4 Chapter 
0 Content 2 Background 2.1 Cosmological 2.1.11 Mathematics 2.1.11.12 Analysis 
Chapter 6 
5.1
The Definite Integral as the Limit of a Sum 5.1.1 Integral sum 5.1.2 The definite Integral Items 1501  1507 Exercises, Answers 5.2 Evaluating Definite Integrals by Means of Indefinite Integrals 5.2.1 A definite Integral with variable upper limit 5.2.2 The NewtonLeibnitz formula Items 1508  1545 Exercises, Answers 5.3 Improper Integrals 5.3.1 Integrals of unbounded functions generalized antiderivative 5.3.2 Integrals with Infinite limits probability integral Items 1546  1575 Exercises, Answers Exercises 5.4 Change of Variable in a Definite Integral Items 1576  1598 Exercises, Answers 5.5 Integration by Parts Items 1599  1609 Exercises, Answers 5.6 MeanValue Theorem 5.6.1 Evaluation of Integrals 5.6.2 The mean value of a function Items 1610  1622 Exercises, Answers 5.7 The Areas of Plane Figures 5.7.1 Area in rectangular coordinates 5.7.2 Area in polar coordinates Items 1623  1664 Exercises, Answers 5.8 Arc Length of a Curve 5.8.1 Arc length in rectangular coordinates 5.8.2 Arc length of a curve represented parametrically 

5.1.1 Integral sum: Let a function f(x) be defined on an interval a £ x £ b, and a = x_{0 }< x_{1} < ··· < x_{n}.= b an arbitrary partition of this interval into n subintervals (Fig. 37). A sum of the form
where
is called the integral sum of the function f(x) on (a, b). Geometrically speaking, S_{n} is the algebraic area of a steplike figure (Fig. 37).
5.1.2 The definite Integral: The limit of the sum S_{n}, provided that the number of subdivisions n tends to infinity and the largest of them, Dx_{i}, 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 subintervals and is independent of the choice of points x_{ i} in these subintervals. 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 xaxis are positive and those below the xaxisare 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 S_{n} for the function
on the interval [1, 10] by dividing the interval into n equal parts and choosing the points x_{i} which coincide with the left endpoints of the subintervals [x_{i}, x_{i+l}]. Find the limit
Solution: We have Dx_{i} = (10  1)/n = 9/n and x_{i} = x_{i} = x_{0 }+ iDx_{i }= 1 + 9i/n, whence f(x_{i}) = 1 + 1 + 9i/n = 2 + 9i/n (Fig.38)
Example 2: Find the area bounded by an arc of the parabola y = x², the xaxis 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 y_{k} by its base Dx = a/n (Fig, 39). Summing, we get the area of the steplike figure
Using the formula for the sum of the squares of integers
we find
and, passing to the limit,
Evaluate the following definite Integrals, treating them as limits of the appropriate integral sums:
Hints and Answers 1501  1507
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 antiderivative of the function f(x), i.e.,
5.2.2 The NewtonLeibnitz formula: If F'(x) = f(x), then
The antiderivative F(x) is computed by finding the indefinite integral
Example 1. Find the integral
Solution:
Hints and Answers 1508  1545
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 antiderivative), 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 < x_{1 }< 1, whence, for x ® 1,
Since the integral
so does (6).
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 (betafunction)
converges when p > 0 and q > 0.
1575*. Prove that the Euler integral of the second kind (gammafunction)
converges for p > 0.
Hints and Answers 1546  1575
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
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
If the functions u(x) and v(x) are continuously differentiable on the interval [a, b], then
Evaluate by parts the integrals:
Hints and Answers 1599  1609
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.
1610*. Determine the signs of the following integrals without evaluating them:
Hints and Answers 1610  1622
5.7.1 Area in rectangular coordinates: If a continuous curve is defined in rectangular coordinates 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 xaxis 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 xaxis (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 yaxis (Fig. 42).
Solution: The roles of the coordinate axes are here interchanged and so the required area is expressed by the integral
where the of integration limits y_{1} = 2 and y_{2} = l are the ordinates of the points of intersection of the curve with the yaxis.
In the more general case, if the area S is bounded by two continuous curves y=f_{1}(x, y=f_{2}(x) and by two vertical lines x=a and x=b, where f_{1}(x) £ f_{2}(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 x_{1}= 1 and x_{2} = 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 xaxis is given by the integral
where t_{1} and t_{2} 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 t_{1 }= p/2 and t_{2} = 0, whence
and hence S = pab.
5.7.2 Area in polar coordinates: If a curve is defined in polar coordinates 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 j_{1 }= a and j_{2} = 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:
Hints and Answers 1623  1664
5.8.1 Arc length in rectangular coordinates: 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 x^{2/3} + y^{2/3} = a^{2/3} (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 t_{1} and t_{2} 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 t_{1} = 0 and t_{2} = 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 coordinates, 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
Hints and Answers 1685  1684
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 xaxis and two vertical lines x=a and x=b about the x and yaxes are expressed by
* The solid is formed by the revolution about the yaxis 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 yaxis a rectangle with sides y and dx at a distance x from the yaxis. Then, the volume element dV_{Y }= 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 xaxis about a) the xaxis, b) the yaxis.
Solution:
The volume of a solid formed by revolution about the yaxis of a figure bounded by the curve x=g(y), the yaxis and by two parallel lines y = c and y = d, may be determined from the formula
obtained from (1) above by interchanging the coordinates x and y.
If the curve is defined in a different form (parametrically, in polar coordinates, 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 yaxes of a figure bounded by the curves y_{1} =f_{1}(x) and y_{2} =f_{2}(x) [where f_{1}(x) £ y_{2} = f_{2}(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
about the xaxis (Fig. 52).
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 crosssections: If S = S(x) is the crosssectional area, cut off by a plane perpendicular to some straight line (which we take to be the xaxis) at a point with abscissa x, then the volume of the solid is
where x_{1} and x_{2} are the abscissae of the extreme crosssections 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 xaxis the diameter of the base along which the plane intersects the base and as yaxis 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
Hints and Answers 1685  1713
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 S_{X} is obtained from (1) by an appropriate change of variables.
Example 1. Find the area of the surface formed by rotation about the xaxis 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 yaxis by the distance p a, we find
Changing to the variable t, we obtain
Hints and Answers 1714  1726
5.11.1 Static moment: The static moment relative to the laxis of a material point A with mass m and at a distance d from the laxis is
The static moment relative to the laxis of a system of n material points with masses m_{1}, m_{2}, ... , m_{n}, lying in the plane of the axis and at distances d_{1}, d_{2}, ... , d_{n} from it is the sum
where the distances of the points lying on one side of the laxis 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 xyplane, then the static moments M_{X} and M_{Y} about the x and yaxes 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 xaxis and two vertical lines x=a and y=b, we obtain
Example 1. Find the static moments about the x and yaxes 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 laxis of a material point of mass m at a distance d from the laxis is the number I_{l }= md². The moment of inertia, about an laxis of a system of n material points with masses m_{1},m_{2},...,m_{n} is the sum
where d_{1},d_{2},...,d_{n} are the distances of the points from the laxis. 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 xaxis as the base of the triangle , the yaxis 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 coordinates of the centre of gravity of a plane figure (arc or area) of mass M are given by
where M_{X} and M_{Y} 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 coordinates 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 coordinates 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 coordinates of the centre of gravity of a volume.
Example 3. Find the centre of gravity of an arc of the semicircle
Solution. We have
whence
Thus,
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.
Hints and Answers 1727  1750
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 [t_{1}, t_{2}] 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 xaxis, then the work of this force over an interval [x_{1}, x_{2}] 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 x_{m} 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 m_{1}, m_{2}, ··· , m_{n}, having the velocities v_{1}, v_{2}, ··· , v_{n}, 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 semicircle 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 higherorder 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
Hints and Answers 1751  1771
Hints and Answers 1772  1781