1000 Solved Problems in Modern Physics

1.3 Solutions 55

= 2

∫

ydx

= 2

∫ 2

1

1

x

dx=2ln2

= 1 .386 units

Fig. 1.14Area enclosed
between the curvesy= 1 /x
andy=− 1 /xand the lines
x=1andx= 2

1.51

∫

1

x^2 − 18 x+ 34

dx=

∫

1

(x−3)^2 + 25

dx

=(1/5) tan−^1

(

x− 3

5

)

1.52

∫ 1

0

x^2 tan−^1 xdx=(x^3 /3) tan−^1 x|^10 − 1 / 3

∫ 1

0

x^3

(x^2 + 12 )

dx

=

π

12

−

1

3

∫ 1

0

[

x−

x

(x^2 +1)

]

dx

=

π

12

−

x^2

6

∣

∣

∣

∣

1

0

+

(

1

6

)

ln(x^2 +1)

∣

∣

∣

∣

1

0

=

π

12

−

1

6

+

1

6

ln 2

1.53 (a) The required area is for the figure formed by ABDGEFA. This area is equal

to the area under the curvey=x^2 +2, that is ACEFA, minusΔBCD, plus

ΔDGE (Fig 1.15a)

=

∫ 2

− 1

ydx−

1

2

BC.CD+

1

2

DE.EG

=

∫ 2

− 1

(x^2 +2)dx−

1

2

. 2. 2 +

1

2

. 1. 1

= 7 .5 units

(b) The required volumeV =Volume of cylinder BDEC of heightHand

radiusrand the cone ABC. (Fig 1.15b)

V=πr^2 H+

1

3

πr^2 h=πr^2

(

H+

h

3

)