**Chapter 6**

February 20 2017

**Chapter 6 Definite Integrals**
**6.1 Introduction**
- at the beginning of chapter 2 we described calculus as the study of methods for calculating two important quantities associated with curves
- slopes to tangent lines and areas of regions bounded by curves
- of course, this description gives an oversimplified view of the subject
- for it emphasises calculus as a tool in the service of geometry but says nothing about its indispensable role in the study of science
- nevertheless, it explains the traditional division of calculus into two distinct parts
- differential calculus, which deals with slopes of tangent lines
- and integral calculus, which is concerned with areas
- the problem of areas was of great interest to the ancient Greeks
- they knew a good deal about the areas of triangles, circles, and related configurations
- but any other figure presented a new and usually insoluble problem
- Newton and Leibniz showed that if an area can be calculated by exhaustion, then it can also be computed much more easily by using anti-derivatives
- this crucial discovery is called the Fundamental Theorem of Calculus
- it binds together the two parts of the subject, and is undoubtedly the most important single fact in the whole of mathematics
**6.2 The Problem of Areas**
- every rectangle and every triangle has a number associated with it called its area
- the area of a rectangle is defined to be the product of its height and its base
- and the area of a triangle is 1/2 of this
- since a polygon can always be decomposed into triangles, its area is the sum of the areas
- the circle is a more difficult figure
- it can be estimated as a polygon, yielding A = pi r2
- A polygon = 1/2 hb + 1/2 hb + ... = 1/2 h (b + b + ...) = 1/2 h p (perimeter)
- now let c be the circumference of the circle
- so as the number of sides of the polygon increases, h approaches r and p approaches c
- so A polygon = 1/2 hp > 1/2 rc = pi r2
- this is the "method of exhaustion", as the area of the circle is "exhausted" by the area of the inscribed polygons
- similarly, Archimedes used the method to show a parabola's area is 4/3 the area of the inscribed triangle
- the general problem before us is that of finding the area of a region with a curved boundary
- however, most of our work will be concentrated on a special case of this general area problem, namely, finding the area under the graph of a function y = f(x) between two vertical lines x = a and x = b
- such a region has a boundary that is curved only along its upper edge, and is therefore much easier to work with
- a knowledge of this special case is often enough to enable us to cope with more complicated regions
- in section 6.4 and thereafter, we will denote this area by the standard symbol
- Sab f(x) dx
- which is read "the definite integral from a to b of f(x) dx"
- the reason for this notation will become clear in section 6.4
- for the present, do not confuse with the indefinite integral (or anti-derivative)
- S f(x) dx
- in spite of the fact that these two integrals have the same family name and look very much alike
- they are totally different entities
- the definite integral is a number
- the indefinite integral is a function or a collection of functions
- at first sight it might appear that the problem of calculating areas is a matter of geometry and nothing more
- interesting to mathematicians, perhaps, but with no practical uses in the real world outside of mathematics
- this is not the case at all
- it will become clear in the next chapter that many important concepts and problems in physics and engineering depend on exactly the same kind of ideas as those used in calculating areas
**6.3 The Sigma Notation and Certain Special Sums**
- in order to clarify our discussion of definite integrals in the next section
- we introduce here a standard mathematical notation used for abbreviating long sums
- this is called the "sigma notation" E
- if a1, a2 ... an are any given numbers, their sum is denoted by
- E1n ak
- this symbol is read "the sum from k = 1 to n of ak"
- the idea compressed in this symbol is that we are to write down each of the numbers ak as the subscript k varies from 1 to n
- and then add all these numbers together
- the letter k used as the subscript here is called the index of summation
- for example
- E1n (2k - 1) = 1 + 3 + ... + (2n - 1) is the sum of the first n odd numbers
- the following are some formulas from elementary algebra that are needed in the next section
- E1n k = 1 + 2 + ... + n = n(n-1) / 2
- E1n k2 = 12 + 22 ... + n2 = n(n + 1)(2n + 1) / 6
- E1n k3 = 13 + 23 + ... + n3 = [ n(n + 1) / 2]2
- these formulas are usually proved by the method of mathematical induction
- another easy way to prove Ek2 is by writing in reverse order and adding
- another way to prove these is by noting (k + 1)2 = k2 + 2k + 1 or (k + 1)2 - k2 = 2k + 1
- and writing each iteration vertically and adding and then solving for Ek
**6.4 The Area Under a Curve. Definite Integrals**
- we begin by restating the problem we are trying to solve
- let y = f(x) be a given non-negative function defined on a closed interval a <= x <= b
- how do we calculate the area under the graph, above the x-axis, and between the vertical lines x = a and x = b?
- closed intervals like the one mentioned here will occur quite often in the discussion, so we use the briefer notation [a, b]
- also, most of the functions we study will be continuous
- this means that from an intuitive point of view, the graph consists of a single piece, with no gaps or holes
- and more precisely, for each point c in [a, b], we must have lim x>c f(x) = f(c)
- such a function has several basic properties that we wish to recognize explicitly
- it is bounded, in the sense that there exists a constant k such that | f(x) | <= k for all x in [a, b]
- and it assumes maximum and minimum values in the sense that the graph has a highest point and a lowest point
- with the specific assumption that y = f(x) is continuous on [a, b]
- and the fact that only the upper edge is curved
- then the method of exhaustion suggests the following approximation procedure
- let n be a positive integer and divide the interval [a, b] into n equal subintervals
- using each subinterval as a base, construct the tallest rectangle that lies entirely under the graph
- write down the sum sn of the area of all these rectangles
- this sum approximates the area under the graph
- the approximation is improved by taking larger values of n
- or equivalently, by dividing [a, b] into a larger number of small subintervals
- finally, calculate the exact area under the graph by finding the limiting value approached by the approximating sums sn as n approaches infinity
- Area of region = lim n>inf sn
- w now describe this idea with greater precision by introducing some suitable notation
- again, let n be a positive integer and divide the interval [a, b] into n equal subintervals
- by inserting n - 1 equally spaced points of division x1, x2, ... xn-1 between a and b
- if we denote a by x0 and b by xn, then the endpoints of these subintervals are:
- a = x0 < x1 < x2 < ... < xn-1 < xn = b
- and the subintervals themselves are [x0, x1], [x1, x2] ... [xn-1, xn]
- we denote the length of the kth subinterval by dxk
- so that dxk = xk - xk-1
- since the subintervals are equal in length
- it is clear that dxk = (b - a)/n
- let mk denote the minimum value of f(x) on th kth subinterval [xk-1, xk]
- then this minimum value is assumed at some point mxk in the subinterval
- f(mxk) = mk, xk-1 <= mxk <= xk
- mxk is easily seen to be the left endpoint of the subinterval when the curve is rising, and the right when it is falling
- since the area of each inscribed rectangle is the product of its height and its base
- the approximating sum sn of the area s of all these rectangles is clearly
- sn = f(mx1)dx1 + f(mx2)dx2 + ... + f(mxn)dxn
- if we use the sigma notation to abbreviate this sum, we get
- sn = E k=1 to n f(mxk) dxk
- and the area of the region becomes
- Area of Region = lim n>inf E k=1 to n f(mxk) dxk
- this formula is all right as far as it goes
- but from several points of view it is inconvenient and unduly restrictive
- we broaden its scope and deepen its meaning with a series of remarks
- it is not necessary that the subintervals be equal in length
- we must now require that the length of the largest subinterval approach zero
- Area = lim maxdk>0 E k=1 to n f(mxk) dxk
- we had been calculating the lower sum because we inscribed rectangles from below
- we can also approximate the area from above
- roughly speaking, we now use each subinterval as a base and construct the shortest rectangle whose top lies entirely above the curve
- to express this in simbols, let Mk denote the maximum value of f(x) on the kth subinterval [xk-1, xk]
- as before, this maximum value is assumed at some point mxk in the subinterval whre
- f(mxk) = Mk, xk-1 <= mxk <= xk
- the sum of the areas of the circumscribed rectangles is therefore
- Sn = E k=1 to n f(mxk) dxk
- this is called the upper sum because it approximates the area of the region from above
- geometric intuition tells us that the area of our region can just as well be obtained as the limit of upper sums
- so we have
- Area = lim maxdk>0 E k=1 to n f(mxk) dxk
- it can be proved that both limits exist and have the same value for any continuous function
- further, if xk is taken to be any point in the kth subinterval [xk-1, xk]
- then we clearly have
- sn <= E k=1 to n f(mxk) dxk <= Sn
- and it follows from the theorem above that both limits can be replaced by
- Area = lim maxdk>0 E k=1 to n f(sxk) dxk
- where the only restriction placed on sxk is that xk-1 <= sxk <= xk
- this limit is symbolized by the Leibniz notation
- Sab f(x) dx
- which is read (as we said in section 6.2) "the definite integral from a to b of f(x) dx"
- if we write down the definition of the definite integral
- Sab f(x) dx = lim maxdk>0 E k=1 to n f(sxk) dxk
- then every part of the symbol on the lefthand side is intended to remind us of the corresponding part of the approximating sum on the right
- the integral sign S is an elongated letter S, as in "sum"
- chosen because of the similarity between a definite integral and a sum of small quantities
- the passage to the limit is suggested by replacing the letter E by the symbol S
- also, the usual symbol d for an increment is replaced by the letter d to remind us of this limit operation
- just as in the Leibniz notation dy/dx for the derivative
- the numbers a and b attached to the integral sign are called the lower and upper limits of integration
- limits of integration are always present in a definite integral
- and help distinguish it from the similar-appearing but very different indefinite integral S f(x) dx
- the function f(x) is called the integrand, the thing being integrated
- and the variable x is the variable of integration
- the role of the dx as an important intuitive component of definite integrals will become much clearer in the next chapter
**6.5 The Computation of Areas as Limits**
- the concepts discussed in section 6.4 suggest an actual procedure for calculating areas
- we now examine how this procedure works in a few specific cases
- consider the function y = f(x) = x on the interval [0, b]
- the region under this graph is a triangle with height b and base b
- so its area is obviously b2/2
- however, it is of some interest to verify that our limit process gives the same result
- let n be a large positive integer and divide the interval [0, b] into n equal subintervals by means of n-1 equally spaced points
- x1 = b/n, x2 = 2b/n ... xn-1 = (n-1)b/n
- the bases of the rectangles are dxk = b/n
- and if we use upper sums, then the heights of the rectangles are
- f(x1) = b/n, f(x2) = 2b/n ... f(xn) = nb/n
- and we have
- Sn = (b/n)(b/n) + (2b/n)(b/n) + ... + (nb/n)(b/n) = b3/n2(1 + 2 + ... + n) = b3/2(1+1/n)
- we therefore conclude that
- area of region = lim n>inf Sn = b2/2
- which we knew at the beginning
- in the notation of definite integrals, this result is
- S0b x dx = b2/2
- now consider the function y = f(x) = x2 on the interval [0, b]
- let n be a large positive integer and again divide the interval [0, b] into n equal subintervals of length dxk = b/n
- we again use upper sums Sn, so the heights of the successive rectangles are easily seen to be
- f(x1) = (b/n)2, f(x2) = (2b/n)2, f(xn) = (nb/n)2
- Sn = (b/n2)(b/n) + (2b/n)2(b/n) + ... + (nb/n)2(b/n) = b3/n3 (12 + 22 + ... + n2)
- Sn = b3/n3 n(n+1)(2n+1)/6 = b3/6 (1+1/n)(2+1/n)
- area of region = lim n>infSn = b3/3
- or equivalently S0b x2 dx = b3/3
- this calculation produces a result which we did not know at the beginning
- it is natural to conjecture that
- S0b xn dx = bn+1/n+1
- which Fermat proved
**6.6 The Fundamental Theorem of Calculus**
- as our main achievement so far in this chapter
- we have formulated a rather complicated definition of the definite integral of a continuous function as a limit of approximating sums
- Sab f(x) dx = lim max dxk>0 E1n f(dxk) dxk
- we have also considered several examples of the use of this definition in calculating the values of certain simple integrals
- such as S0b x dx = b2/2
- these calculations have had two purposes
- to emphasize the essential nature of the integral by giving students some direct experience with approximating sums
- and also to suggest the severe limitations of this method as a practical tool for evaluating integrals
- thus, for example, how can we possibly use limits of sums to find the numerical values of complicated integrals
- this is clearly out of the question, so where do we go from here?
- what is evidently needed is a much more efficient and powerful method of computing integrals
- and we find this method in the ideas of Newton and Leibniz
- the Newton-Leibniz approach to the problem of calculating the integral depends on an idea that seems paradoxical at first sight
- in order to solve this problem, we replace it by an apparently harder problem
- instead of asking for the fixed area, we ask for the variable area produced where the right-hand border is considered moveable
- if this area function is denoted A(x)
- then clearly A(a) = 0 and A(b) is the area between a and b
- our aim is to find an explicit formula for A(x), and then to determine the desired fixed area by setting x = b
- there are several steps in this process, which we consider separately for the sake of clarity
- Step 1
- we begin by establishing the crucial fact that dA/dx = f(x)
- this says that the rate of change of the area A with respect to x is equal to the length of the right edge of the region
- Step 2
- the equation dA/dx = f(x) makes it possible for us to achieve our goal of finding a formula for the area function A(x)
- since A(x) is one of the antiderivatives of f(x)
- but F(x) is any antiderivative of f(x)
- then we know that A(x) = F(x) + c
- at x = a, A(a) = F(a) + c, but since A(a) = 0, then c = -F(a)
- so A(x) = F(x) - F(a)
- Step 3
- all that remains is to observe that
- Sab f(x) dx = A(b) = F(b) - F(a)
- we summarize our conclusions by formally stating the Fundamental Theorem of Calculus
- if f(x) is continuous on a closed interval [a, b], and if F(x) is any antiderivative of f(x), so that (d/dx) F(x) = f(x)
- or equivalently S f(x) dx = F(x)
- then Sab f(x) dx = F(b) - F(a)
- this theorem transforms the difficult problem of evaluating definite integrals b calculating the limits of sums into the much easier problem of finding antiderivatives
- to find the value of Sab f(x) dx, we therefore no longer have to think about sums at all
- we merely find an antiderivative F(x) in any way we can - by inspection, by routine calculation, ingenious calculation, or looking it up in a book
- and then compute the number F(b) - F(a)
- for instance, in section 6.5 we used a good deal of algebraic ingenuity to obtain the formula Sab x dx = b2/2
- now, with the aid of the Fundamental Theorem, we see these formulas as obvious consequences of the following simple fact
- S x dx = x2/2
- more generally, for any exponent n > 0, we clearly have
- Sab xn dx = bn+1/n+1 - an+1/n+1 because S xn dx = xn+1/n+1
- in the process of working problems, it is often convenient to use the bracket symbol
- F(x) ]ab = F(b) - F(a)
- by using this notation, the Fundamental Theorem can be written in the form Sab f(x) dx = F(x) ]ab
- the Fundamental Theorem establishes a strong connection between definite integrals and antiderivatives
- this connection has made it customary to use the integral sign to denote an antiderivative
- and to replace the word "antiderivative" by the term "indefinite integral"
- from this point on we will often drop the adjective (indefinite, definite) and use the word "integral" alone to refer to either
- relying on the context to avoid confusion
- as an infallible aid in keeping track of which is which, we emphasize the fact that a definite integral always has limits of integration
- and that an indefinite integral never has such limits
- from our experience in chapter 5, we know many indefinite integrals, so many definite integrals are now within our reach
**6.7 Properties of Definite Integrals**
- the algebraic area is the region bounded by the curve and the x-axis
- counting areas below the x-axis as negative
- the geometric area is the actual total area bounded by the curve and the x-axis
- to find the geometric area, we must sketch the graph, locating crossing points, and calculate each integral separately and adding their absolute values
- if we drop the condition a < b and instead assume that a > b
- we can still retain the purely mathematical definition of the definite integral
- the only change is that as we traverse the interval from a to b, the increments dxk are negative
- this yields the equation
- Sab f(x) dx = - Sba f(x) dx
- also, since Saa f(x) dx = 0
- then Sab f(x) dx = Sac f(x) dx + Scb f(x) dx
- and Sab c f(x) dx = c Sab f(x) dx
- Sab [f(x) + g(x)] dx = Sab f(x) dx + Sab g(x) dx
- if f(x) <= g(x) on [a, b] then Sab f(x) dx <= Sab g(x) dx
- we have used x as the "variable of integration" in writing the definite integral Sab f(x) dx
- however, the definite integral is a fixed number whose value does not depend on which letter is used for this variable
- letters used in this way are often called dummy variables
- in most situations it doesn't matter what letters are used, as long as the ideas are clearly understood
- however, sometimes we wish to construct a new function F(x) by integrating a given function f(x) from a fixed lower limit to a variable upper limit
- as in F(x) = Sax f(x) dx
- it is evident that this usage can be confusing
- because the letter x is used with two different meanings on the right
- as the upper limit of integration above the integral sign
- and as a dummy variable behind the integral sign
- for this reason, it is customary to write
- F(x) = Sax f(t) dt
- the function F(x) as defined above has two properties that make it important
- first, it exists whenever the integrand is continuous on the interval between a and x
- second, we proved in section 6.6 that the derivative of this function is simply the value of the integrand at the upper limit
- d/dx F(x) = d/dx Sax f(t) dt = f(x)
- this provides a satisfactory theoretical solution of the problem of finding an indefinite integral for a given continuous function f(x)
- as a practical matter, it may be very difficult, or even impossible, to calculate
- S f(x) dx = F(x)
- in any recognizable form involving familiar functions
- but even if we can't find a formula for F(x)
- it is at least some consolation to know that in principle an indefinite integral always exists, namely
- F(x) = Sax f(t) dt