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Hints/Solutions for Tutorial Problem Set - 5
1. If P = {x0, x1,...,xn} is any partition of [−1, 1], then clearly mi = inf {f (x) : x ∈[xi−1, xi]} = 0 and M i = {f (x) : x ∈ [xi−1, xi]} ≥ 0 for i = 1, 2,...,n and so
L(f, P ) = 0 and U (f, P ) ≥ 0. Hence1
−1
f (x) dx = 0 and1
−1
f (x) dx ≥ 0. Let
ε > 0. There exists n0 ∈ N such that1
n0 < ε2 . We choose u, v and sk, tk fork = 2, 3,...,n0 such that 1
n0+1< u < sn0 < 1
n0< tn0 < · · · < s2 < 1
2< t2 < v < 1
and also 1 − v < ε2n0
and tk − sk < ε2n0
for k = 2, 3,...,n0. Then the partition
P 0 = {−1, 0, u , sn0, tn0,...,s2, t2, v, 1} of [−1, 1] is such that U (f, P 0) < ε. It follows
that 0 ≤1
−1
f (x) dx ≤ U (f, P 0) < ε and so1
−1
f (x) dx = 0. Thus1
−1
f (x) dx =
1
−1
f (x) dx = 0. Therefore f is Riemann integrable on [−1, 1] and1
−1
f (x) dx = 0.
As above we can see that F (x) = 0 for all x ∈ [−1, 1]. Hence F is differentiable and
F
(0) = 0 = f (0). But f is not continuous at 0, because
1
n → 0 but f (
1
n) → 1 (sincef ( 1n
) = 1 for all n ∈ N).
(Alternative method of showing F (x) = 0 for all x ∈ [−1, 1]: Since f (t) ≥ 0 for all
t ∈ [−1, 1], we have 0 ≤ F (x) ≤ F (x) +1
x
f (t) dt =1
−1
f (t) dt = 0 for all x ∈ [−1, 1].
Hence F (x) = 0 for all x ∈ [−1, 1].)
2. Assume that f (c) = 0 for some c ∈ (a, b), so that f (c) > 0. Since f is continuous atc, there exists δ > 0 such that|f (x) − f (c)| < 1
2f (c) for all x ∈ (c − δ, c + δ). This
implies that f (x) >
1
2f (c) for all x ∈ (c − δ, c + δ). So
b
a f (x) dx =
c−δ/2
a f (x) dx +
c+δ/2
c−δ/2
f (x) dx +b
c+δ/2
f (x) dx ≥ 1
2δf (c) > 0, a contradiction. Almost similar arguments
work if c = a or c = b.
3. We have 1+x−x2
1+x4≤ 1+x−x2 for all x ∈ [0, 1] ⇒
1
0
1+x−x2
1+x4dx ≤
1
0
(1+x−x2) dx = 7
6< 5
4.
Also, 1+x−x2
1+x4≥ 1
2for all x ∈ [0, 1], since 2(x − x2) + (1 − x4) ≥ 0 for all x ∈ [0, 1].
Hence1
0
1+x−x2
1+x4dx ≥
1
0
1
2dx = 1
2.
4. Applying integration by parts, we find that |b
a
f (x)cos nxdx| ≤ 1
n(|f (a)| + |f (b)| +
M (b − a)) → 0 as n → ∞, where M = max{|f (x)| : x ∈ [a, b]}. Hence the resultfollows.
5. Let F (u) =u
0
f (t) dt for all u ∈ [0, 1]. Then LHS =x
0
F (u) · 1 du. Integration by parts
gives the RHS. (Note that F (u) = f (u) for all u ∈ [0, 1].)(Alternative: We differentiate both sides with respect to x.)
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