Math 2260 Written HW #1 Solutions
1. Consider the curves y = x1 , y =
1
,
x2
and x = 6.
(a) Sketch the region enclosed by these curves.
Answer:
2.0
1.5
1.0
0.5
1
2
3
4
5
6
(b) Find the exact value (i.e., not a decimal approximation) of the area of this region.
Answer: To find the area of the region, we just need to integrate the height of the
region from x = 1 to x = 6. Of course, the height of the region at x is given by the
difference of the values of the functions (being careful to subtract the bottom curve from
the top curve):
1
1
− 2.
x x
Therefore, the area of the region is
Z
Area =
6
1
1
−
dx
x x2
1
1 6
= ln(x) +
x 1
1
1
= ln(6) +
− ln(1) +
6
1
1
= ln(6) + − 1
6
5
= ln(6) − .
6
So we conclude that the area of the region is
7
6
+ ln(6) ≈ 2.95.
2. In general, if T is a random variate, then X obeys some probability distribution which says
where T is likely to be. The most famous example is the so-called normal or Gaussian
distribution. A random variate has a certain probability of taking on values in an interval
which is given by probability density function. Specifically, if the random variable T is chosen
1
from a probability distribution with probability density function f (x), then the probability
that T lies between the real numbers a and b (with a < b) is given by
Z
b
Pr[a ≤ T ≤ b] =
f (x) dx.
a
Pictorially, this means that the area under the graph of the probability density function
between x = a and x = b is precisely the probability that T is between a and b.
Now, suppose T is a random variate distributed according to the Kuramaswamy(2,7) distribution, which has the probability density function f (x) = 14x(1 − x2 )6 graphed below:
2.0
1.5
1.0
0.5
0.2
0.4
What is the probability that T is between
0.6
1
4
0.8
1.0
and 21 ?
Answer: By definition of the probability density function, the probability that T is between
1
1
4 and 2 is exactly
Z 1/2
Z 1/2
1
1
≤T ≤
=
f (x) dx =
14x(1 − x2 )6 dx
Pr
4
2
1/4
1/4
which is to say, the area of the following region under the curve:
2.0
1.5
1.0
0.5
0.2
0.4
0.6
2
0.8
1.0
To evaluate this integral, we can do a u-substitution where u = 1 − x2 . Then du = −2x dx,
so the above integral is equal to
Z
1/2
Z
2 6
2x(1 − x ) dx = −7
−7
1/4
3/4
u6 du
15/16
7 3/4
u
7 15/16
" 7 #
3 7
15
=−
−
4
16
= −7
=
157 37
− ,
167 47
which is fine to leave as is, or you could expand out to get the probability as the single fraction
1
135, 025, 567
1
≤T ≤
=
≈ 0.503.
Pr
4
2
268, 435, 456
3