Calculus II
Class Notes, 1/25/99
We show that the integrals of sin(`theta) / [ 1 +
cos^5(`theta) ] and 1 / [ x + x (ln x) ^ 5 ] in fact are variations
of the same integral.
- For the first integral we use the obvious substitution u =
cos(`theta), as shown.
- The substitution required for the second integral is less
obvious. However, as in any case where the substitution is not obvious, we opt to
attack the simplest composite function, which is in this case (
ln(x) ^ 5).
- Letting u be the innermost function of the composite, we set u
= ln x, obtaining du = 1 / x dx.
- We obtain the expression in the third line, and note that we can factor
1/x out of the fraction.
- In the fourth line we note that we have du = 1/x dx within
the integral and we immediately obtain the integral in the last line.
We see that the integral obtained in both cases is the same.
To perform the integral in the figure below, we begin by completing the square on
the denominator:
- We see that a perfect square containing x^2 + 6x must
be a square of (x + 3), since squaring this term gives
us the required x^2 + 6x in the square x^2 + 6x + 9.
- We therefore see that x^2 + 6x + 10 = (x^2 + 6x + 9) + 1 = (x+3)^2
+ 1.
We rewrite the integral as shown below and use the substitution
u = x + 3.
- We obtain the integral shown in the third line and proceed as usual.
- We note that the arcTangent function might have something to do with
the integral.
If a whittler of heirloom washers is paid on a varying
scale for washers whittled in a year, receiving
- $1.5 apiece for
all washers from the 14,000th to the 30,000th, but
- paid only $.10 for
the first, and up to the 14,000th paid according to the linear
function in the first line below, and
- paid on a decreasing
exponential scale for washers over the 30,000th, as
indicated in the third line below,
how much does the whittler make for whittling some number wf > 30,000
washers?
We are given a function specifying pay rate per unit
produced.
- Since the product of
pay rate per unit and number of units gives us the amount
of pay, to get the total pay we must integrate the
pay rate function with respect to the number w of units
produced.
A sketch of the pay rate function is shown below,
with wf indicated at some point above 30,000.
- The total pay is represented by the area under the
curve.
- The integral is evaluated by integrating each function over the
appropriate domain. We integrate
- the linear function from 0 to 14K (14K meaning 14,000), then
- the constant function fro 14K to 30K, then t
- he exponential from 30K to wf.
- The first two integrals are very straightforward.
- The third integral requires the substitution u = 2 - w / 15K, and is easily evaluated.
Given the flow rate function below, we analyze the flow amount
from t = 0 to t = 5 as follows:
- We first determine the flow amount from some clock time t to
clock time t + `dt, where `dt is a short time interval.
- Over a short time interval the flow rate will not change appreciably
and the approximate flow amount will be rate * time interval =
r(t) `dt.
- We then write a Riemann some of the flow amounts, using
a uniform time interval `dt.
- We think of dividing the
interval from 0 to 5 into n equal increments, each of
length `dt = 5 / n.
- We take ti to be
the right-hand endpoint (5 / n) * i of interval
# i (the end of the first interval will be at t = 5 / n; the end of the second at
t = (5 / n) * 2; . . . ; the end of the ith at t = (5 / n) * i).
- The rate at t = ti is r(ti).
- We thus obtain the sum of the terms r(ti) `dt, as
shown in the last line below.
The figure below shows the subdivision of the interval into
equal increments, and indicates the area of increment
# i.
- The indicated area of the single increment is the local
contribution, in the vicinity of t = ti, to the total.
- The Riemann sum totals all the local contributions.
- In the limit as n approaches infinity, so that `dt -> 0,
the Riemann sum approaches the integral as its limit.
- The integral is thought of as the global, as opposed
to local, total.
The integral is evaluated below.
- We use the substitution u = e^( t `sqrt(b) ) + e^(-t `sqrt(b)), hoping
that somehow having the denominator equal to u will
magically lead us to something we can integrate.
- Miraculously, but somehow not surprisingly, we see that du contains the
numerator.
- Proceeding in the usual manner, we obtain the integral shown, with the
result that the total flow is `sqrt(b) * ln(u), evaluated
between t = 0 and t = 5.
- The remaining details are straightforward.
Below we integrate z `sqrt(7 + z), using integration by parts (a
u substitution would work, but we want to illustrate the process of deciding how to break
down an integral for integration by parts).
- We might try u = `sqrt(7 + z), v' = z, but as shown this leads us to an
expression for we have to integrate z^2 / `sqrt(7 + z).
- Comparing this with the original integral, it is
pretty clear the we have not made any progress with this
substitution.
- So instead we try u = z, v' = `sqrt(7 + z).
- This time, as shown, we obtain an integral
of u' v that we can evaluate.
- The rest of a problem is easily
completed.
The details of evaluating the integral are shown below.
