March 9, 2006
Lecturer: Dr Martin Kurth
Hilary Term 2006
Problem Sheet 5 – Solutions
1. First, we have to find the points where the graphs of f and g intersect.
f (x) = g(x)
1
1
3
= − x+
x
2
2
leading to
x2 − 3x + 2 = 0
with solutions
a = 1,
b = 2.
Now we apply our formula for washer cross sections:
Z b ³
´
2
2
π [g(x)] − [f (x)] dx,
V =
a
which in our case (after a bit of rearranging) gives
¶
Z 2 µ
9
1
1 2 3
x − x + − 2 dx
V =
π
4
2
4 x
1
¶¯2
µ
1 ¯¯
1 3 3 2 9
= π
x − x + x+
12
4
4
x ¯1
π
=
.
12
2. As the functions f and g are the same as in question 1, we still have
a = 1,
b = 2.
The formula for the volume generated by revolving the area between the
x-axis and the graph of g about the line x = 1/2 is
¶
Z b µ
1
2π x −
g(x)dx,
Vg =
2
a
and the corresponding formula for f
¶
Z b µ
1
f (x)dx.
Vf =
2π x −
2
a
To get the volume V of the solid generated by revolving the area between
the graphs, we have to subtract Vf from Vg , which gives
¶
Z 2 µ
1
V =
[g(x) − f (x)] dx
2π x −
2
1
1
=
=
=
=
2
¶·
¸
1
1
3 1
− x+ −
dx
2π x −
2
2
2 x
1
¶
Z 2 µ
7 1
7
dx
π −x2 + x − +
2
2 x
1
µ
¶¯2
¯
1
7
7
π − x3 + x2 − x + ln |x| ¯¯
3
4
2
1
µ
¶
7
π − + ln 2 .
12
Z
µ
3. We start by defining a function f by
f (x) = x − 1 − ln x.
Taking its derivative, we get
f 0 (x) = 1 −
1
,
x
which for x ≥ 1 means
f 0 (x) ≥ 0.
Now we write f (x) as
f (x) = f (1) +
Z
x
f 0 (t)dt,
1
and as the integrand is ≥ 0, so is the integral (by the domination rule), ie
f (x) ≥ f (1) = 0,
proving the statement.
2