Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PRELIMINARY CALCULUS

2.2.12 Integral inequalities

Consider the functionsf(x),φ 1 (x)andφ 2 (x) such thatφ 1 (x)≤f(x)≤φ 2 (x)for

allxin the rangea≤x≤b. It immediately follows that

∫b

a

φ 1 (x)dx≤

∫b

a

f(x)dx≤

∫b

a

φ 2 (x)dx, (2.39)

which gives us a way of estimating an integral that is difficult to evaluate explicitly.

Show that the value of the integral

I=

∫ 1

0

1

(1 +x^2 +x^3 )^1 /^2

dx

lies between 0. 810 and 0. 882.

We note that forxin the range 0≤x≤1, 0≤x^3 ≤x^2 .Hence

(1 +x^2 )^1 /^2 ≤(1 +x^2 +x^3 )^1 /^2 ≤(1+2x^2 )^1 /^2 ,

and so
1
(1 +x^2 )^1 /^2

≥

1

(1 +x^2 +x^3 )^1 /^2

≥

1

(1+2x^2 )^1 /^2

.

Consequently,
∫ 1

0

1

(1 +x^2 )^1 /^2

dx≥

∫ 1

0

1

(1 +x^2 +x^3 )^1 /^2

dx≥

∫ 1

0

1

(1+2x^2 )^1 /^2

dx,

from which we obtain

[

ln(x+

√

1+x^2 )

] 1

0

≥I≥

[

√^1

2 ln

(

x+

√

1

2 +x

2

)] 1

0

0. 8814 ≥I≥ 0. 8105

0. 882 ≥I≥ 0. 810.

In the last line the calculated values have been rounded to three significant figures,
one rounded up and the other rounded down so that the proved inequality cannot be
unknowingly made invalid.

2.2.13 Applications of integration

Mean value of a function

The mean valuemof a function between two limitsaandbis defined by

m=

1

b−a

∫b

a

f(x)dx. (2.40)

The mean value may be thought of as the height of the rectangle that has the

same area (over the same interval) as the area under the curvef(x). This is

illustrated in figure 2.10.