1.3 Solutions
55
= 2
ydx
= 2
2
1
1
x
dx = 2 ln 2
= 1.386 units
Fig. 1.14 Area enclosed
between the curves y = 1/x
and y = −1/x and the lines
x = 1 and x = 2
1.51
1
x 2 − 18x + 34
dx =
1
(x − 3) 2 + 25
dx
= (1/5) tan
−1
x − 3
5
1.52
1
0
x
2 tan
−1 x dx = (x
3
/3) tan
−1 x|
1
0 − 1/3
1
0
x
3
(x 2 + 1 2 )
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 curve y = x
2
+ 2, that is ACEFA, minus ΔBCD, 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 volume V = Volume of cylinder BDEC of height H and
radius r and the cone ABC. (Fig 1.15b)
V = πr
2 H +
1
3
πr
2 h = πr
2
H +
h
3