8/7/2019 Tut5_Soln

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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.)