1000 Solved Problems in Modern Physics

1.3 Solutions 85

1.103

<x>=

∫∞

0

xf(x)dx/

∫∞

0

f(x)dx

=

∫∞

0

x^2 e−

xλ

dx/

∫∞

0

xe−

xλ

dx

=

2 λ^3

λ^2

= 2 λ

Most probable value ofxis obtained by maximizing the functionxe−x/λ

d

dx

(xe−x/λ)= 0

e−

xλ(

1 −

x

λ

)

= 0

∴x=λ

x(most probable)=λ

1.3.14 NumericalIntegration

1.104 The trapezoidal rule is

Area=

(

1

2

y 0 +y 1 +y 2 +···+yn− 1 +

1

2

yn

)

Δx

Given integral is

∫ 10

1 x

(^2) dx.Dividex=1tox=10 into 9 intervals.
Thusb−na=^109 −^1 = 1 =Δx
Substituting the abscissas in the equationy=x^2 , we get the ordinatesy=
1 , 4 , 9 , 16 ,···100.
area=

(

1

2

+ 4 + 9 + 25 + 36 + 49 + 64 + 81 +

1

2

× 100

)

= 334. 5

This may be compared with the value obtained from direct integration,[

x^3

3

] 10

1

=333.

Theerroris0.45%.

1.105 For Simpson’s rule
take 10 intervals
Hereb−na=^1010 −^0 = 1 =Δx
The area under the curvey=x^2 is given by
Δx
3

(y 0 + 4 y 1 + 2 y 2 + 4 y 3 + 2 y 4 +···+ 4 yn− 1 +yn)