MATH 105 – 951
Written Assignment #2
Solutions
1. Use methods of integration to show that
Z
sec(x)dx = ln | tan(x) + sec(x)| + C.
Hint#1: First, multiply the numerator
Z
Z
1
sec(x)dx =
cos(x)
and denominator by cos(x):
Z
Z
cos(x)
cos(x)
cos(x)
dx.
·
dx =
dx =
cos(x)
cos2 (x)
1 − sin2 (x)
Hint#2: It might be helpful to note that
1 + sin(x)
1 + sin(x) 1 + sin(x)
(1 + sin(x))2
=
·
=
=
1 − sin(x)
1 − sin(x) 1 + sin(x)
1 − sin2 (x)
Solution: From the hint we have that
Z
Z
sec(x)dx =
1 + sin(x)
cos(x)
2
.
cos(x)
dx.
1 − sin2 (x)
Let u = sin(x). Then du = cos(x)dx and we get that
Z
Z
Z
1
1
sec(x)dx =
du =
du.
1 − u2
(1 − u)(1 + u)
To calculate this integral we need to use the method of partial fractions to find numbers A and B such
that
1
A
B
=
+
.
(1 − u)(1 + u)
(1 − u) (1 + u)
Multiplying both sides by (1 − u)(1 + u) gives
1 = A(1 + u) + B(1 − u).
Plugging in u = 1 we see that 1 = 2A, so A = 1/2. Plugging in u = −1 we see that 1 = 2B, so B = 1/2
as well. Therefore
Z
Z 1/2
1/2
1
1
sec(x)dx =
+
du = − ln |1 − u| + ln |1 + u| + C.
1−u 1+u
2
2
Substituting back into the variable x, and using the second hint and properties of logarithms we get
2 Z
1 1 + sin(x) 1 1 + sin(x) sec(x)dx = ln + C = ln + C = ln | sec(x) + tan(x)| + C.
2
1 − sin(x) 2 cos(x)
2. Due to the construction on SW Marine Drive, the 480 bus is often behind schedule and leaves UBC
late. The normal distribution is the probability distribution of the probability density function
(x−4)2
1
f (x) = √ e− 2 ,
2π
and can be used to predict the amount of time the bus is delayed (in min).
(a) Use a left Riemann sum with n = 10 to approximate the probability that the bus will be between
1 and 5 minutes behind schedule.
Solution: We want to use a Riemann sum to approximate the probability
Z 5
(x−4)2
1
√ e− 2 dx.
P(1 ≤ X ≤ 5) =
2π
1
Therefore a = 1, b = 5, and ∆x =
5−1
10
= 25 , and so the left Riemann sum approximation is
P(1 ≤ X ≤ 5) ≈
10
X
i=1
((1+(i−1) 2 )−4)2
1
2
5
2
√ e−
· .
5
2π
(b) If you want to take the 480 from UBC, how long do you expect to wait at the bus stop?
Hint#1: Note that
(x−4)2
(x−4)2
x
(x − 4) − (x−4)2
4
2
√ e− 2 = √
e
+ √ e− 2 .
2π
2π
2π
Z ∞
Hint#2: Since f (x) is a probability density function,
f (x)dx = 1.
−∞
Solution: The amount of time we’d expect to wait at the bus stop is
Z ∞
(x−4)2
x
√ e− 2 dx.
E(X) =
2π
−∞
From the first hint, this should be the same as
Z ∞
Z ∞
(x−4)2
(x − 4) − (x−4)2
4
2
√
√ e− 2 dx.
E(X) =
e
dx +
2π
2π
−∞
−∞
By the second hint, we can see that the second integral in the above equation is simply 4. That is,
Z ∞
Z ∞
Z ∞
(x−4)2
(x−4)2
4
1
√ e− 2 dx = 4
√ e− 2 dx = 4
f (x)dx = 4.
2π
2π
−∞
−∞
−∞
2
. Then du = (x − 4)dx, and we
To calculate the first integral, we make the substitution u = (x−4)
2
have that
Z ∞
Z x=∞
(x − 4) − (x−4)2
1
2
√
√ e−u du
e
dx =
2π
2π
−∞
x=−∞
x=∞
−1 −u =√ e 2π
x=−∞
2 ∞
(x−4)
−1
= √ e− 2 = 0.
2π
−∞
Therefore we’d expect to wait E(X) = 4 minutes at the bus loop.
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3. Use the precise definition of the definite integral to show that the function
(
0 if x =
6 3e
f (x) =
1 if x = 3e
is integrable on [0, 1].
Solution: We need to show that, for any choice of sample points x∗1 , x∗2 , . . . , x∗n
lim
n→∞
n
X
f (x∗i )∆x
i=1
exists and is equal
Pn (for every choice of sample points). First let’s try to approximate the (finite)
Riemann sum i=1 f (x∗i )∆x. Since f (x) ≥ 0, certainly the sum of the areas of the rectangles approximating the area beneath the graph of the function must also be greater or equal to 0. That
is,
n
X
0≤
f (x∗i )∆x
i=1
Now, since e/3 is an irrational number, the point x = e/3 lies in exactly one of the intervals
[0, n1 ], [ n1 , n2 ], . . . , [ n−1
n , 1]. On that one interval containing x = e/3, the maximum value of the function
f (x) is 1, so the maximum height of the rectangle approximating the area beneath the curve is 1. The
function is 0 at every point on every other interval. Therefore there is at most one rectangle with
non-zero height, so
n
X
1
0≤
f (x∗i )∆x ≤ 1 ·
n
i=1
(here 1/n is the width of each rectangle, so the maximum sum of the areas of the rectangles is 1 · 1/n).
Note that this inequality holds regardless of how we choose our sample points. Now, taking limits as
n → ∞ we get
n
X
1
0 = lim 0 ≤ lim
f (x∗i )∆x ≤ lim 1 · = 0.
n→∞
n→∞
n→∞
n
i=1
Therefore,
0 ≤ lim
n→∞
and so we must have that
lim
n→∞
n
X
f (x∗i )∆x ≤ 0,
i=1
n
X
f (x∗i )∆x = 0
i=1
Therefore we’ve shown that, regardless of how we choose our sample points, the limit exists and is
equal (to 0).
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