**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* = *x*_{0 }< *x*_{1}
< ··· < *x*_{n}.= *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, *S*_{n} is the
algebraic area of a step-like 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, *D**x*_{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 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 *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 end-points of the subintervals [*x*_{i}*,*
x_{i+l}]. Find the limit

**Solution:** We have D*x*_{i}
= (10 - 1)/*n = *9/*n* and *x*_{i}
= *x*_{i} = *x*_{0 }+ *i*D*x*_{i
}= 1 + 9*i*/*n*, whence *f*(*x*_{i})
= 1 + 1 + 9*i*/*n =* 2 + 9*i*/*n*
(Fig.38)

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

**Solution:** Partition
the base *a* into *n* equal parts D*x = 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 D*x = 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,

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

**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:**** **

**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* < *x*_{1 }<
1, whence, for *x* ® 1,

Since the integral

so does (6).

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.**

**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

**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 **

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

Evaluate by parts the integrals:

**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:

**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 =* **x*²/2*,*
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 - y*² 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 *y*_{1}*
= -*2 and *y*_{2} = 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=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 *x*-axis 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 = a*cos*t* and then *x
= a,* we get the limits of integration *t*_{1 }=
*p*/2 and *t*_{2}* = *0*, *whence

and hence *S = **p**ab.*

**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 *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:

**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 *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} = 2*p* 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