initial point x = a is zero. Hence, substituting into 10),


            0 = F(a) + c


or


            c = - F(a) .


Replacing c by - F(a) in 10) we have the expression for the area from x = a to any other point x :


11)      A(x) = F(x) - F(a)


or

              ole25.gif


Finally, the area under the curve from x = a to x = b is found by replacing x by b so


ole26.gif

or


            A = F(b) - F(a) .


We see that 12) is the same equation as 7) above and we have derived the Fundamental Theorem of Integral Calculus.


The area function A(x) of some function f(x) is conceived of as the measure of an area generated by an ordinate of variable length which starts at x = a and moves to the right, its upper end always on the curve. The area generated when the moving ordinate has reached any point x is a quantity which depends on x; i.e. it is a function of x.


Problem. Given an arbitrary function f(x), one not necessarily expressed in terms of formulas or analytically, perhaps one in tabular or graphical form. Does a primitive for it exist? If so, exhibit it.


Solution. The area function A(x) representing the area under the curve extending from an arbitrary starting point “a” up to the value x constitutes a primitive for it. It can indeed be proved that if a function f(x) is continuous (and even if it is discontinuous, but Lebesgue-summable) a primitive F(x) satisfying 7) above exists.



Properties of definite integrals. If f(x) and g(x) are continuous on the interval of integration a ole27.gif x ole28.gif b the following hold:


ole29.gif

ole30.gif

ole31.gif


ole32.gif


ole33.gif


6. The First Mean Value Theorem:


             ole34.gif


ole35.gif



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