Advanced Calculus III

Unit 6.3: Problems 1, 2b, 3, 5a, 5b, 9, 11f, 12, 13, 16

Megan Bryant

September 16, 2013

6.3.1

Let f ∈ R[a, b]. For x ∈ [a, b], set F (x) =∫ xa f . Prove that F is continuous

on [a, b].

Since f is bounded, |f(x) ≤M for all x ∈ [a, b].

If x, y ∈ [a, b] with x<y then |F (y)− F (x)| = |∫ yx f | ≤

∫ yx |f | ≤M |y − x|.

Therefore F is Lipchitz continuous on [a, b] and is also uniformily continuous.

6.2b

For x ∈ [0, 1], find F (x) =∫ x0 f(t)dt for each of the following gunctions f

defined on [0, 1]. In each case verify that F is continuous on [0, 1], and thatF ′(x) = f(x) at all points where f is continuous.

F1(x) = x for 0 ≤ x ≤ 12 .

For all x, y ∈ [0, 12), for all ε>0, choose δ = ε>0. Then for all |x − y|<δ,|F1(x) − F1(y)| = |x − y|<δ = ε. Thus, F1(x) = x is uniformly continuouson [0, 12).

F ′1(x) = 1 = f(x) for all x ∈ [0, 12 ].

F2(x) = 32 − 2x for x in[0, 12).

For all x, y ∈ [12 , 1], for all ε>0, choose δ = ε>0. Then, for all |x − y|<δ,|F2(x)− F2(y)| = |32 − 2x3

2 + 2y| = | − 2(x− y)|<δ = ε.Thus F2 = 32 − 2x is

uniformly continuous on [12 , 1].

F ′2(x) = −2 = f(x) for all x ∈ [12 , 1].

1

6.3.3a

Let f(t) be defined by

f(t) =

{t 0 ≤ t<1

b− t2 1 ≤ t ≤ 2

and let F (x) be defined by F (x) =∫ x0 f(t)dt, 0 ≤ x ≤ 2 Find F (x).

f(t) =

{t 0 ≤ t<1

b− t2 1 ≤ t ≤ 2

By straightforward computation we see that

F (x) =

{x2

2 0 ≤ x<1

bx− x3

3 1 ≤ x ≤ 2

6.3.3b

Let f(t) be defined by

f(t) =

{t 0 ≤ t<1

b− t2 1 ≤ t ≤ 2

and let F (x) be defined by F (x) =∫ x0 f(t)dt, 0 ≤ x ≤ 2 For what value of b

in the definition of f is F (x) differentiable for all x ∈ [0, 2] .

From above, we know that

F (x) =

{x2

2 0 ≤ x<1

bx− x3

3 1 ≤ x ≤ 2

limx→1−F (1) = limx→1−12

2 = 12 , from the left.

F (1) = b ∗ 1− 13

3 = b− 13 , from the right.

Set the two sides equal to eachother. b− 13 = 1

2

b = 26 + 3

6 = 56

Therefore for b = 56 F (x) is differentiable for all x ∈ [0, 2].

2

6.3.5a

Find F ′(x) where F is defined on [0, 1] as follows: F (x) =∫ x0

11+t2

dt.

F (x) =∫ x0

11+t2

dt = tan−1(x)− tan−1(0) = tan−1(x)− 0 = tan−1(x)

Therefore, F ′(x) = 11+x2

.

6.3.5b

Find F ′(x) where F is defined on [0, 1] as follows: F (x) =∫ x0 cost

2dt.

F (x) =∫ x0 cost

2dt = 12(x+ sin(x)cos(x))|10

= 12(x+ sin(x)cos(x))− 1

2(0 + sin(0)cos(0)) = 12(x+ sin(x)cos(x))

Therefore, F ′(x) = cos2(x)

6.3.9

Let f be a continuous real-valued function on [a, b],g ∈ R[a, b] with g(x) ≥ 0

for all x ∈ [a, b]. Prove that there exists c ∈ [a, b] such that∫ ba f(x)g(x)dx =

f(c)∫ ba g(x)dx.

Let m = min{f} and M = max{f} for f on [a, b].

g(x) ≥ 0 for all x.

Therefore, m ∗ g(x) ≤ f(x)g(x) ≤M ∗ g(x). (1)

For∫ ba g>0, we know from (1) that m ≤

∫ ba fg∫ ba g≤M .

When∫ ba g = 0, we know |

∫ ba g| = c

∫ ba g =

∫ ba cg ≤

∫ ba |f |, for any c ∈ [a, b].

Thus, when∫ ba g = 0,

∫ ba fg = 0.

Therefore, there exists c ∈ [a, b] such that∫ ba f(x)g(x)dx = f(c)

∫ ba g(x)dx.

6.3.11f

Evaluate the following integral. Justify each step.∫ 41

1x√x+1

dx.∫ 41

1x√x+1

dx =∫

1(√x+1−1)(

√x+1+1)

√x+1

3

Let u =√x+ 1 and du = 1

2√x+1

.

= 2∫

1(u−1)(u+1)du = 1

∫1

2(u−1) −1

2(u+1)du

= ln(|u− 1|)− ln(|u+ 1|) = ln(|√x+ 1− 1|)− ln(|

√x+ 1 + 1|)

Which implies∫ 41

1x√x+1

= (ln(|√x+ 1− 1|)− ln(|

√x+ 1 + 1|))|41

= ln(√

5− 1)− ln(√

5 + 1) + ln(√

2 + 1)− ln(√

2− 1).

6.3.12

Suppose f is continuous on [0, 1]. Prove that limn→∞∫ 10 f(xn)dx = f(0).

limn→∞∫ 10 f(xn)dx = limn→∞ f(xn)|10 −

∫ 10 f′(xn)nxndx

= limn→∞[f(1)− 0−∫ 10 f′(u) ∗ u

1ndu], where u = xn and du = nxn−1dx.

= f(1)−∫ 10 f′(u)(limn→∞ u

1n )du

= f(1)−∫ 10 f′(u) ∗ 1du = f(1)− (f(1)− f(0)) = f(0).

Thus, limn→∞∫ 10 f(xn)dx = f(0).

6.3.13

Suppose f : [a, b] → R is continuous. Let M = max{|f(x)| : x ∈ [a, b]}.Show that limn→∞(

∫ ba |f(x)|ndx)

1n = M .

Since f is continuous, for all ε>0, there exists δ such that f(x)>M − ε forall x ∈ (x0 − δ, x0 + δ).∫ ba f(x)ndx >

∫ x0+δx0−δ f(x)ndx >

∫ x0+δx0−δ (M − ε)ndx = (M − ε)n2δ

Therefore, for any ε>0, (∫ ba f(x)ndx)

1n>((M − ε)n2δ)

1n

limn→∞(∫ ba f(x)n)

1n ≥ limn→∞(M − ε)2δ

1n = (M − ε).

However, since ε is arbitrary, we can conclude thatlimn→∞(

∫ ba |f(x)|ndx)

1n = M .

4

6.3.16

Cauchy-Schawrz Inequality for Integrals: Let f, g ∈ R[a, b]. Prove that

|∫ ba f(x)g(x)dx|2 ≤ (

∫ ba f

2(x)dx)(∫ ba g

2(x)dx).

0 ≤∫ ba (f(x)− αg(x))2dx

=∫ ba f

2(x)dx− 2α∫ ba f(x)g(x)dx+ α2

∫ ba g

2(x)dx

Let α =∫ ba f(x)g(x)dx∫ b

a g2(x)dx

, which is a constant.

This implies 0 ≤∫ ba f

2(x)dx− 2(∫ ba f(x)g(x)dx)

2∫ ba g

2(x)dx+

(∫ ba f(x)g(x)dx)

2(∫ ba g

2(x)(dx))

(∫ ba g

2(x)dx)2

=∫ ba f

2(x)dx− (∫ ba f(x)g(x)dx)

2∫ ba g

2(x)dx.

This implies that(∫ ba f(x)g(x)dx)

2∫ ba g

2(x)dx≤

∫ ba f

2(x)dx.

Thus, |∫ ba f(x)g(x)dx|2 ≤ (

∫ ba f

2(x)dx)(∫ ba g

2(x)dx)

5