January 26, 2006
Lecturer: Dr Martin Kurth
Hilary Term 2006
Problem Sheet 2 – Solutions
1. (a) If f is a continuous function on [a, b], then the function F defined by
Z x
F (x) =
f (t)dt
a
is differentiable on (a, b) and
d
F (x) = f (x).
dx
(b) If f is continuous on [a, b] and F is an antiderivative of f , then
Z b
f (x)dx = F (b) − F (a).
a
2. There are different ways to tackle these integrals, for example one can use
the substitution rule for definite integrals:
Z b
Z g(b)
f (g(x))g 0 (x)dx =
f (u)du
a
g(a)
with
du = g 0 (x)dx.
u = g(x),
(a)
u = g(x) = x + 3,
Z
2
du = dx
1
(x + 3)dx
= 2
0
Z
4
udu
3
1
= 2 u2 |43
2
= 16 − 9
= 7
(b)
u = x + 2,
Z
du = dx
1
(x + 2)2 dx =
Z
3
u2 du
2
0
=
=
=
1
1 33
u |
3 2
1
(27 − 8)
3
19
3
(c)
u = x2 ,
−
Z √π/2
0
du = 2xdx
x sin(x2 )dx = −
1
2
Z
π/2
sin(u)du
0
1
π/2
cos(u)|0
2
³π´
´
1³
=
cos
− cos(0)
2
2
1
=
(0 − 1)
2
1
= −
2
=
3. First let’s write
Z
π
2
Z
π
¢
1 − cos2 (x) dx
Z0 π
Z π
=
1dx −
cos2 (x)dx
0
Z π 0
π
cos2 (x)dx
= x|0 −
0
Z π
= π−
cos2 (x)dx
sin (x)dx =
0
¡
0
Using
cos(x) = sin(x + π/2)
we get
Z
π
2
cos (x)dx
=
0
Z
π
Z
3π/2
sin2 (x + π/2)dx
0
=
sin2 (x)dx.
π/2
From
sin(x + π) = − sin(x)
we know that
sin2 (x + π) = sin2 (x),
which means that sin2 is periodic with period π. As we integrate over a
whole period of sin2 it doesn’t matter where we start our integration, as
long as we integrate over an interval of length π. So we have
Z π
Z 3π/2
sin2 (x)dx,
sin2 (x)dx =
0
π/2
2
and
Z
0
π
sin2 (x)dx = π −
Z
π
sin2 (x)dx.
0
Bringing the integral from the right hand side to the left hand side and
deviding by 2 we get
Z π
π
sin2 (x)dx = .
2
0
3