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
)
