**Chapter 5**

October 24 2016

**Chapter 5 Indefinite Integrals and Differential Equations**
**5.1 Introduction**
- our work in the preceding chapters was concerned with the problem of tangents - given a curve, find the slope of its tangent; or equivalently, given a function, find its derivative
- in addition to launching the full-scale study of derivatives, Newton and Leibniz also discovered that many problems in geometry and physics depend on "backwards differentiation," or "antidifferentiation."
- this is sometimes called the inverse problem of tangents: given the derivative of a function, find the function itself
- in this chapter we work with the same derivative rules as in Chapter 3
- here, however, these rules are read backwards, and lead in particular to the "integration" of polynomials
- even these relatively simple procedures have some remarkable applications
**5.2 The Notation of Differentials**
- as we know, the definition of the derivative f'(x) of a function y = f(x) can be stated as f'(x) = lim dx > 0 dy / dx
- it is understood here that dx is a nonzero change in the independent variable x
- and that dy = f(x + dx) - f(x) is the corresponding change in y
- we had also introduced the equivalent notation dy/dx for this derivative
- we had emphasized that dy/dx is a single symbol and not a fraction
- however, it is certainly true dy/dx looks like a fraction, and in some circumstances it even acts like one
- the most important example of this is the chain rule dy/du du/dx = dy/dx
- where the correct formula for the derivative of a composite function is obtained by cancelling as if the derivatives were fractions
- our present purpose is to give individual meaning to the pieces dy and dx
- in such a way that their quotient is indeed the derivative f'(x)
- our reasons for doing so are impossible to explain in advance
- suffice it to say that this notational device is a necessary prelude to the powerful computational methods introduced in this chapter
- integration by substitution, and the solution of certain differential equations by separating the variables
- we begin by considering the special case in which y is a linear function of x, y = mx + b
- let P = (x,y) be a point on this line
- if x is given an increment dx and if the corresponding increment in y is dy, then the slope of the line is
- m = change in y / change in x = dy / dx
- when working in this way with increments a long a straight line, we denote them by the symbols dx and dy so that
- dy = dx and dy = dy, and call them differentials
- with this notation, dy = m dx becomes dy = m dx
- now consider an arbitrary function y = f(x)
- and assume that this function has a derivative at x
- then the tangent at P is the straight line PR with slope m = f'(x)
- by the differentials dx and dy arising from y = f(x) we refer to the increments in the variables x and y that are associated with this tangent line
- to state this more precisely, the differential d of the independent variable x is any increment dx in x, as shown, dx = dx
- and the differential dy of the dependent variable y is the corresponding increment in y along the tangent line, dy = f'(x) dx
- thus, if dx = dx = PQ is any change in x, then dy = QS and dy = QR are the corresponding changes in y along the curve and along th tangent line
- we observe that dy = f'(x) dx reduces to dy = m dx when f(x) = mx + b
- if dx != 0, then we can divide dy = f'(x) dx by dx to obtain dy/dx = f'(x)
- up to this point dy/dx = f'(x) has been trivially true because its two sides have been merely two different ways of writing the same thing
- namely, the derivative of the function y = f(x)
- the new feature in our present discussion is that now the Leibniz symbol on the left not only looks like a fraction, but is a fraction
- dy/dx = differential of y / differential of x
- the Leibniz notation for derivatives makes it particularly easy to produce the differential formula dy = f'(x) dx when the function y = f(x) is given, by computing the derivative and multiplying by dx
- for example, for y = x2, dy/dx = 2x, so dy = 2x dx
- a little experience with the use of this notation makes us realise that we can proceed directly from y = x2 to the formula dy = 2x dx without bothering to write the intermediate step of dy/dx = 2x
- it is often convenient to write df(x) instead of dy
- so, for example, d(x2) = 2x dx, d(5x4) = 20x3 dx, d(1/x) = (-1/x2)dx = -dx / x2
- our standard rules for calculating derivatives can now be given useful equivalent formulations in the notation of differentials
- for example, if we multiply the product rule by dx, it becomes the differential product rule
- d/dx(uv) = u dv/dx + v du/dx dx, d(uv) = udv + vdu
- the same applies to the other rules
- d/dx c = 0, dc = 0
- d/dx xn = n xn-1, d(xn) = nxn-1 dx
- d/dx (cu) = c du/dx, d(cu) = c du
- d/dx (u + v) = du/dx + dv/dx, d(u + v) = du + dv
- d/dx(uv) = u dv/dx + v du/dx dx, d(uv) = udv + vdu
- d/dx (u/v) = v du/dx - u dv/dx / v2, d(u/v) = vdu - udv / v2
- d/dx un = nun-1 du/dx, d(un) = nun-1 du
- the differentials for d(xn) and d(un) look the same except for the letters used
- but their connotations are quite different
- for d(xn) we think of x itself as the independent variable, and in d(un) we think of u as some unspecified function of x
- this point is illustrated by the calculations
- d(x4) = 4x3 dx
- d(x2 + 1)4 = 4(x2 + 1)3 d(x2 + 1) = 4(x2 + 1)3 2x dx = 8x(x2 + 1)3 dx
- in order to acquire facility with the differential notation, one must practice using these formulas
- if y = x4 + 3x2 + 7, find dy
- one way of doing this is to find the derivative dy/dx, dy/dx = 4x3 + 6x
- and multiplying by dx: dy = (4x3 + 6x) dx
- we can also use the differential formuluas
- dy = d(x4 + 3x2 + 7) = d(x4) + 3d(x2) + d(7) = 4x3dx + 3 2x dx + 0 = (4x3 + 6x) dx
- we emphasise that a differential on the left side requires that the right side must also contain a differential
- so we can never write dy = 4x3, but instead dy = 4x3 dx
- the method of differentials is particularly useful in implicit differentiation
- assume that y is a differentiable function of x that satisfies the equation x2y3 - 2xy + 5 = 0
- use differentials to find an expression for dy/dx
- by calculating the differential for each term in this equation, using the rules for products, powers, and constants, we get
- x2 3y2 dy + y3 2x dx - 2x dy - 2y dx = 0
- we now collect on the left the terms containing dy and on the right the terms containing dx
- (3x2y2 - 2x) dy = (2y - 2xy3) dx, and so dy/dx = 2y - 2xy3 / 3x2y2 - 2x
- most people who use calculus routinely as a tool in their work think of differentials as very small quantities
- even though the definitions contain no such requirement
- there are several good reasons for this
- one such reason is that the tangent to a curve hugs the curve closely near the point of tangency
- this means that when dx is small, the curve is virtually indistinguishable from its tangent
- and therefore the differential dy, which is comparatively easy to calculate
- provides a very good approximation to the exact increment dy, which may be harder to calculate
**5.3 Indefinite Integrals. Integration by Substitution**
- if y = F(x) is a function whose derivative is known, say, d/dx F(x) = 2x
- can we discover what the function F(x) is?
- it doesn't take much imagination to write down one function with this property, namely F(x) = x2
- moreover, adding a constant term doesn't change the derivative
- so each of the functions x2 + c also works
- are there any others? the answer is no
- the justification for this answer lies in the following principle
- if F(x) and G(x) are two functions having the same derivative f(x) on a certain interval, then G(x) differs from F(x) by a constant
- that is, there exists a constant c with the property that G(x) = F(x) + c, for all x in the interval
- to see why this statement is true, we notice that the derivative of the difference G(x) - F(x) is zero on the interval
- d/dx [ G(x) - F(x) ] = d/dx G(x) - d/dx F(x) = f(x) - f(x) = 0
- it now follows that this difference itself must have a constant value c, so G(x) - F(x) = c, or G(x) = F(x) + c
- this principle allows us to conclude that every solution must have the form x2 + c
- the problem just discussed involved finding an unkwon function whose derivative is known
- if f(x) is given, then a function F(x) such that d/dx F(x) = f(x) is called an antiderivative of f(x)
- and the process of finding F(x) from f(x) is antidifferentiation
- we have seen that f(x) does not have a single, uniquely determined antiderivative
- but, if we can find one antiderivative F(x), then all others have the form F(x) + c
- for example, 1/3 x3 is one antiderivative of x2, and the formula 1/3 x3 + c comprises all possible antiderivatives of x2
- for historical reasons, an antiderivative of f(x) is usually called an integral of f(x)
- and antidifferentiation is called integration
- the standard notation for an integral of f(x) is S f(x) dx
- which is read "the integral of f(x) dx
- the equation S f(x) dx = F(x) is therefore completely equivalent to d/dx F(x) = f(x)
- the elongated S symbol is called the integral sign
- it was introduced by Leibniz in the earliest days of calculus
- its origin will become clear in the next chapter
- to illustrate a point of usage
- we remark that the formulas S x2 dx = 1/3 x3 and S x2 dx = 1/3 x3 + c are both correct
- but the first provides one integral while the second provides all possible integrals
- for this reason the integral S f(x) dx is usually called the indefinite integral
- in contrast to the definite integrals discussed in the next chapter
- the constant c is called the constant of integration and is often referred to as an "arbitrary" constant
- it suffices to find one integral by any method that works, and then to add an arbitrary constant at the end
- every derivative that we have ever calculated can be reversed and rewritten as an integral
- in particular, the power rule d/dx xn = n xn-1 becomes S n xn-1 dx = xn
- for our present purposes, this formula is more convenient version of the power rule
- d/dx xn+1 / n+1 = xn
- this gives the form of the integral that we shall memorise and use
- S xn dx = xn+1 / n+1, n != 1
- in words: to integrate a power, push the exponent up one unit and divide by the new exponent
- the following integrals are all special cases of this rule
- S x3 dx = x4 / 4 = 1/4 x4
- S dx/x5 = S x-5 dx = x-4 / -4 = -1/4x4
- the reader will note that when n = -1, the formula is meaningless
- the treatment of this case, that is, the determination of the integral S dx/x
- is one of the most important and fascinating parts of calculus
- we return to this problem in Chapter 8
- the following additional integration rules are also slightly disguised versions of familiar facts about derivatives
- a constant factor can be moved from one side of the integral sign to the other
- S c f(x) dx = c S f(x) dx
- it is important to understand that this does not apply to variable factors, like x
- the integral of a sum is the sum of the integrals
- this applies to any finite number of terms
- S [ f(x) + g(x) ] dx = S f(x) dx + S g(x) dx
- to verify these rules, it is enough to notice that they are equivalent to the differentiation formulas
- d/dx c F(x) = c d/dx F(x)
- and d/dx [ F(x) + G(x) ] = d/dx F(x) + d/dx G(x)
- when these three rules are combined, they enable us to integrate any polynomial
- for instance
- S (3x4 + 6x2) dx = 3 S x4 dx + 6 S x2 dx = 3/5 x5 + 2 x3 + c
- we can also integrate many non-polynomials that are expressible as linear combinations of powers
- S (2x3 - x2 - 2) / x2 dx = S(2x - 1 - 2x-2) dx = x2 - x + 2/x + c
- the formula S un du = u n+1 / n+1, n != 1
- appears to be a trivial variation of S xn dx = x n+1 / n+1 in which the letter x is replaced by u
- howerver, let us think of u as some function f(x) of x and take du seriously as the differential of u, so that
- u = f(x) and du = f'(x) dx
- then the formula becomes
- S [ f(x) ]n f'(x) dx = [ f(x) ]n+1 / n+1, n != 1
- which is a far-reaching generalisation
- in practice, we usually exploit this idea by explicitly changing the variable in order to reduce a given integral to an integral of the simple form
- for instance
- S (3 x2 - 1) 1/3 4x dx
- we notice that the differential of the expression in the parentheses is 6x dx, which differs from 4x dx only by a constant factor, so we write
- u = 3x2 - 1, du = 6x dx, x dx = 1/6 du
- this enables us to translate the given integral from the x-notation to the u-notation
- S (3 x2 - 1) 1/3 4x dx = S u 1/3 4 1/6 du = 2/3 S u 1/3 du = 2/3 3/4 u 4/3 + c = 1/2 u 4/3 + c
- and by returning to the x-notation we obtain our result
- S (3 x2 - 1) 1/3 4x dx = 1/2 (3 x2 - 1) 4/3 + c
- this method is called integration by substitution
- because it depends on a substitution or change of variable to simplify the problem
- as S [ f(x) ]n f'(x) dx = [ f(x) ]n+1 / n+1 suggests,
- the success of the method depends on having an integral in which one part of the integrand is essentially the derivative of another part
- where "essentially" means "except for a constant factor"
- the integral in the last example was deliberately constructed so that the method of substitution works
- to emphasise this point, we observe that the similar integral
- S (3 x2 - 1) 1/3 dx
- seems to be "simpler" but is actually much more difficult
- if we try the substitution that worked before
- we get S (3 x2 - 1) 1/3 dx = S u 1/3 du / 6x
- and there is no practical way to get rid of the x in the denominator
- in a later chapter we will study deeper methods that succeed in this type of problem
- but just now there is nothing further we can do
**5.4 Differential Equations. Separation of Variables**
- we have seen that the equation S f(x) dx = F(x) is equivalent to d/dx F(x) = f(x)
- this statement can be interpreted in two ways
- in accordance with the explanation in section 5.3, we can think of the symbol S ... dx as operating on the function f(x) to produce its integral, or antiderivative, F(x)
- from this point of view the integral sign and the dx go together as parts of a single symbol
- the integral sign specifies the operation of integration and the only role of the dx is to tell us that x is the "variable of integration"
- a second interpretation is suggested by our treatment of the example in section 5.3
- let us write d/dx F(x) = f(x) in the form d F(x) = f(x) dx
- so that f(x) dx is explicitly seen to be the differential of F(x)
- if we now take dx in S f(x) dx at its face value, as the differential of x
- then the integral sign in S f(x) dx acts on the differential of a function, namely, on f(x) dx and products the function itself
- thus, the symbol S for integration (without considering the dx as part of the symbol), stands for the operation which is the inverse of the operation denoted by the symbol d
- we shall use both interpretations
- however, the second is particularly convenient
- not only for the actual procedures used in computing integrals
- but also for solving certain simple differential equations
- a differential equation is an equation involving an unknown function and one or more of its derivatives
- the order of such an equation is the order of the highest derivative that occurs in it
- in the process of integration we have been solving first-order differential equations of the form dy/dx = f(x)
- where f(x) is a given function
- thus, the equation dy/dx = 3x2 is equivalent to dy = 3x2 dx
- and we integrate to obtain the solution
- S dy = S 3x2 dx or y = x3 + c
- notice that a constant of integration arises on both sides here, y + c1 = x3 + c2
- but this can be written as y = x3 + (c2 - c1) and no generality is lost by replacing (c2 - c1) by c
- we can also handle more complicated differential equations
- let us find find y if dy/dx = -2xy2
- if we set aside the obvious solution y = 0, this can be written as
- -dy/y2 = 2x dx
- integration now yields
- 1/y = x2 + c or y = 1 / x2 + c
- this is called the general solution and different choices of c give different particular solutions
- we were able to solve this by the method of separation of variables
- that is, by isolating the y's from the x's and integrating
- in general, if a first-order differential equation can be written in the form
- g(y) dy = f(x) dx, with its variables separated
- and if we can carry out the integrations, then we have the solution
- S g(x) dy = S f(x) dx + c
- it should be noted that only in very special cases can the variables be separated this way
- for instance, the differential equation dy/dxy = x + y / x - y cannot be solved by this method
- the solutions to these differential equations consist of a family of curves corresponding to various values of the constant c
- the arbitrary constant that appears in the general solution of a first-order equation can be determined by prescribing, as an initial condition, the value of the unknown function y = f(x) at a single value of x, say y = y0 when x = x0
- in geometric language, an initial condition means that the solution curve is required to pass through a specific point in the plane
- we shall see in the next section that this terminology is particularly suitable for mechanical problems, where time is the independent variable and the initial positions or initial velocities of moving bodies are specified
- in the problems just discussed
- S g(y) dy = S f(x) dx + c, was easily solved for y to yield the solution of the differential equation as a family of functions
- it is often convenient not to press this point, and to accept a family of equations as the general solution without demanding explicitly displayed functions
- we illustrate this by finding the most general curve whose normal at each point passes through the origin
- and also the particular curve with this property through the point (2, 3)
- the normal OP has slope y/x, and the slope is the negative reciprocal of this
- so our differential equation is dy/dx = - x/y
- separating variables gives y dy = -x dx, and by integrating we get
- 1/2 y2 = - 1/2 x2 + c
- if we put r2 = 2c, our general solution takes the neater form
- x2 + y2 = r2, and this is the family of all circles with center at the origin
- as the reader may have foreseen, by setting x = 2 and y = 3, we find that r2 = 13, so x2 + y2 = 13 is the particular solution of the problem passing through the point (2, 3)
- it is clearly more reasonable to leave this solution as it is than to insist that it be solved for y
- by rights, differential equations should perhaps be called derivative equations
- however, in the early days of calculus differentials were the primary concepts and derivatives were secondary
- and so the term arose in a natural way, in any case, it has been standard use for hundreds of years and no one would dream of changing it now
- the mathematical description of a physical (or biological or chemical) process is usually given in terms of functions that show how the quantities involved changes as time goes on
- when we known such a function, we can find its rate of change by calculating the derivative
- often, however, we are faced with the reverse problem of finding an unknown function from given information about its rate of change
- this information is usually expressed in the form of an equation involving derivatives of the unknown function
- these differential equations arise so frequently in scientific problems that their study constitutes one of the main branches of mathematics
- we continue with some important applications of the subject in the next section, and return to it from time to time throughout the rest of our work
**5.5 Motion Under Gravity. Escape Velocity and Black Holes**
- much of the original inspiration for the development of calculus came from the science of mechanics
- and these two subjects have continued to be inseparably connected down to the present day
- mechanics rests on certain basic principles that were first laid down by Newton
- the statement of these principles requires the concept of the derivative
- and we shall see that their applications depend on integration and the solution of differential equations
- rectilinear motion is motion along a straight line
- in contrast, motion along a curved path is sometimes called curvilinear motion
- our present purpose is to study the rectilinear motions of a single particle, that is, of a point at which a body of mass is imagined to be concentrated
- the position of our particle is completely determined by its coordinate s with respect to a conveniently chosen coordinate system on the line
- since the particle moves, s is a function of the time t
- as measured from some initial instant t = 0
- we symbolise this by writing s = s(t)
- as we know from the discussion in section 2.4
- the velocity v of the particle is the rate of change of its position
- v = ds/dt
- and the speed is the absolute value of the velocity
- in general, the velocity of a moving particle changes with time, and acceleration is the rate of change of velocity
- a = dv/dt = d/dt (ds/dt) = d2s / dt2
- this is positive or negative as v is increasing or decreasing
- the basic assumption of Newtonian mechanics is that force is required in order to change velocity
- that is, acceleration is caused by force
- the concept of force originates in the subjective feeling of effort that we experience when we change the velocity of a physical object
- for instance, when we push a stalled car or throw a rock
- in the case of rectilinear motion, we assume that a force can be expressed by a number
- which is positive or negative according as the force acts in the positive or negative direction
- Newton's second law of motion states that the acceleration of a particle is directly proportional to the force F acting on it and inversely proportional to its mass m
- a = F/m
- or equivalently, F = ma
- the units of measurement for these quantities are always chosen so that the constant of proportionality in a = F/m has value 1
- thus, if the force is doubled, then by a = F/m, the acceleration is also doubled
- and if the mass is doubled, the acceleration is cut in half
- in this context, the mass of a body can be interpreted as its capacity to resist acceleration
- from one point of view, F = ma can be considered as nothing more than a definition of force
- for the right-hand side is a quantity that can be calculated by measuring the mass and observing the motion
- and this determines the force
- on the other hand, the force F is often known in advance from fairly simple physical considerations
- the innocent-looking equation F = ma then becomes the second-order differential equation m d2s / dt2 = F
- this equation has profound consequences
- for in principle we can find the particle's position at anytime t by solving m d2s / dt2 = F with appropriate initial conditions
- for example, find the motion of a stone of mass m which is dropped from a point above the surface of the earth
- the most important example of a known force is the familiar "force of gravity"
- from direct experimental evidence
- we know that the force of gravity acting on the stone is directed downward and has magnitude F = mg
- where g is the constant acceleration due to gravity near the surface of the earth
- g = 32 ft / s2 or 9.80 m / s2
- if s is the stone's position as measured along a vertical axis, with the positive direction pointing downward and the origin at the initial position of the stone
- then m d2s / dt2 = mg or d2s / dt2 = g
- integrating the equation twice gives
- v = ds / dt = gt + c1
- s = 1/2 gt2 + c1t + c2
- where c1 and c2 are constants of integration
- since the stone is "dropped", that is, released with no initial velocity at time t = 0 from the point chosen as the origin
- the initial conditions are v = 0 and s = 0 when t = 0
- the condition v = 0 when t = 0 gives c1 = 0
- and s = 0 when t = 0 gives c2 = 0
- we therefore have
- v = gt, s = 1/2 gt2
- if we change the situation and require that the stone be thrown downward with an initial velocity v0 from the initial position s = s0 at time t = 0
- then the initial conditions are v = v0 and s = s0 when t = 0
- so the equations become
- v = gt + v0 and s = 1/2 gt2 + v0t + s0
- if distance is measured in feet and time in seconds, so g = 32 ft / s2
- then v = 32t and s = 16t2
- it is clear that the velocity of the stone increases by 32 ft/s during each second of fall
- this of course is what is meant by the statement that acceleration due to gravity is 32 feet per second per second