EXERCISES FOR CHAPTER 2: Riemann Sums
1. (a) By finding a partition of the interval [0,1], using the function f (x) = e x and a
1
2
3
n1 1
n
n
n
suitable Riemann sum show that lim 1+ e + e + e + + e n = e 1. (b)
n n
1
2
3
Rederive this limit by explicitly finding the sum 1+ e n + e n + e n + + e
using L’Hôpital’s rule.
Solution
(a) A Riemann sum for this function on this interval is (choosing c k =
k= n
k= n
f (c )x = e
k
ck
k
k=1
k=1
k= n
= e
k=1
k1
n
n1
n
and
k 1
)
n
1
n
1
n
1
= 1+ e n + e n + e n + + e
n
1
2
3
n1
n
Thus
1
2
3
n 1
1
n
n
n
lim 1 + e + e + e + + e n = e x dx
n n 0
1
= e1
(b)
1
n
2
n
3
n
1+ e + e + e ++ e
n 1
n
=
n
n
(1 e )
1
n
summed a geometric series
1 e
1 e
1
=
we let = 1 e
n
1
2
3
n 1
1
(1 e)
lim 1 + e n + e n + e n + + e n = lim
n n 0 1 e
= (1 e)
= e1
1
1
by L'Hôpital's rule
1
1
2
3
n 2. Show that lim ln(1+ ) + ln(1+ ) + ln(1+ ) + + ln(1+ ) = 2ln2 1.
n n n
n
n
n Solution
14
K. A. Tsokos: Series and D.E.
k
Use the function f (x) = ln x on the interval [1, 2] with c k = 1+ . Then
n
k= n
k= n
1
f (c k )x k = lnc k n
k=1
k=1
k= n
k 1
= ln1+ n n
k=1
1
1
2
3
n = ln(1+ ) + ln(1+ ) + ln(1+ ) + + ln(1+ )
n
n
n
n
n Thus
1
2
3
n 1
lim ln(1+ ) + ln(1+ ) + ln(1+ ) + + ln(1+ ) =
n n n
n
n
n 2
ln x dx
1
2
= x ln x x 1
= (2ln2 2) (1)
= 2ln2 1
2
3
(n 1) 2
1 + sin
+ + sin
3. Show that lim sin + sin
= .
n n n
n
n
n Solution
Use the function f (x) = sin x with ck =
k 1
on the interval [0, ] . A Riemann sum
n
is then
k=n
n
k=n
f (ck )xk = sin(ck )
k =1
k =1
k 1 = sin n n
k =1
k=n
=
2
3
(n 1) + sin
+ + sin
sin + sin
n
n
n
n
n Thus
2
3
(n 1) 1 = sin x dx = cos x 0 = 2
lim sin + sin
+ sin
+ + sin
n n n
n
n
n 0
2
3
(n 1) 2
1 lim sin + sin
+ sin
+ + sin
=
n n n
n
n
n 4. Show that lim
n 1
n
k +1
(1
k
+ 2 k + 3k + + n k ) =
1
.
k +1
Solution
Use the function f (x) = x k with ck =
k
on the interval [0,1] . A Riemann sum is then
n
15
K. A. Tsokos: Series and D.E.
k=n
k=n
k =1
k =1
f (ck )xk = f (ck )
1
n
k
k=n
k 1
= n
k =1 n
1 1 k 2 k
3 k
n k
( ) + ( ) + ( ) + + ( ) n n
n
n
n
1
= k +1 1k + 2 k + + n k
n
=
(
)
Thus
lim
n
1
n k +1
(
1
)
1k + 2 k + + n k = x k dx
0
1
x k +1
=
k +1 0
=
5. Show that lim
n 1
1 1
+
+
2+ n
n 1+ n
1
k +1
1
1 ++
= 2( 2 1) .
3+ n
2n Solution
Use the function f (x) =
1
k
with ck = on the interval [0,1] . A Riemann sum is
n
1+ x
then
k=n
k=n
1
f (c )x = f (c ) n
k =1
k
k
k =1
k=n
=
k =1
k=n
k
1
1
k n
1+
n
n 1
n+k n
k =1
k=n
1
1
=
n+k n
k =1
1 1
1
1
1 =
+
+
++
n 1+ n
2+n
3+ n
2n =
Thus
16
K. A. Tsokos: Series and D.E.
1
1
1 1 1
lim
+
+
++
=
n
2+n
3+ n
2n n 1+ n
x +1
1/ 2
=
1
0
1
0
=2 22 1
= 2( 2 1)
6. Evaluate
x
2
0
dx
.
+ a2
Solution
1
x
dx
dx
=
lim
lim arctan
0 x 2 + a 2 0 x 2 + a 2 = a a
7. Evaluate
e
3x
=
0
1
lim arctan =
a a 2a
dx .
0
Solution
e
3x
0
dx = lim e
0
8. Evaluate
3x
xe
5x
e3x
dx = lim
3
0
e3 1 1
= lim
+ = .
3
3 3
dx .
0
Solution
xe
0
5 x
dx = lim xe
5 x
0
5 x e
25
= 0 lim
9. Evaluate
1
x
4
xe5 x
dx = lim
5
0
=
0
1
+ lim e5 x dx
5 0
1
25
dx .
1
Solution
1
1
1
lim 4 dx = lim
1 x 4 dx = 3x 3
x
1
10. Evaluate
1
ln x
dx .
x2
1 1
1
+ = .
3
3
3 3
= lim
1
1
dx
x +1
17
K. A. Tsokos: Series and D.E.
Solution
Integrate by parts to get
1 ln x
ln x
1
lim 2 dx = lim + lim 2 dx
1 x 2 dx = x ln x 1 1 x
x
1
ln 1
ln x 1
= lim + lim = lim + lim + 1
x 1 x 1 =1
1
ln = lim = 0 by L’ Hôpital’s rule.
since lim
1
1
11. Determine whether the improper integral dx exists.
x
2
Solution
1
1
lim dx = lim ln x 2 = lim ln ln 2 , integral does not exist.
2 x dx = x
2
12. Determine whether the improper integral sin 2 x dx exists.
0
Solution
1 cos 2x
x sin 2x lim sin x dx = lim dx = lim and so the
0 sin x dx = 2
4 0
2
0
0
integral does not exist.
2
2
13. Evaluate
x
2
x
dx .
1
Solution
x2 x
x
x
dx
=
lim
x2
dx
=
lim
1 2 x
ln 2
1
+
0
1
lim 2 x dx
ln 2 1
1
1
1
1 1
(lim ) (lim )
=
2
2 (ln 2) 2
2
ln 2 2
1
1
1
1
lim +
+
=
ln 2 2 ln 2 2 ln 2 2(ln 2)2
1
1
=
+
2 ln 2 2(ln 2)2
ln 2 + 1
=
2(ln 2)2
18
K. A. Tsokos: Series and D.E.
14. Consider the improper integral (n) = x n 1e x dx that defines a function called the
0
gamma function. Here n is a positive integer. (a) By using integration by parts show
that (n) = (n 1)(n 1). (b) Show that (1) = (2) = 1. (c) Hence or otherwise
establish that (n) = (n 1)! .
Solution
(a)
x
n 1 x
e
0
dx = lim x
n 1 x
e
0
(
= lim (
dx = lim x
n 1 e
)
n 1 x e
0
+ (n 1) lim x n 2 e x dx
) + (n 1)(n 1)
0
= (n 1)(n 1)
(
because lim n 1 e
)
n 1 = lim = 0 by repeated application of L’ Hôpital’s rule.
e
(b)
(
(2) = (2 1)(1) = e x dx = lim e x dx = lim e x
0
0
)
0
= 1.
(c) (n) = (n 1)(n 1) = (n 1)(n 2)(n 2) = = (n 1)!(1) = (n 1)!
15. (a) By careful diagrams establish that when f (n) is a decreasing function
1
n =1
1
f (n)dn < f (n) < f (1) + f (n)dn . Hence show that
2 n 3
< ne <
e
e n =1
(b) Consider F( ) = e n . Show that F( ) =
n =1
Hence find the exact value of
ne
n
1
dF( )
and
that
=
ne n .
e 1
d =1 n =1
. (c) How many terms must we keep in the sum so
n =1
6
that the error is less than 5 10 ?
Solution
(a) The discussion of the inequalities is given in the text. We must do the integral
x
x
x x
x =
lim
xe
dx
=
lim
xe
dx
=
lim
xe
+
e
dx
(x
+
1)e
= 2e1
1
1
1
1
1
and so 2e < ne
1
n =1
n
2 n 3
< e + 2e , i.e. < ne < as required.
e
e n =1
1
1
19
K. A. Tsokos: Series and D.E.
(b) F( ) = e
+e
2 e 1
+ =
= since we have a geometric series.
e 1
1 e
dF( ) d n
=
e = ne n , which gives our series when = 1 .
d n =1
d
n =1
d
e
1
=
d e 1 =1 (e 1)2
e
0.9206735942 . (This is
(e 1)2
n =1
=1
3
2
within the bounds of the inequality in (a), i.e. = 0.736 < 0.921 < 1.104 = .)
e
e
Hence
ne
n
=
=
(c) In keeping N terms the error made is at most RN = xe x dx = e N (N + 1) . We
N
N
6
demand that e (N + 1) < 5 10
and so N 15 . With 15 terms the sum is
approximately 0.9206706422 with an error therefore of 2.952 10 6 , i.e. less than
5 10 6 as expected.
16. By considering the function f (n) =
9 + 4 3
3 1
1
< 2
<
show
that
.
2
9
36
n +3
n =1 n + 3
Solution
The required integral is
dn
dn
1 lim 2
=
arctan
arctan
1 n 2 + 3 = n +3
3
3
1
Hence
3 1
< 2
<
9
n =1 n + 3
1 1 3
( )=
=
9
3
3 2 6
1 3
+
4
9
9 + 4 3
3 1
< 2
<
9
36
n =1 n + 3
1
17. Estimate the sum of the infinite series by keeping just the first 10 terms.
2
n =1 1 + n
Give an estimate of the error involved. (The exact sum is 1.07667.)
Solution
10
1
1 + n 2 0.981793 . The error is less than
n =1
R10 =
x
10
2
1
dx = arctan arctan10 0.0996687
+1
The actual error is 1.07667 0.981793 = 0.0948812 which is less than our estimate of
the error as expected.
20
K. A. Tsokos: Series and D.E.
n
n
n +1
1
k =1
1
18. Show that if f (x) is an increasing function f (1) + f (x)dx < f (k) <
Hence find out how many terms must be kept so that the sum
n2
f (x)dx .
n
equals
n =1
approximately 4.1 million.
Solution
The proof of the inequalities is a straightforward application of the rectangle trick. For
the function f (x) = x2 x , the integral to be performed is
x
x2 dx = x
2x
1
2x
2x
2x
x
2
dx
=
x
=
(x ln 2 1)
ln 2 ln 2 ln 2 (ln 2)2 (ln 2)2
and so we have the inequalities
2N
2
2+
(ln 2 1) < 4.1 10 6
( N ln 2 1) 2
2
(ln 2)
(ln 2)
2 N +1
2
4.1 10 <
(ln 2 1)
((N + 1)ln 2 1)) 2
(ln 2)2
(ln 2)
6
The first inequality states that
2 N ( N ln 2 1) < (4.1 10 6 2)(ln 2)2 + 2(ln 2 1) = 2.0 10 6 i.e. N 17 (by trial and
error with the GDC).
The second inequality is
2 N +1 ( (N + 1)ln 2 1)) > (4.1 10 6 )(ln 2)2 + 2(ln 2 1) = 2.0 10 6 i.e. N 17 .
We must thus choose 17 terms. The computer gives the sum of the first 17 terms as 4.19
million. (The sum of the first 16 is about 1.97 million and that of the first 18 about 8.9
million.)
19. Apply the inequalities of the previous problem on the increasing function
(n + 1)n +1
nn
n!
. Deduce that lim n = 0 and that
f (x) = ln x to show that n 1 < n! <
n
n
e
e
n
n
n! 1
lim
= .
n n
e
Solution
n
n
1
k =1
From f (1) + f (x)dx < f (k) <
n +1
1
f (x)dx we have that
21
K. A. Tsokos: Series and D.E.
n +1
ln1 + (x ln x x) 1 < ln1 + ln 2 + + ln n < (x ln x x) 1
n
n ln n n + 1 < ln(1 2 3 n) < (n + 1)ln(n + 1) n
ln n n (n 1) < ln(n!) < ln(n + 1)(n +1) n
ln(n n e(n 1) ) < ln(n!) < ln((n + 1)(n +1) e n )
Hence
nn
e(n 1)
(n + 1)(n +1)
< n! <
en
as required.
Thus,
n! (n + 1)(n +1)
<
nn
n n en
n! (n + 1)(n +1)
< n <
n
n
n n +1en
1
<
e(n 1)
1
(n 1)
e
1
e
and
so
n + 1
lim n n theorem,
lim
n
n +1
1
e
(n 1)
<
(n 1)
n! n + 1 <
nn n n!
n + 1
lim n lim n n
n n n +1
n +1
n
.
en
n
en
Because
lim
n
n!
=0.
nn
n +1
n! n + 1 n
< n <
we have that
n en
n
lim
Further, taking the nth root of
1
e
(n 1)
1
e
and taking limits lim
n
lim
1
e
(n 1)/n
= lim
n
1
e
(n 1)/n
(n 1)/n
n! n + 1 <
<
n n
n
(n +1)/n n
n!
n + 1
lim
lim n n
n n n
n
e
(n +1)/n n
n
. Because
e
1
1
=
1/n
e
ee
and
n + 1
lim n n e
=0
and
n
n!
= e 0 = 0 we have that 0 lim n 0 and so by the squeeze
n
n n
e
n
n
1
(n 1)
(n +1)/n n
1
n 1
= lim 1 + e n n
e
we have by the squeeze theorem that
1+
1
n
lim n n =
n
1
1
11 =
e
e
22
K. A. Tsokos: Series and D.E.
n
lim
n
n! 1
= ,
e
n
as required.
20. The value of
1
k
k =1
2
is to be approximated by keeping just the first n terms. What
should n be so that the error is less than 5 10 5 , i.e. so that we have agreement to 4
2
decimal places? The exact answer for the infinite sum is
.
6
Solution
1
1
1
1
The error involved is 2 < 2 dx = = . So we need
x
xn n
k = n +1 k
n
1
1
< 5 10 5 n >
= 2 10 4
5
5 10
n
The sum of the first 2 10 4 + 1 terms is 1.64488 1.6449 to 4 decimal places. The
exact sum is 1.64493 1.6449 giving agreement to 4 decimal places.
b
21. If the functions f (x) and g(x) satisfy 0 f (x) g(x) then
a
b
f (x) dx g(x) dx .
a
Use this fact to determine whether the improper integrals (a)
x
1
3
x
sin x
dx and (c) dx exist.
3
+ x + x +1
x
1
1
Solution
(a) We may compare,
dx
1
dx
dx
dx
1
lim 4 = lim 3 =
1 x 4 + 1 < 1 x 4 and 1 x 4 = 3x
3
x
1
1
so the original integral exists.
(b) We may compare,
3
x
dx
x3
dx
1
lim 2 = lim
=1
1 x 5 + x 3 + x + 1 dx 1 x 5 dx = 1 x 2 = x
x
1
1
so the original integral exists.
(c) By integration by parts,
x
5
cos x
cos x
sin x
1 x dx = x 1 1 x 2 dx
cos x
cos x
= lim
2 dx
x 1 1 x
dx
+1
4
(b)
23
cos1
The
first
term
is
and
cos x
1
cos x
1 x 2 dx 1 x 2 dx 1 x 2 dx = 1 . Hence
exists and equals .)
2
K. A. Tsokos: Series and D.E.
the
integral
is
finite
since
sin x
sin x
1 x dx exists. (In fact also 0 x dx