Solution of Term Exam 3
Problem 1.
 Compute
.
 Compute
.
 For , compute
.
Solution. (Graded by Shay Fuchs)
 To use the second fundamental theorem of calculus we are looking for a
function for which
. The most obvious guess is
, but this is off by a factor of , for
. So a good answer would be
. Now
 Likewise choose
to get
, and so using
the second fundamental theorem of calculus,
 Let
and let . Using the
first fundamental theorem of calculus,
. So using the chain
rule,
Problem 2.
 Perhaps using L'Hôpital's law, compute
and
.
 Use these results to give educated guesses for the values of
and (no calculators, please).
Solution. (Graded by Shay Fuchs)
 is differentiable at 0 and . So
(L'Hôpital's law also works and gives the same result).
The second limit is of the form so we can use
L'Hôpital:
 is close to 0, so
. Multiplying both
sides by we get
.
Likewise,
, so
,
so
.
Problem 3.
 State the ``one partition for every '' criterion of the
integrability of a bounded function defined on an interval
.
 Let be an increasing function on and let be the
partition defined by , for
. Write simple formulas
for and for .
 Under the same conditions, write a very simple formula for
.
 Prove that an increasing function on is integrable.
Solution. (Graded by Derek Krepski)
 A bounded function defined on an interval is integrable
iff for every
there is a partition of for which
.
 As is increasing,
and
. Thus
Likewise,
.

using telescopic summation this is
 Since is increasing, is bounded (with upper bound and
lower bound ). So using the criterion of part 1, to show that is
integrable it is enough to show that for every
there is a
partition of for which
. Indeed, let
be given. Choose so big so that
, and then the partition of before
satisfies
, as required.
Problem 4.
 Show that the function
is monotone on the interval
.
 Deduce that for every
the equation has a
unique solution in the range
.
 For
, let be the unique in the range
for which . Write a formula for and
simplify it as much as you can. Your end result may still contain in
it, but not , or .
Solution. (Graded by Brian Pigott)

. On we know that , so
. So is increasing on .
 By the theorem about the existence of inverses of monotone
functions, has an inverse on and it is defined on
. This precisely means that for
the
equation (which defines ) has a unique solution
with in the range
.
 By the theorem about the derivative of an inverse function,
The results. 67 students took the exam; the average
grade was 77.7 and the standard deviation was about 22.
Dror BarNatan
20050207