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
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.
and the integral diverges.
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
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