1000 Solved Problems in Modern Physics

(Romina) #1

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

)
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