Description

Devansh Jain
IIT Bombay
Questions to be Discussed
Sheet 4
2 (a), (b) 3 (ii), (iv) 4 (b) (i), (ii) –
6 –
2. (a) Let P = {x0,x1,…,xn} be any partition of [a,b]. For i = 1,2,…,n, we have
Mi(f ) = sup f (x) ≥ 0.
x∈[xi−1,xi ]
We have used that the supremum of a set of non-negative real numbers is nonnegative.
(Why?)
Thus, U(P,f ) ≥ 0. As f is given to be Riemann integrable on [a,b], there exists a
Z b
sequence (Pn) of partitions of [a,b] such that U(Pn,f ) → f (x)dx. But
a
U(Pn,f ) ≥ 0 for all n. (Shown above) Z b
Thus, f (x)dx = lim U(Pn,f ) ≥ 0. a n→∞
Note that here we have used the fact that the limit of a sequence of nonnegative real numbers, if it exists, is nonnegative.

To prove the next part, let us prove the contrapositive. That is, if f (x) 6= 0 for some
Z b
x ∈ [a,b], then f (x)dx 6= 0.
a
Suppose c ∈ [a,b] is the number such that f (c) 6= 0. As f (x) ≥ 0 for all x ∈ [a,b], we have it that f (c) > 0. Let .
As f is continuous, there is a δ > 0 such that if x ∈ [a,b] and |x − c| < δ, then
2 which implies that .
Now, let us take the partition P := {x0,x1,x2,x3} with x0 = a, x1 = c − δ, x2 = c + δ and x3 = b. If it is the case that x1 < x0, then discard x1. If it is the case that x2 > x3, discard x2. Relabel if required.
Now, there exists xi ∈ P such that inf f .
x∈[xi−1,xi ]
Thus, L(P,f ) > 0. As f is Riemann integrable,
Z b
f (x)dx = sup{L(P,f ) : P is a partition of [a,b]} > 0 as we have found a partition
a that has a strictly positive lower sum.
2. (b) Let a = 0,b = 2 and f : [a,b] → R be defined as
0 ;x 6= 1
f (x) =
1 ;x = 1
Show that f is actually Riemann integrable on [0,2] with the integral equal to 0.
3. (ii) Note that
n n
X n 1 i i − 1
Sn = 2 + n . i
i=1 =1 n
Define f : [0,1] → R by f (x) := tan−1 x. Then, we have that f .
As f 0 is continuous and bounded, it is (Riemann) integrable.
For n ∈ N, let Pn := {0,1/n,…,n/n} and ti := i/n for i = 1,2,…,n. Then, Sn = R(Pn,f 0). Since ||Pn|| = 1/n → 0, it follows that
1
R(Pn, .
By the Fundamental Theorem of Calculus (Part 2), we have it that
lim Sn.
n→∞
3. (iv) Note that
Sn . n n n n n i=1 i=1
Define f : [0,1] → R by f . Then, we have that f 0(x) = cos(πx).
As f 0 is continuous and bounded, it is (Riemann) integrable.
For n ∈ N, let Pn := {0,1/n,…,n/n} and ti := i/n for i = 1,2,…,n. Then, Sn = R(Pn,f 0). Since ||Pn|| = 1/n → 0, it follows that
Z 1 Z 1
R(Pn,f 0) → cos(πx)dx = f 0(x)dx.
0 0
By the Fundamental Theorem of Calculus (Part 2), we have it that
lim Sn.
n→∞
4. (b) Let u and v be differentiable functions defined on appropriate domains.
Z x
Let g be a continuous function. Define G(x) := g(t)dt. Then G0(x) = g(x), by
a
Fundamental Theorem of Calculus (Part 1). Note that
u(x) g(t)dt = G(v(x)) − G(u(x)). Thus, by the Chain Rule, one has d Z v(x) g(t)dt = G0(v(x))v0(x) − G0(u(x))u0(x) = g(v(x))v0(x) − g(u(x))u0(x).
dx u(x)
We can now easily solve the question.
(i)
2
Given, Fdt
dF dx
= 2cos(4×2).
(ii)
Z x2 Given, F(x) = cos(t)dt
0
dF
∴ = cos x dx
= 2x cos(x2).
6.
1 Z x g(x) =f (t)sinλ(x − t)dt λ 0
1 Z x
=f (t)(sinλx cosλt − cosλx sinλt)dt λ 0
x
f (t)sinλtdt
Now, we can differentiate g using product rule and Fundamental Theorem of Calculus (Part 1).
Z x Z x
∴ g0(x) = cosλx f (t)cosλtdt + sinλx f (t)sinλtdt
0 0
It is easy to verify that both g(0) and g0(0) are 0. We can differentiate g0 in a similar way and get,
Z x Z x
g00(x) = −λsinλx f (x)cosλtdt + f (x)cos2 λx + λcosλx f (t)sinλtdt
0 0
+ f (x)sin2 λx

=⇒ g00(x) + λ2g(x) = f (x)

References