Chapter V


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


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


EXERCISES 1508 - 1545

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 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.


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 1510 - 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

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