Math 3210, Fall 2013
Instructor: Thomas Goller
20 November 2013
Learning Celebration #4
Example (10 points)
Give an example of each of the following or state that no example exists:
(1) A function f defined on [2, 5] that is not integrable on [2, 5]. (2 points)
(
1
if x 2 [2, 5] is rational;
f (x) =
3 if x 2 [2, 5] is irrational
(2) A continuous function f defined on a closed bounded interval [a, b] that does not have
an antiderivative on (a, b). (2 points)
No example exists
(3) A nonzero function on R that is its own antiderivative. (2 points)
exp
(4) A sequence of real numbers {ak }1
k=1 such that lim ak = 0 but
points)
⇢ 1
1
k k=1
P1
k=1
ak diverges. (2
(5) A non-geometric series that converges. (2 points)
1
X
1
k=1
2k + 1
Computation (4 points)
(6) Let a, r 2 R such that r 6= 1. Show that
Since
(1
Pn
k
k=0 ar =
a(1 rn+1 )
. (4 points)
1 r
r)(a + ar + ar2 + · · · + arn ) = a + ar + ar2 + · · · + arn
ar ar2 · · · arn
= a arn+1 ,
dividing both sides by 1
r yields
a + ar + ar2 + · · · + arn =
a(1 rn+1 )
.
1 r
arn+1
Math 3210, Fall 2013
Instructor: Thomas Goller
20 November 2013
Precision (6 points)
(7) State the First Fundamental Theorem of Calculus. (3 points)
Let f be a continuous function on [a, b] that is di↵erentiable on (a, b). Assume f 0 is
integrable on [a, b]. Then
Z b
f 0 (x) dx = f (b) f (a).
a
(8) Define the natural logarithm ln(x) for all x > 0. (1 point)
Z x
1
ln(x) =
dt
1 t
P
(9) What does it mean for a series 1
k=1 ak to diverge? (2 points)
P1
P
A series k=1 ak diverges if the sequence of partial sums {sn = nk=1 ak }1
n=1 does not
converge.
Proof (20 points)
(10) Let f (x) = 2x. For every b > 0, use the definition of the integral to prove that
exists and is equal to b2 . (5 points)
Proof. Let Pn = {0 < nb < 2b
< ··· <
n
subintervals of equal length. Then
(n 1)b
n
Rb
0
f (x) dx
< b} be the partition of [0, b] into n
✓ ◆
n
n
X
kb b
2b2 X
2b2 n(n + 1)
n+1
U (f, Pn ) =
f
= 2
k=
·
= b2 ·
n n
n k=1
n
2
n
k=1
and
✓
n
X
(k
L(f, Pn ) =
f
k=1
1)b
n
◆
n
b
2b2 X
= 2
(k
n
n k=1
2b2 (n 1)n
n 1
1) =
·
= b2 ·
.
n
2
n
Since
lim U (f, Pn )
the integral
Rb
0
✓
2
L(f, Pn ) = lim b ·
(n + 1)
(n
n
f (x) dx exists, and thus
Z
b
f (x) dx = lim U (f, Pn ) = b2 .
0
1)
◆
= lim
2b2
= 0,
n
Math 3210, Fall 2013
(11) Let f (x) = x
p
Instructor: Thomas Goller
20 November 2013
for some real number p > 1. Prove that
1 p
R1
f (x) dx =
1
1
.
p 1
(5 points)
The function x1 p is an antiderivative for f on [1, 1). Thus for any s > 1, we can use
FTC1 to compute
Z s
s1 p
11 p
1
1
p
x dx =
=
+
.
1 p 1 p
(1 p)sp 1 p 1
1
Since p
1 > 0, lims!1 sp 1 = 1, and thus
✓
Z 1
1
p
x dx = lim
s!1
(1 p)sp
1
1
+
1
p
1
◆
=
1
p
1
.
(12) Let f be an integrable function on a closed bounded interval [a, b]. Assume f (x)
Rb
all x 2 [a, b]. Prove that a f (x) dx 0. (5 points)
0 for
Proof. Let P = {a = x0 < x1 < · · · < xn = b} be any partition of [a, b]. Since f (x)
for all x 2 [a, b], inf [xk 1 ,xk ] f 0 for all 1  k  n. Thus
L(f, P ) =
n
X
(inf [xk
1 ,xk ]
f )(xk
xk 1 )
0
0
k=1
since every term in the sum is non-negative. Since the integral is bounded below by
every lower sum,
Z b
f (x) dx L(f, P ) 0.
a
(13) Let f, g be continuous functions on a closed bounded interval [a, b] that are di↵erentiable
on (a, b). Assume f 0 · g and f · g 0 are integrable on [a, b]. Prove the integration by parts
Z b
Z b
0
formula:
f (x)g (x) dx = f (b)g(b) f (a)g(a)
f 0 (x)g(x) dx. (5 points)
a
a
Proof. By FTC1,
Z b
a
(f · g)0 (x) dx = (f · g)(b)
(f · g)(a) = f (b)g(b)
f (a)g(a).
By the product rule for derivatives, (f · g)0 = f · g 0 + f 0 · g, so by the linearity of the integral
we can write the integral on the left side as
Z b
Z b
Z b
0
0
(f · g) (x) dx =
f (x)g (x) dx +
f 0 (x)g(x).
a
a
a
Thus we have proved that
Z b
Z b
0
f (x)g (x) dx +
f 0 (x)g(x) = f (b)g(b)
and subtracting
Rb
a
a
f (a)g(a),
a
f 0 (x)g(x) dx from both sides gives the integration by parts formula.