Solution of Term Exam 3
Problem 1. Suppose that is nondecreasing on .
Notice that is automatically bounded on , because
for any in .
- If
is a partition of , write formulas
for and in as simple terms as possible.
- Suppose that
for each . Show that
.
- Prove that is integrable.
Solution. Notice that as is nondecreasing,
and
. Hence,
- The upper and lower sums are given by
- If
then
and likewise
. Hence
. By telescopic
summation, the last sum is
.
- Setting
we get partitions with
. This can be made arbitrariry
small, and hence
can be made
arbitrarily small, hence is integrable.
Problem 2. In each of the following, is a
continuous function on .
- Show that
- Characterize the functions that have the property that
Solution.
- If is a primitive of , i.e., , then
. Hence using the
second fundamental theorem of calculus,
- Differentiate both sides of the equality
using the first fundamental theorem
of calculus and get that
and thus that for all .
Problem 3.
- Prove that if two functions and both satisfy
the differential equation and if they have the same value
and the same first derivative at 0, then they are equal.
- Use the above to show that
for all
. (Do not use the formula for the of a sum!)
Solution.
- Set . As
, it follows that
and clearly
. Now set
and
calculate
. Hence is a constant. But
hence is the constant 0. But as
the two terms composing a non-negative, this forces each of
them to be 0, and in particular and thus
.
- Set
and
. Then
,
,
and
. Hence and both
satisfy the differential equation . In addition,
and
. Hence the requirements for
the theorem of the previous part of this problem are met, and hence its
conslusion is satisfied. So or
.
Problem 4.
- Compute
- Use your result to estimate the difference between and
. Warning: a 10 digit answer obtained with your calculator may
contribute negatively to your grade. You shouldn't use any
calculating device and your derivation of the answer should be simple
enough that it be clear that you didn't need any machine help.
Solution.
- Using L'Hôpital's rule twice, we get
- From the above we get that
is
approximately if is small. Hoping that is small
enough, we find that
, meaning
that
.
Problem 5. Evaluate the following integrals in terms of
elementary functions:
-
-
(cancelled)
-
-
Solution.
- We integrate by parts twice:
- Not required, though still,
.
- Take and then
and hence
and
so
- Take and then and so
The results. 83 students took the exam; the average grade
is 65.31, the median is 65 and the standard deviation is 25.17.
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Dror Bar-Natan
2003-02-17