MATH 138 Assignment
Assignment
MATH 138 Assignment
Question 1. [10 marks] Consider the functions f1, f2, . . . with domain [0,∞) given by:
fn(x) =
1
n
if x ∈ [0, n)
− 1
n
x + 1 + 1
n
if x ∈ [n, n + 1)
0 if x ∈ [n + 1,∞)
Prove the following:
(a) The sequence {fn}∞n=1 converges to the function f : [0,∞)→ R given by f(x) = 0 for all x ∈ [0,∞).
i.e. for all x ∈ [0,∞), we have lim
n→∞
fn(x) = f(x).
(b) Each integral
∫ ∞
0
fn(x) dx is convergent, and the integral
∫ ∞
0
f(x) dx is also convergent.
(c) lim
n→∞
(∫ ∞
0
fn(x) dx
)
6=
∫ ∞
0
f(x) dx. That is, the limit of the integrals is NOT the same as the
integral of the limit.
Fun fact: the functions f1, f2, . . . given above actually converge to f in a much stronger sense called uni-
form convergence (which you will learn in the third-year real analysis courses). And uniform convergence
on bounded intervals [a, b] does imply convergence of the integrals on that interval. But as we saw above,
this fails when we work with unbounded intervals.
MC Questions
2. Suppose there is an m such that the line y = mx divides the area under the curve y = x(1− x) and
above the x-axis into two regions of equal areas. In what interval does m lie:
(a) m ∈ [0, 1
5
]
(b) m ∈ (1
5
, 2
5
]
(c) m ∈ (2
5
, 3
5
]
(d) m ∈ (3
5
, 4
5
]
(e) m ∈ (4
5
, 1]
3. Consider the integral I =
∫ 1
0
1
x + x2
dx.
Which of the following can be concluded by the Comparison Theorem?
(a) I diverges by comparison with
∫ 1
0
1
x
dx
(b) I diverges by comparison with
∫ 1
0
1√
x
dx
(c) I diverges by comparison with
∫ 1
0
1
2x
dx
(d) I diverges by comparison with
∫ 1
0
1
1 + x2
dx
(e) none of the above
4. The integral
∫ ∞
1
x2/3 + x5/3
x2
dx:
(a) diverges by comparison with
∫ ∞
1
1
x4/3
dx
(b) converges by comparison with
∫ ∞
1
1
x4/3
dx
(c) diverges by comparison with
∫ ∞
1
1
3
√
x
dx
(d) converges by comparison with
∫ ∞
1
1
3
√
x
dx
(e) none of the above
5. Given the following descriptions of a region R along with expressions involving integrals over x and
y respectively, determine which pairs of integrals represent the area of the same region.
(a) R is the region lying below y = −x3 and above y = −4x for x ∈ [0, 2].∫ 2
0
−x3 + 4x dx,
∫ 8
0
−y
4
+ 3
√
y dy
(b) R is the region lying below y = 4x− x2 and above y = 2x for x ∈ [0, 2].∫ 2
0
(4x− x2)− 2x dx,
∫ 4
0
y
2
− (2−
√
4− y) dy
(c) R is the region lying above the x-axis, below y = x on x ∈ [0, 1], and below y = √2− x on
[1, 2]. ∫ 1
0
x dx +
∫ 2
1
√
2− x dx,
∫ 1
0
2− y2 − y dy
(d) R is the region in the first quadrant bounded by the x-axis, y =
√
x and y =
1
2
x− 3
2
.
∫ 3
0
√
x dx +
∫ 9
3
√
x−
(
1
2
x− 3
2
)
dx,
∫ 3
0
2y + 3− y2 dy
(e) none of the above
6. Which of the following are true:
(a) If
∫ ∞
1
xp dx converges then
∫ 1
0
xp dx diverges.
(b) If
∫ ∞
a
f(x) dx diverges then lim
t→∞
∫ t
a
f(x) dx = ±∞
(c) If a < b and
∫∞
b
f(x) dx converges then
∫∞
a
f(x) dx converges
(d) If
∫ ∞
1
xp dx diverges then
∫ 1
0
xp dx converges.
(e) none of the above
7. Which of the following are true:
(a) If lim
x→∞
xf(x) = A, where A > 0 is a constant, then
∫ ∞
a
f(x) dx diverges.
(b) If lim
x→∞
f(x) exists and
∫ ∞
0
f(x) dx converges then lim
x→∞
f(x) = 0
(c) If lim
x→∞
xf(x) = 0, then there is an a ∈ R such that
∫ ∞
a
f(x) dx converges.
(d)
∫ ∞
−∞
x3 dx = 0
(e) none of the above
8. Assume n ∈ N (i.e. n = 1, 2, 3 · · · )
Let A(n) represent the area in the first quadrant bound by y = x and y = xn. Which of the following
are true:
(a) A(n) = A( 1
n
)
(b) There is an n such that A(n) = 1
10
(c) lim
n→∞
A(n) = 1
(d) There is an n such that A(n) = 5
11
(e) none of the above
9. Let A represent the area of the “triangular” region bounded by the functions y = ln(x), y = ln(1−x)
and y = b for b ≥ 0. Which of the following are true:
(a) A = 2eb − b− ln(2e).
(b) It is possible to calculate A by evaluating a single integral.
(c) To find A we must compute an improper integral
(d) A = 2
∫ b
− 1
2
e−t dt
(e) none of the above