Math 1102: Calculus I (Math/Sci majors)
MWF 3pm, Fulton Hall 230
Homework 10 Solutions
Please write neatly, and show all work. Caution: An answer with no
work is wrong!
Problems from the book: Chapter IX p. 307 (1,2,7), p. 311(2); Chapter
X p. 317 (1,2,3,4,10), p. 325 (1,2,3,4,5,10)
Chapter IX, p. 307: Write out the lower and upper sums for the following functions and intervals. Use a partition such that the length of
each small interval is (a) 1/2, (b) 1/3, (c) 1/4, (d) 1/n.
1. f (x) = x2 in the interval [1, 2].
3
Solution (a). The desired partition is Pa = 1, , 2 . We have
2
1 9
+ ·
2 4
9 1
U(Pa , f ) = · + 4 ·
4 2
L(Pa , f ) = 1 ·
1
13
= ,
2
8
1
25
= .
2
8
and
4 5
Solution (b). The desired partition is Pb = 1, , , 2 . We have
3 3
1 16 1 25
+
· +
·
3
9 3
9
16 1 25 1
U(Pb , f ) =
· +
· +4·
9 3
9 3
L(Pb , f ) = 1 ·
1
50
= ,
3
27
1
77
= .
3
27
Solution (c). The desired partition is Pc =
1 25 1 9 1 49
+
· + · +
·
4 16 4 4 4 16
25 1 9 1 49 1
U(Pc , f ) =
· + · +
· +4·
16 4 4 4 16 4
L(Pc , f ) = 1 ·
and
5 3 7
1, , , , 2 . We have
4 2 4
1
126
63
=
= ,
4
64
32
1
174
87
=
= .
4
64
32
and
n+1 n+2
2n − 1
Solution (d). The desired partition is Pd = 1,
,
,...,
,2 .
n
n
n
We have
2
n−1 X
n+i−1
1
L(Pd , f ) =
· , and
n
n
i=1
2
n−1 X
1
n+i
· .
U(Pd , f ) =
n
n
i=1
2. f (x) = 1/x in the interval [1, 3].
3
5
Solution (a). The desired partition is Pa = 1, , 2, , 3 . We have
2
2
19
1 1 1 1
L(Pa , f ) = + + + = , and
3 4 5 6
20
1 1 1 1
77
U(Pa , f ) = + + + = .
2 3 4 5
60
4 5
7 8
Solution (b). The desired partition is Pb = 1, , , 2, , , 3 . We
3 3
3 3
have
2509
1 1 1 1 1 1
, and
L(Pb , f ) = + + + + + =
4 5 6 7 8 9
2520
1 1 1 1 1 1 1
341
U(Pb , f ) = + + + + + + =
.
3 4 5 6 7 8 9
280
9 5 11
5 3 7
Solution (c). The desired partition is Pc = 1, , , , 2, , , , 3 .
4 2 4
4 2 4
We have
1 1 1 1 1
1
1
1
28271
L(Pc , f ) = + + + + +
+
+
=
, and
5 6 7 8 9 10 11 12
27720
1 1 1 1 1 1
1
1
32891
U(Pc , f ) = + + + + + +
+
=
.
4 5 6 7 8 9 10 11
27720
n+1 n+2
3n − 1
Solution (d). The desired partition is Pd = 1,
,
,...,
,3 .
n
n
n
We have
3n 3n
X
X
n
1
1
· =
, and
L(Pd , f ) =
i
n i=n+1 i
i=n+1
U(Pd , f ) =
3n−1
X
i=n
3n−1
X1
n 1
· =
.
i
n
i
i=n
7. Let f (x) = log x. Let n be a positive integer. Write out the upper
and lower sums, using the partition of the interval between 1 and n
consisting of the integers from 1 to n, i.e. the partition (1, 2, . . . , n).
Solution 7. Let Pn indicate the partition {1 ≤ 2 ≤ . . . ≤ n}. Since
log x is an increasing function (as (log x)0 = 1/x > 0 for x > 0, and by
Theorem A), the minimum of log x on the interval [j, j + 1] is x = j,
and the maximum is x = j + 1. Thus the lower and upper sums are
given by:
L(Pn , f ) =
n−1
X
f (j) · 1 =
n−1
X
j=1
log j
j=1
= log(n − 1)!, and
U(Pn , f ) =
n−1
X
f (j + 1) · 1 =
j=1
n−1
X
log(j + 1)
j=1
= log n!.
p. 311, 2. Let f be continuos on the interval [a, b]. Prove that there
exists some number c in the interval such that
Z b
f (c)(b − a) =
f (t)dt.
a
Solution 2. Let F (x) be defined by
Z x
F (x) =
f (t)dt,
a
Rb
so that F (a) = 0, and F (b) = a f (t)dt. By the Mean Value Theorem,
there is a c between a and b so that
F (b) − F (a)
F 0 (c) =
.
b−a
By the Fundamental Theorem of Calculus, F 0 (x) = f (x), for all x in
(a, b). Thus the above equation implies that
Rb
f (t)dt − 0
.
f (c) = a
b−a
Multiplying by (b − a), we find the desired result:
Z b
f (t)dt.
f (c)(b − a) =
a
p. 317, For problems 1-4, find the desired integral.
1.
R2
1
x5 dx.
Solution 1. Since (x6 )0 = 6x5 , the function x6 /6 is an anti-derivative
for 6x5 . Thus by the Fundamental Theorem of Calculus,
Z
2
1
2.
R1
−1
x6
x dx =
6
5
2
=
1
26 16
63
21
−
=
= .
6
6
6
2
x1/3 dx.
Solution 2. Since (x4/3 )0 = 4x1/3 /3, the function 3x4/3 /4 is an antiderivative for x1/3 . Thus by the Fundamental Theorem of Calculus,
Z
1
x
1/3
−1
3.
Rπ
−π
3x4/3
dx =
4
1
=
−1
3(−1)4/3 3 · 14/3
−
= 0.
4
4
sin xdx.
Solution 3. Since (cos x)0 = − sin x, the function − cos x is an antiderivative for sin x. Thus by the Fundamental Theorem of Calculus,
Z π
sin xdx = − cos x]π−π = − cos(−π) − cos π = 0.
−π
4.
Rπ
0
cos xdx.
Solution 4. Since (sin x)0 = cos x, the function sin x is an antiderivative for cos x. Thus by the Fundamental Theorem of Calculus,
Z π
cos xdx = sin x]π0 = sin π − sin 0 = 0.
0
10. For this exercise verify first that if we let F (x) = x log x − x, then
F 0 (x) = log x.
(a) Evaluate the integral
Z
n
log xdx.
1
(b) Compare this integral with the upper and lower sum associated
with the partition P = {1, 2, . . . , n} of the interval [1, n].
(c) Prove the following inequality:
(n − 1)! ≤ nn e−n e ≤ n!
Solution 10(a). Note that, using the product rule for derivatives,
1
F 0 (x) = 1 · log x + x · − 1 = log x + 1 − 1 = log x.
x
Thus F (x) is an anti-derivative of log x, and we may use the Fundamental Theorem of Calculus:
Z n
log xdx = x log x − x]n1
1
= (n log n − n) − (1 log 1 − 1) = n log n − n + 1.
Solution 10(b). Let Pn be the partition {1 ≤ 2 ≤ . . . ≤ n} of the
interval [1, n]. We’ve computed the upper and lower sums U(Pn , f ) and
L(Pn , f ) in problem 7 on p. 307, where we found:
U(Pn , f ) = log n!
L(Pn , f ) = log(n − 1)!
By the definition of the Riemann integral, we have
log(n − 1)! ≤ n log n − n + 1 ≤ log n!.
Solution 10(c). Since ex is increasing (as (ex )0 = ex > 0, and by
Theorem A), the inequality
A≤B≤C
implies that
eA ≤ eB ≤ eC .
Thus the inequality at the end of 10(b) implies that
elog(n−1)! ≤ en log n−n+1 ≤ elog n! .
Simplifying, we find that
(n − 1)! ≤ nn e−n e ≤ n!
as desired.
p. 325, In problems 1-5, find the given integral.
1.
R
4x3 dx
Solution 1. Since (x4 )0 = 4x3 , we have
Z
4x3 dx = x4 + C.
2.
R
(3x4 − x5 )dx
Solution 2. Since (x5 )0 = 5x4 and (x6 )0 = 6x5 , we have
0
3 5 1 6
x − x
= 3x4 − x5 .
5
6
Thus
Z
3
1
3x4 − x5 dx = x5 − x6 + C.
5
6
R
3. (2 sin x + 3 cos x)dx
Solution 3. Since (cos x)0 = − sin x and (sin x)0 = cos x, we have
(−2 cos x + 3 sin x)0 = 2 sin x + 3 cos x.
Thus
Z
(2 sin x + 3 cos x)dx = −2 cos x + 3 sin x + C.
4.
R
(3x2/3 + 5 cos x)dx
Solution 4. Since (x5/3 )0 = 5x2/3 /3 and (sin x)0 = cos x, we have
0
9 5/3
x + 5 sin x = 3x2/3 + 5 cos x.
5
Thus
Z
9
(3x2/3 + 5 cos x)dx = x5/3 + 5 sin x + C.
5
R
1
x
5.
5e + x dx
Solution 5. Since (ex )0 = ex and (log x)0 = x1 , we have
1
(5ex + log x)0 = 5ex + .
x
Thus
Z 1
x
5e +
dx = 5ex + log x + C.
x
10. Find the area between the curves y = x and y = x2 from 0 to their
first point of intersection for x > 0.
Solution 10. The curves y = x and y = x2 intersect when x = x2 .
Collecting terms on one side and factoring, we find
x(1 − x) = 0,
so that x = 0 or x = 1. Thus the only intersection point for x > 0
2
occurs when x = 1. For x between 0 and
R 1 1, we 2have x < x, so that
the area between the curves is given by 0 (x − x )dx. We calculate
1
Z 1
x2 x3
1 1
1
2
(x − x )dx =
−
= − = .
2
3 0 2 3
6
0
Problem A. Let f be a function on [a, b], let P be a partition of the
interval [a, b] given by
P = {a = x0 ≤ x1 ≤ . . . ≤ xn = b},
0
and let P be a partition obtained by adding one point x̄ to P. Show
that the upper sums U(P, f ) and U(P 0 , f ) satisfy
U(P 0 , f ) ≤ U(P, f ).
(Hint: This is the other half of Theorem 4.1, as discussed in class).
Solution A. Suppose x̄ is in the piece [xj , xj+1 ] of P. Let sj be the
maximum of f on [xj , xj+1 ], let u be the maximum of f on [xj , x̄], and
let v be the maximum of f on [x̄, xj+1 ].
The only terms that differ between U(P, f ) and U(P 0 , f ) are
f (sj )(xj+1 − xj )
in U(P, f ) and
f (u)(xj+1 − x̄) + f (v)(x̄ − xj )
in U(P , f ). Since each of the intervals [xj , x̄] and [x̄, xj+1 ] is inside
[xj , xj+1 ], we have
0
f (sj ) ≥ f (u),
f (sj ) ≥ f (v).
and
That implies that
f (sj )(xj+1 − xj ) = f (sj )(xj+1 − x̄ + x̄ − xj )
= f (sj )(xj+1 − x̄) + f (sj )(x̄ − xj )
≥ f (u)(xj+1 − x̄) + f (v)(x̄ − xj ),
so that
U(P, f ) ≥ U(P 0 , f )
as desired.
Problem B. The goal of this problem is to compute
Z b
x dx
a
directly, using Riemann’s definition of the definite integral. Let f (x) =
x for the rest of this problem.
(1) Suppose Pn = {a = x0 ≤ x1 ≤ . . . ≤ xn = b} is an evenlyspaced partition of the interval [a, b]. Find the width of this
partition, and an expression for xj . (Compare with p. 297).
(2) Compute explicitly the lower sum L(Pn , f ).
(3) Compute the limit
lim L(Pn , f ),
n→∞
and, using Theorem 4.4, deduce the value of
Z b
x dx.
a
(4) When b ≥ a ≥ 0, interpret the region under the graph of f
between a and b as a trapezoid, and compute the area of this
region directly.
Solution B(1). If the width of the partition Pn is w, then it is evident
that
x1
x2
x3
xn
= x0 + w,
= x1 + w = x0 + 2w,
= x2 + w = x0 + 3w, . . . ,
= xn−1 + w = x0 + nw.
Since xn = b and x0 = a, we must have
a + nw = b,
so that
b−a
, and
n
xj = a + jw.
w=
Solution B(2). Since f (x) = x is increasing, the minimum of f on
[xj , xj+1 ] is xj . Thus L(Pn , f ) is given by
L(Pn , f ) =
n−1
X
f (xj )(xj+1 − xj )
i=0
=
n−1
X
xj (xj+1 − xj )
i=0
n−1 X
b−a b−a
=
a+j
n
n
i=0
2
n−1
X
b−a
b−a
=
a
+j
n
n
i=0
!
2 X
n−1
b−a
b−a
=a
·n+
j
n
n
i=0
!
2 X
n−1
b−a
j .
= a(b − a) +
n
i=0
Since
n−1
X
i=0
j=
(n − 1)n
,
2
we have
b−a
n
2 (n − 1)n
L(Pn , f ) = a(b − a) +
2
2
n −n
2
2
.
= ab − a + (b − a)
2n2
Solution B(3). We have
n(n − 1)
lim L(Pn , f ) = lim ab − a + (b − a)
n→∞
n→∞
2n2
1
1
2
2
= ab − a + (b − a) lim
−
n→∞
2 2n
1
= ab − a2 + (b − a)2 ·
2
1
= ab − a2 + (b2 − 2ab + a2 ) ·
2
1 2
= (b − a2 ).
2
b−a
Note that the width
goes to 0 as n goes to ∞. Thus, by Theorem
n
4.4 we have
Z b
1
x dx = lim L(Pn , f ) = (b2 − a2 ).
n→∞
2
a
2
2
Solution B(4). If b ≥ a ≥ 0, the region under the graph of f is a
trapezoid bounded by the vertical lines x = a and x = b. The base of
the trapezoid is b − a, and the two vertical sides have heights f (a) = a
and f (b) = b. The area of this trapezoid is given by
1
a+b
= (b2 − a2 ).
(b − a) ·
2
2