4 Chapter 0 Content 2 Background 2.1 Cosmological 2.1.11 Mathematics 2.1.11.12 Analysis Chapter 6
 Contents Chapter 1 2 3 4 5 6 7 8 9 10 - Answers 1 2 3 4 5 6 7 8 9 10 Appendix Index All

## 5.1 The Definite Integral as the Limit of a Sum

5.1.1 Integral sum: Let a function f(x) be defined on an interval a £ x £ b, and a = x0 < x1 < ··· < xn.= b an arbitrary partition of this interval into n sub-intervals (Fig. 37). A sum of the form

where

is called the integral sum of the function f(x) on (a, b). Geometrically speaking, Sn is the algebraic area of a step-like figure (Fig. 37).

5.1.2 The definite Integral: The limit of the sum Sn, provided that the number of subdivisions n tends to infinity and the largest of them, Dxi, to zero, is called the definite integral of the function f(x) within the limits from x = a to x = b, i.e.,

If the function f(x) is continuous on (a, b), it is integrable on [a, b], i.e., the limit of (2) exists and is independent of the mode of partition of the interval of integration [a, b] into sub-intervals and is independent of the choice of points x i in these sub-intervals. Geometrically speaking, the definite integral (2) is the algebraic sum of the areas of the figures that make up the curvilinear trapezoid aABb, in which the areas of the parts located above the x-axis are positive and those below the x-axisare negative (Fig. 37).

The definitions of integral sums and definite integrals are naturally generalized to the case of an interval [a, b], where a > b.

Example 1. Form the integral sum Sn for the function

on the interval [1, 10] by dividing the interval into n equal parts and choosing the points xi which coincide with the left end-points of the subintervals [xi, xi+l]. Find the limit

Solution: We have Dxi = (10 - 1)/n = 9/n and xi = xi = x0 + iDxi = 1 + 9i/n, whence f(xi) = 1 + 1 + 9i/n = 2 + 9i/n (Fig.38)

Example 2: Find the area bounded by an arc of the parabola y = x², the x-axis and the ordinates x=0, x = a (a > 0).

Solution: Partition the base a into n equal parts Dx = a/n. Choosing the value of the function at the beginning of each subinterval, we have

The areas of the rectangles are obtained by multiplying each yk by its base Dx = a/n (Fig, 39). Summing, we get the area of the step-like figure

Using the formula for the sum of the squares of integers

we find

and, passing to the limit,

## Exercises 1501 - 1507

Evaluate the following definite Integrals, treating them as limits of the appropriate integral sums:

Hints and Answers 1501 - 1507

## 5.2 Evaluating Definite Integrals by Means of Indefinite Integrals

5.2.1 A definite Integral with variable upper limit: If a function f(t) is continuous on an interval [a, b], then the function

is the anti-derivative of the function f(x), i.e.,

5.2.2 The Newton-Leibnitz formula: If F'(x) = f(x), then

The anti-derivative F(x) is computed by finding the indefinite integral

Example 1. Find the integral

Solution:

## Exercises 1508 - 1545

Hints and Answers 1508 - 1545

## 5.3 Improper Integrals

5.3.1 Integrals of unbounded functions: If a function f(x) is unbounded in any neighbourhood of a point c of an interval [a,b] and is continuous for a £ x < c and c < x £ b, then, by definition, we write

If the limits on the right hand side of (1) exist and are finite, the improper integral is said to converge, otherwise it diverges. When c = a or c = b, the definition is correspondingly simplified.

If there is a continuous function F(x) on [a, b] such that F '(x) = f(x), when x ¹ c (generalized anti-derivative), then

If f(x) £ F(x) when a £ x ~J~b and

converges, then Integral (1) also converges (comparison test).

If f(x) ³ 0 and

as x ® c, then a) the integral (1) converges for m < l, b) the integral (1) diverges for m ³ l.

5.3.2. Integrals with Infinite limits: If the function f(x) is continuous when a £ x < ¥, then we assume that

and, depending on whether there is a finite limit or not on the right hand side, the respective integral is said to converge or diverge.

Similarly,

If f(x) £ F(x) and the integral

converges, then Integral (3) also converges.

If f(x) ³ 0 and

as x ® ¥, then a) Integral (3) converges for m > 1, b) for m<l Integral (3) diverges for m £ 1.

Example 1.

and the integral diverges.

Example 2.

Example 3. Test the convergence of the probability integral

Solution: We set

The first of the two integrals on the right hand side is not improper, while the second one converges, since e-x² £ e-x when x £ l and

whence Integral (4) converges.

Example 4. Test the convergence of the integral

Solution: When x ® +¥ , we have

Since the integral

converges, Integral (5) likewise converges.

Example 5: Test the convergence of the elliptic integral

Solution: The point of discontinuity of the integrand is x = l. Applying Lagrange's formula, we find

where x < x1 < 1, whence, for x ® 1,

Since the integral

so does (6).

## Exercises 1546 - 1575

Evaluate the improper Integrals or establish their divergence

):

1546.~. D Test the convergence of the integrals

1574*. Prove that the Euler integral of the first kind (beta-function)

converges when p > 0 and q > 0.

1575*. Prove that the Euler integral of the second kind (gamma-function)

converges for p > 0.

Hints and Answers 1546 - 1575

## 5.4 Change of Variable in a Definite Integral

If a function f(x) is continuous over a £ x £ b and x = j(t) is a function which is continuous together with its derivative j'(t) over a £ t £ b, where a = j(a) and b = j(b), and f[j(t)] is defined and continuous on the interval a £ t £ b, then

Example 1. Find

Solution. We set

Then, t = arsin x/a and, consequently, we can take a = arsin 0 = 0, b = arsin l = p/2, whence

## Exercises 1576 - 1598

1576. Can the substitution x = cos t be made in the Integral

Transform the definite integrals using the indicated substitutions:

Find an integral linear substitution

as a result of which the limits of integration will be 0 and 1, respectively. Applying the indicated substitutions, evaluate the integrals:

Evaluate by means of appropriate substitutions the integrals:

Evaluate the integrals:

1595. Prove that if f(x) is an even function, then

But, if f(x) is an odd function, then

Hints and Answers 1576 - 1598

## 5.5 Integration by Parts

If the functions u(x) and v(x) are continuously differentiable on the interval [a, b], then

## Exercises 1599 - 1609

Evaluate by parts the integrals:

Hints and Answers 1599 - 1609

## 5.6 Mean-Value Theorem

5.6.1 Evaluation of Integrals: If f(x) £ F(x) for a £ x £ b, then

If f(x) and j(x) are continuous for a £ x £ b and, besides, j(x) ³ 0, then

where m is the smallest and M the largest value of the function f(x) in the interval [a, b].

In particular, if j(x) = l, then

Inequalities (2) and (3) may be replaced, respectively, by the equivalent equalities

where c and x are certain numbers lying between a and b.

Example 1. Evaluate the integral

Solution: Since 0 £ sin² x £ 1, we have

5.6.2 The mean value of a function: The number

is called the mean value of the function f(x) over the interval a £ x £ b.

## Exercises 1610 - 1622

1610*. Determine the signs of the following integrals without evaluating them:

Hints and Answers 1610 - 1622

## 5.7 The Areas of Plane Figures

5.7.1 Area in rectangular co-ordinates: If a continuous curve is defined in rectangular co-ordinates by the equation

the area of the curvilinear trapezoid, bounded by this curve, by two vertical lines at the points x = a and x = b and by the segment of the x-axis a £ x £ b (Fig. 40), is given by

Example 1. Compute the area bounded by the parabola y = x²/2, the straight lines x = 1 and x = 3, and the x-axis (Fig.41).

Solution: The area is expressed by the integral

Example 2. Evaluate the area bounded by the curve x = 2 - y - y² and the y-axis (Fig. 42).

Solution: The roles of the co-ordinate axes are here interchanged and so the required area is expressed by the integral

where the of integration limits y1 = -2 and y2 = l are the ordinates of the points of intersection of the curve with the y-axis.

In the more general case, if the area S is bounded by two continuous curves y=f1(x, y=f2(x) and by two vertical lines x=a and x=b, where f1(x) £ f2(x) when a £ x £ b (Fig. 43), we will then have

Example 3. Evaluate the area S contained between the curves

Solution: Solving Equations (3} simultaneously, we find the integration limits x1= -1 and x2 = l. By (2), we obtain

If the curve is defined by equations in parametric form: x = j(t), y = y(t), the area of the curvilinear trapezoid bounded by this curve, by two vertical lines (x = a and x = b) and by a segment of the x-axis is given by the integral

where t1 and t2 are determined from the equations

Example 4. Find the area of the ellipse (Fig. 45) by using its parametric equations

Solution: Due to the symmetry, it is sufficient to compute the area of one quadrant and then multiply the result by four. If we first set x = 0 in the equation x = acost and then x = a, we get the limits of integration t1 = p/2 and t2 = 0, whence

and hence S = pab.

5.7.2 Area in polar co-ordinates: If a curve is defined in polar co-ordinates by the equation r = f(j), then the area of the sector AOB (Fig. 46), bounded by an arc of the curve and by two radius vectors OA and OB, which correspond to the values j1 = a and j2 = b, is given by the integral

Example 5. Find the area contained inside Bernoulli's lemniscate

Solution. By virtue of the symmetry of the curve, we determine first one quadrant of the required area:

## Exercises 1623 - 1664

Hints and Answers 1623 - 1664

## 5.8 Arc Length of a Curve

5.8.1 Arc length in rectangular co-ordinates: The arc length s of a curve y=f(x), contained between two points with abscissae x = a and x = b is

Example 1. Find the length of the astroid x2/3 + y2/3 = a2/3 (Fig. 49).

Solution: Differentiating the equation of the astroid, we get

whence we have for the arc length of a quarter of the astroid:

5.8.2 Arc length of a curve represented parametrically: If a curve is represented by equations in the parametric form x=j(t), y = y(t), then the arc length s of the curve is given by

where t1 and t2 are the values of the parameter which correspond to the extremities of the arc.

Example 2. Find the length of one arc of the cycloid (Fig. 50)

Solution: We have

The limits of integration t1 = 0 and t2 = 2p correspond to the extreme points of the arc of the cycloid.

If a curve is defined by the equation r = f(j) In polar co-ordinates, then the arc length s is

where a and b are the values of the polar angle at the extreme points of the arc.

Example3. Find the length of the entire curve

It is described by a point as j ranges from 0 to 3p.

Solution: We have

whence the entire arc length of the curve is

## Exercises 1665 - 1684

Hints and Answers 1685 - 1684

## 5.9 Volumes of Solids

5.9.1 Volume of a solid of revolution: These volumes, formed by the revolution of a curvilinear trapezoid [bounded by the curve y = f(x)], the x-axis and two vertical lines x=a and x=b about the x- and y-axes are expressed by

* The solid is formed by the revolution about the y-axis of a curvilinear trapezoid bounded by the curve y = f(x) and the straight lines x = a, x = b and y = 0. As volume element, we take the volume of that part of the solid, formed by revolving about the y-axis a rectangle with sides y and dx at a distance x from the y-axis. Then, the volume element dVY = 2pxydx, whence

Example 1. Compute the volumes of solids formed by the revolution of a figure bounded by a single lobe of the sinusoidal curve y = sinx and by the segment 0 £ x £ p of the x-axis about a) the x-axis, b) the y-axis.

Solution:

The volume of a solid formed by revolution about the y-axis of a figure bounded by the curve x=g(y), the y-axis and by two parallel lines y = c and y = d, may be determined from the formula

obtained from (1) above by interchanging the co-ordinates x and y.

If the curve is defined in a different form (parametrically, in polar co-ordinates, etc.), then we must change in the foregoing formulae the integration variable in an appropriate fashion.

In the more general case, the volumes of solids, formed by revolution about the x- and y-axes of a figure bounded by the curves y1 =f1(x) and y2 =f2(x) [where f1(x) £ y2 = f2(x)] and the straight lines x = a and x = b are, respectively, equal to

Example 2. Find the volume of a torus formed by rotation of the circle

Solution: We have

whence

(the latter integral is taken with the substitution x = a sin t).

The volume of a solid, obtained by rotation about the polar axis of a sector formed by an arc of the curve r = F(j) and by the two radius vectors j = a, j =b, is computed using

The same formula is conveniently used when seeking the volume obtained by rotation about the polar axis of some closed curve defined in polar coordinates.

Example 3. Find the volume formed by the rotation of the curve r = asin 2j about the polar axis.

Solution:

5.9.2 Computing volumes of solids from known cross-sections: If S = S(x) is the cross-sectional area, cut off by a plane perpendicular to some straight line (which we take to be the x-axis) at a point with abscissa x, then the volume of the solid is

where x1 and x2 are the abscissae of the extreme cross-sections of the solid.

Example 4. Determine the volume of a wedge cut off from a circular cylinder by a plane passing through the diameter of its base and inclined to the base at an angle a. The radius of the base is R (Fig. 53 above).

Solution: We take as x-axis the diameter of the base along which the plane intersects the base and as y-axis the diameter of the base perpendicular to it. The equation of the circumference of the base is x² + y² =R².

The area of the section ABC at a distance x from the origin O is

Hence, the required volume of the wedge is

## Exercises 1685 - 1713

Hints and Answers 1685 - 1713

## 5.10 Area of a Surface of Revolution

The area of a surface, formed by rotation about the x axis of an arc of the curve y=f(x) between the points x =a and x = b, is expressed by

(ds is the differential of the arc of the curve).

If the equation of the curve is represented differently, the area of the surface SX is obtained from (1) by an appropriate change of variables.

Example 1. Find the area of the surface formed by rotation about the x-axis of a loop of the curve

Solution. For the upper part of the curve, where 0 £ x £ 3, we have v = [(3 - x)Öx]/3. whence the differential of the arc ds = (x + 1)dx/3Öx. By (1), the area of the surface is

Example 2. Find the area of the surface formed by rotation of one arc of the cycloid x = a(t—sin t), y = a(1 - cos t) about its axis of symmetry (Fig. 55 above).

Solution. The required surface is formed by rotation of the arc OA about the straight line AB with the equation x = p a. Taking y as the independent variable and noting that the axis of rotation AB is displaced relative to the y-axis by the distance p a, we find

Changing to the variable t, we obtain

## Exercises 1714 -1726

Hints and Answers 1714 - 1726

## 5.11 Moments. Centres of Gravity. Guldin's Theorems

5.11.1 Static moment: The static moment relative to the l-axis of a material point A with mass m and at a distance d from the l-axis is

The static moment relative to the l-axis of a system of n material points with masses m1, m2, ... , mn, lying in the plane of the axis and at distances d1, d2, ... , dn from it is the sum

where the distances of the points lying on one side of the l-axis are postive sign, those on the other side are negative. In a similar manner, we define the static moment of a system of points relative to a plane.

If the masses fill continuously the line or figure in the xy-plane, then the static moments MX and MY about the x- and y-axes are expressed by integrals and not by Sums (1). For geometric figures, the density is considered to equal unity.

In particular,

1) for the curve x = x(s), y = y(s), where the parameter s is the arc length,

(ds = Ö(dx)² + (dy)² is the differential of the arc);

2) for a plane figure, bounded by the curve y = y(x), the x-axis and two vertical lines x=a and y=b, we obtain

Example 1. Find the static moments about the x- and y-axes of a triangle bounded by the straight lines

Solution: Here, y = b(1 - x/a) . Applying (3), we obtain

5.11.2 Moment of inertia: The moment of inertia, about an l-axis of a material point of mass m at a distance d from the l-axis is the number Il = md². The moment of inertia, about an l-axis of a system of n material points with masses m1,m2,...,mn is the sum

where d1,d2,...,dn are the distances of the points from the l-axis. In the case of a continuous mass, a corresponding integral replaces the sum.

Example 2. Find the moment of inertia of a triangle with base b and altitude h about its base.

Solution: We take the x-axis as the base of the triangle , the y-axis as its altitude (Fig. 58 above).

Divide the triangle into infinitely narrow horizontal strips of width dy, which become the elementary masses dm. Using the similarity of triangles, we obtain

whence

5.11.3 Centre of gravity: The co-ordinates of the centre of gravity of a plane figure (arc or area) of mass M are given by

where MX and MY are the static moments of the mass M. Jn the case of geometric figures, the mass M is numerically equal to the corresponding arc or area.

We have for the co-ordinates of the centre of gravity of an arc of the plane curve y = f(x) (a£x£b), connecting the points A[a, f(a)] and B[b, f(b)],

The co-ordinates of the centre of gravity of the curvilinear trapezoid a £ x £ b, 0 £ y £ f(x) may be computed from the formulae

where

is the area of the figure.

There are similar formulae for the co-ordinates of the centre of gravity of a volume.

Example 3. Find the centre of gravity of an arc of the semi-circle

Solution. We have

whence

Thus,

5.11.4 Guldin's theorems:

Theorem 1. The area of a surface, obtained by rotation of an arc of a plane curve about some axis lying in the same plane as the curve and not intersecting, it is equal to the product of the length of the curve and the circumference of the circle described by the centre of gravity of the arc of the curve.

Theorem 2. The volume of a solid, obtained by rotation of a plane figure about some axis lying in the plane of the figure and not intersecting it, is equal to the product of the area of the figure by the circumference of the circle described by the centre of gravity of the figure.

## Exercises 1727 -1750

Hints and Answers 1727 - 1750

## 5.12 Applying Definite Integrals to the Solution of Physical Problems

5.12.1 Path traversed by a point: If a point is in motion along some curve and the absolute value of the velocity v = f(t) is a known function of the time t, then the path traversed by the point in a time interval [t1, t2] is

Example 1. The velocity of a point is

Find the path s covered by the point in the time interval T= 10 sec following the commencement of the motion. What is the mean velocity of motion during this interval?

Solution. We find

5.12.2 Work of a force: If a variable force X = f(x) acts in the direction of the x-axis, then the work of this force over an interval [x1, x2] is

Example 2. What is the work which must be performed to stretch a spring 6 cm long, if a force of 1 kg stretches it by 1 cm?

Solution: According to Hooke's law, the force of X kg which stretches the spring by xm equals X=kx, where k is a proportionality constant. Setting x = 0.0l m and X = l kg, we get k = 100, whence X = l00x and the required work is

5.12.3 Kinetic energy: The kinetic energy of a material point of mass m at velocity u is defined as

The kinetic energy of a system of n material points with masses m1, m2, ··· , mn, having the velocities v1, v2, ··· , vn, respectively, is equal to

In order to compute the kinetic energy of a solid, the latter is appropriately partitioned into elementary particles (which take the part of material points); then, by summing the kinetic energies of these particles, we get in the limit an integral in place of the sum (1).

Example 3. Find the kinetic energy of a homogeneous circular cylinder of density d with base radius R and altitude h rotating about its axis with the angular velocity w.

Solution: We take for the elementary mass dm the mass of a hollow cylinder of altitude h with internal radius r and wall thickness dr (Fig. 60). We find

Since the linear velocity of the mass dm is equal to v = rw, the element of kinetic energy is

whence

5.12.4 Pressure of a liquid: In order to compute the force of liquid pressure, we use Pascal's law, which states that the force of the pressure of a liquid on an area S at an immersion depth h is

where g is the specific weight of the liquid.

Example 4. Find the pressure experienced by a semi-circle of radius r submerged vertically in water so that its diameter is flush with the water surface (Fig. 61).

Solution: We partition the area of the semi- circle into elements - strips parallel to the surface of the water. The area of one such element (ignoring higher-order infinitesimals) located at a distance h from the surface is

The pressure experienced by this element is

where y is the specific weight of the water equal to unity.

Hence the entire pressure is

## Exercises 1751 - 1771

Hints and Answers 1751 - 1771

## Miscellaneous Exercises 1772 - 1781

Hints and Answers 1772 - 1781

4 Chapter                            Chapter 6

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