MATH-035-04 - MTWF 1:15--2:05 Reiss 264 MATH-035-05 - MWF 2:15--3:05 Reiss 264; T 2:15--3:05 Reiss 283
See my index page for info on office hours. Also check it out occasionally for updates and announcements.
You can reach me by phone (or voice-mail) at 7-2703 or by e-mail kainen at georgetown.edu or drop by St. Mary's D315 during my office hours. See my index page.
Information is available on syllabus and course mechanics . I suggest you review this material which also explains homework, quiz policies, and the exam schedule.
Math assistance center, St. Mary's 3rd floor, Sun. - Thurs. evenings - open 6 to 9 pm
**** Last updated ****
Sept. 3, 2007
Announcements:
Third Midterm is postponed.
Instead of Tues. April 24th in the evening,
the exam will be in class on Monday April 30.
Our final exam is scheduled for May 10 at 9am.
There will be two review sessions during the break:
Monday May 7 9:15 pm room TBD - latter third of course,
we'll go over the midterm from the 30th of April as well
Tuesday May 8 9:15 pm room TBD - review of the earlier
material: from Chapters 5,6,7.
An opportunity to show extra effort!
3rd floor lab (East end of bldg): Sat. 1 to 5, Sun. 1 to 3
Both 21-22 and 28-29 April. Also reviews in Reiss 284 on
Sunday at 5pm both weekends.
This weekend and next weekend, come to St. Mary's
computer lab (on the 1st floor in the center) at
1:00 pm on Sat. I will lead us up to the 3rd floor
(otherwise locked) and open the laboratory there.
If you come late, go to the stairway in St. Mary's
down the hallway to the East from the central area
and the door on the 3rd floor should be propped
open. Anytime between 1:00 and 5:00 will be ok.
Same deal on Sunday but only from 1:00 to 3:00.
On Sunday at 5pm, I'll give a review in Reiss.
For next weekend, the 28th and 29th, we'll follow
the same schedule. As the midterm is on Monday
April 30, the review on Sunday the 29th will be
extended until 6:30.
A list of topics for the Final will appear later.
In class Monday, we went over 1,3,7, and 13. Later,
I'll go over 17,33,39,41,49. In reading section 8.7,
you can skip the part about _dividing_ two power
series (but you'll need to know about multiplying
them to do problems 45 and 48).
For Tues. try 8.7: #6 and 12 as practice problems,
and later, try 18,20,22,24,34,42,45. I may ask
some of you to provide answers (or some indication
you tried them) regarding these practice problems.
So be sure to try them!
For HW due Fri., 4/20, 8.7: #4,8,14,36,40,44,48,50
For Fri. 3/30, hw 8.4: #4,6,8 - hint: the answers are div,div,conv but why?
- ans: test for div for 4 and 6, while 8 converges by the AST since
(1/n)ln(n) is decreasing for n at least 3.
No Sunday review till 4/22.
Homework due on Friday 13 April - to be reviewed next week and after
Easter - 8.4: 12,26,32; 8.5: 8,16,20; 8.6: 6,8,16,22,30
On Tues., Wed., and Friday of next week (10, 11, 13 April), classes will involve a review of the practice problems listed below. If you try these practice problems, it will make the class review much more helpful and this will enable you to do the homework more easily. Although I'll be away, one of my colleagues will be taking the class. He will give you at least one quiz (and maybe more). The homework given below is due on Friday April 13.
Here are some practice problems, in addition to the homework listed above. Try 8.4: 14,27,34; 8.5: 10,19; 8.6: 2,4,9,14,24,27,32. Don't forget the odd-numbered problems, too. PS = power series
Here are some answers and hints - but you should understand _why_.
8.4: #14 number of terms needed is 6, #34 series converges for k >= 2
8.5: #10 R = 5, Interval of convergence = [-5,5]
8.6: #2 Same radius of convergence (R = 2) but might differ on endpts
#4 f(x) = sum_(n=0 to oo) 3*x^(4n); I = (-1,1)
#14 Take PS for 1/(1 - 2x) and differentiate it; R = 1/2
#24 See example 7.
For practice on 7.5, try #7,8, p. 543.
For #7, the differential equation is dy/dt = ky(1-y), where y(t) is
the fraction of the population which has heard the rumor at time t,
where t is the number of hours since 8:00 a.m. The solution y is
y = y_0/[y_0 + (1 - y_0)e^(-kt). The rest of the problem is overly
complicated in terms of the arithmetic - try #8 which is simpler.
answer to #8 of 7.5: P(t) = 10,000/[1 + 24*(11/36)^t]
so 5000 = P(t) implies that 24*(11/36)^t = 1;
hence, t*ln(11/36) = ln(1/24), or t is approximately 2.68 years.
Ok, here are more practice problems.
In the review problems for Chapter 6, try #6,7,11--14,17,18.
ans. #6(a) 5/12, (b) 41*pi/105, (c) 13*pi/30; #12 64/15; #14 pi/3;
#18 124/5
In the review problems for Chapter 7, try #5--9, 13,18,19.
ans. #6 x = -1 + B*e^(t - t^2/2); #8 hint: you should get the
following after separating variables and solving the integrals
ln |1 + e^y| = - ln |1 + cos(x)| + C; #18 y = 10(1 - e^(t/10).
More answers are given below. Also, try the following:
#1
Find the volume obtained by rotating a 1m x 1m square about
the diagonal. Your answer should be in cubic meters.
#2
What is the area of the connected region R contained between
the curves y = pi*x - x^2 and y = - sin(x)?
#3
For the same region R as in #2, find the volume obtained by
rotating the region about the y-axis (i.e., the line x = 0).
#4
Again for R as above, find the volume obtained by rotating
about the line y = -2.
#5
What is the integral of x sin(x)?
#6
What is the integral of x ln(x)?
#7
Solve the differential equation dy/dx = -6xy.
#8
Solve the differential equation dx/dt = a*x + b,
where a and b are given constants and a is non-zero.
#9
What figure is produced when a line segment is
(a) rotated about a line parallel to the line segment?
(b) rotated about a line perpendicular to the line segment?
#10
Returning to the region R given in #2 above, if a solid
intersects the x-y plane in the region R and each cross-section
perpendicular to the x-axis is a square, set up the integral
which gives the volume of this solid.
... some answers given below ...
**********************
I suggested that you try #34 of 7.3 - the answer to (c) is t = 200 ln 10 days.
Since 200 ln 10 is approximately equal to 460.5, it takes just over 15 months.
I'll give the main equation later. Let x(t) the amount of new money at time
t be given in $Billion (US) and time t in days. Note that rate-out (of new
money) is zero. Also note that x(0) = 0 since initially there is no new
money in circulation.
You could also try #36 which is a standard-type mixing problem. Can you
answer the last part of the problem without doing anything? Try your
common-sense guess for this part. BUt now go ahead and write down the
formula which will tell you how long you must wait before the long term
solution is effectively reached. What is the connection of this with
the idea of an asymptote? Answer is p(t) = .05 + (1/10)e^(-t/90).
Practice: 7.3 2,6,16,24; Homework for 3/23 - 7.3 #4,10,14,26
answers: #14 L = (kt - kt*ln(t) - k - 1)^(-1); #16 y = 2x^2/(x^2 + 1);
#26 y = (C - x^3)^(1/3)
Some selected answers for the review at the top of this page (3/24):
(note - it is possible that I made some errors in these "answers";
if you find one, let me know to get extra points ;-)
#1 Volume of rotation is pi/3sqrt(2)
#2 Area of region R is (pi^3)/6 + 2
#3 Vol(R,x=0) = (1/6)pi^5 + 2pi^2
#4 Vol(R,y=-2) = (1/30)pi^6 + (2/3)pi^4 + (17/2)pi^2 - 8pi
#7 y = Ae^(-3x^2), A an arbitrary constant
#8 Solution is x = (-b/a) + B*e^(at), where B is arb real constant.
Note that when b=0, this is the same as our previous solution for
the standard problem x' = ax.
#10 int_(from 0 to pi) (sin x + pi*x - x^2)^2 dx
Here are some answers from other problems in Chapter 6:
6.1 #6 A = e^(pi/2) - 2
6.2 #8 V = (3/4)pi, #14 V = (10/21)pi, #34 V = sqrt(3)/2,
#40a V = 128/3sqrt(3)
6.3 #8 L = 6 + ln(2)/4, #14 L = 12sqrt(3)
6.4 #2 g_(avg) = 26/9
For homework due Fri. 3/16, please do 9a, 12,14a, and 18 pp. 494--495
(Chap. 6 Review). Answer to #4 below: pi[25/12 + (1/4)(e^(-4))].
Try the review problems 1--4, 6--7, 9, 11, 12--15, 17--18.
I'll put some selected answers (for the even-numbered
problems) on the web later. These are the problems on
pp. 494--495 in our much beloved text.
In case you forgot your book, here are a few of them.
For #1,2, find the area of the region bounded by the given curves:
#1: y = e^x - 1, y = x^2 - x, x = 1
#2: x + y = 0, x = y^2 + 3y
For #3, you'll need the formula on p. 445 of the text.
To do #1 and 2 without graphing calculators, the second
equation for both problems can be simplified by "completing
the square". So, e.g., y = x^2 - x is equivalent to
y + (1/4) = x^2 - x + (1/4) = (x - (1/2))^2, and hence y
is graphed as the standard parabola y = x^2 but shifted
one-half unit to the right and one-quarter unit upward.
#9 Describe the solid whose volume is given by the integral
(a) int_(from 0 to pi/2) 2pi cos^2(x) dx
(b) int_(from 0 to 1) pi [(2 - x^2)^2 - (2 - sqrt(x))^2] dx
#12 The base of a solid is the region bounded by the parabolas
y = x^2 and y = 2 - x^2. Find the volume of the solid if the
cross-sections perpendicular to the x-axis are squares with
one side lying along the base.
#15 Find the length of the curve with parametric equations
x = 3t^2, y = 2t^3, 0 <= t <= 2 (i.e., t from 0 to 2).
We'll start on Monday after the break with these five and
then go on to the others for Tuesday and Wed. Homework for
Friday after break will be posted later.
For the week of Feb. 26--Mar. 2: sections 6.2 through 6.4
Reading: Section 6.2 (rest of section), 6.3, and 6.4. Homework: 6.2 #12, 36; 6.3 #6, 10; 6.4 #4,10 Practice: 6.2 #10,34,52 (ans. pi/6 by either integrating slices or shells); 6.3 #8, 14 (ans. is 12 sqrt(3). Finding (dx/dt)^2 + (dy/dt)^2 = (3 + 3t^2)^2, integrate the sqrt of this from t = -sqrt(3) to t= +sqrt(3)); 6.4 #2, 6 (a) f_avg = (1/2)int_(from 1 to 3) ln x dx = (1/2)[(3 ln 3 - 3) - (1 ln 1 - 1)] = (3/2)ln3 - 1 (b) f_avg = f(c) if and only if (3/2)ln3 - 1 = ln c Now take the exponential function of both sides to get c = e^[(3/2)ln3 - 1] = e^(-1) * 3^(3/2)
For Tues. Feb. 20 -- Feb. 23, we covered 6.1 and the beginning of section 6.2 (rotating areas around lines). The homework assignment due 2/23 was 6.1: #6,12; 6.2: #4,8.
For #2 of 6.2, a cross-section is a disk with radius (1 - x^2),
and you should integrate from -1 to 1. So the integral I is
I = int_(from -1 to 1) pi (1 - x^2)^2 dx
= 2pi int_(from 0 to 1) 1 - 2x^2 + x^4 dx
= 2pi [x - (2/3)x^3 + (1/5)x^5]_(eval from 0 to 1)
= 2pi [1 - (2/3) + (1/5) = 2pi [(15-10+3)/15] = 16 pi/15
For #4 of 6.2, the easier choice is the variable y, from
1 to 2, and that means you are thinking of the volume as
as an integral of disks. So you need to see how x is a
function of y. One can find the volume also as an integral
of shells with x as the variable for x between 0 and e^2,
but the heights of the shells require a different formula
for x in [0,e] and x in [e,e^2].
More details on above:
For #4 of 6.2, the easier choice is the variable y, from
1 to 2, and that means you are thinking of the volume as
as an integral of disks. So you need to see how x is a
function of y.
V = int_(from 1 to 2) pi (e^y)^2 dy
= pi int_(from 1 to 2) e^(2y) dy
= (pi/2) e^(2y)|_(eval from 1 to 2)
= (pi/2) (e^4 - e^2)
One can find the volume also as an integral
of shells with x as the variable for x between 0 and e^2,
but the heights of the shells require a different formula
for x in [0,e] and x in [e,e^2].
V = int_(from 0 to e) (2pi)x(1) dx
+ int_(from e to e^2) (2pi)x(2 - ln x) dx
The first integral is just the volume of the cylinder of
radius e and height 1, while the second could be evaluated
using integration by parts (or other methods).
For #8 of 6.2, the answer is V = (3/4)pi
The area of an annulus (washer) with inner radius y^(3/2)
and outer radius 1 is A(y) = ... = pi(1 - y^3), for y in [0,1]
so V = int_(y from 0 to 1) A(y) dy
= pi(y - (1/4)y^4)|_(eval from 0 to 1) = pi(3/4).
For #2 of 6.1, a cross-section is a disk with radius (1 - x^2),
and you should integrate from -1 to 1. So the integral I is
I = int_(from -1 to 1) pi (1 - x^2)^2 dx
= 2pi int_(from 0 to 1) 1 - 2x^2 + x^4 dx
= 2pi [x - (2/3)x^3 + (1/5)x^5]_(eval from 0 to 1)
= 2pi [1 - (2/3) + (1/5) = 2pi [(15-10+3)/15] = 16 pi/15
#6 of 6.1 ans. is e^(pi/2) - 2
#12 Find pts of intersection: Given
(1) 4x + y^2 = 12 and
(2) x = y.
Then x^2 + 4x - 12 = 0 so x = 2 or x = -6 (by factoring or q.f.).
Looking at the drawing, integrating wrt y will be easier than x.
Hence, find corresp. y at pts of intersection: y = 2 or y = -6.
Also, solve (1) for x in terms of y: x = 3 - (1/4)y^2
A = int_(from -2 to 6) [3 - (1/4)y^2 - y] = 64/3
For practice, try 6.1: #4,8,10,14 - and for more challenge, 38,40(a).
Then for 6.2: #2,6,10 (you might want to use a calculator for #6).
We'll start with 6.1 on Tuesday. Exams will be returned and reviewed
on Wed.
Answers for 6.1: #4 A = 9, #10 A = 8/3; #38 A = 1/12 (note that
by integrating with respect to y, one only needs one integral,
but in order to integrate wrt x, we would need _two_ integrals.
Answers for 6.2:
#2 (2pi)(8/15), #10 (2pi)[(ln 3) + (1/3)]
Homework is how you learn to _do_ calculus and so it is an
essential part of the course. It is _not sufficient_ to just
write down the answer. You must show your work.
Most Sundays, there will be a review session (open to both sections) in
Reiss 284 from 5 to 6 pm.
Note that the exam will be held on Thurs. night at 9:30 pm in Reiss 264.
You can bring a sheet of ordinary 8 1/2 x 11 paper
using both sides with your notes to the exam.
Notes can consist of whatever you like - but no microdot
recordings of the textbook are allowed!
Here are some additional problems you can try out for practice
(listed Mon. 2/12):
More problems to try out: pp. 433--436:
T-F problems 1--12 - give a reason or an example.
On pp. 435--436, #12,18,20,22,26,28,30,32,34,
54,58,60,70,74.
For Chap. 8, try the following on pp. 631--633:
Concept Problems #1--4, 5(a,b,c,d); T-F #1--3,
9,11,16; Exercises (pp.632--633) #2,4,6 (and
same question for the corresponding series),10,14.
Here are answers to the even-numbered problems below
T-F problems 1--12 - give a reason or an example.
2 F, 4 F, 6 T, 8 F, 10 T, 12 T
On pp. 435--436, #12,18,20,22,26,28,30,32,34,
54,58,60,70,74.
12 1/10 18 2/3pi 20 (-1/3)ln(4) 22 IBP (u=ln(x)) 4ln(2) - 15/16
26 2/81 28 -sin(cos x) + C 30 (1/2)arcsin(x^2) + C
32 IBP u=arctan(x) 34 zero (the integrand is an odd function)
54 at least 0 and at most 1/5 - since the integrand is at most x^4
58 f(x) = 1/(2 - 3x) has a vertical asymptote at x=2/3; the
integral from 0 to 2/3 is an improper integral and diverges. Hence,
the original integral diverges. 60 40/3 70 a=pi/6 74 2/3
NB #26 is obtained by the sub u = 2t+1; #60 by u = sqrt(y-2).
For Chap. 8, try the following on pp. 631--633:
Concept Problems #1--4, 5(a,b,c,d); T-F #1--3,
9,11,16;
T-F 2 F (sin 1 < 1), 16 T
Exercises (pp.632--633) #2,4,6 (and
same question for the corresponding series),10,14.
2 sequence converges to zero, series is geometric series so converges
4 sequence is 0 for n odd and +/-1 for n even so divergent; also div series
6 seq converges since the corresp fcn is ln(x)/(x^(1/2)) which has limit = 0
as x --> oo using l'H rule; series diverges by compar. test
with sum(1/(n^(1/2)) which diverges by p-test p = 1/2 <= 1.
10 diverges by limit compar test with the harmonic series sum(1/n)
14 limit of the sequence of terms is 1/3 so series diverges by test for div.
The first midterm covers sections 5.6 (IBP), 5.7 (trig and PF),
5.9 (approximate integration), 5.10 (improper integrals), 8.1
(sequences), 8.2 (series), and 8.3 (convergence tests, see below).
I decided to leave out 5.8 on tables and computer-aided algebra
systems. For approximate integration, we are only doing the
Midpoint and Trapezoid rules. Be sure you can use the error
formulas and also that you can figure out how large n must be
to ensure that the error is not larger than some given amount.
Here is a problem dealing with the last question: suppose that
f''(x) cannot exceed 20 on the interval [a,b] = [5,7]. How large
must n be to guarantee an error no worse than 0.01 for the
Midpoint Rule when estimating the value of the integral of f(x)
on the interval [5,7]?
(20 * (7 - 5)^3)/(24 n^2) <= .01 if and only if
24 n^2 >= 100 * 20 * 8 if and only if n^2 >= 2000/3 = 666 2/3
Now 25^2 = 625 and 26^2 = 676 so n = 26 is the least integer n
for which the midpoint rule will estimate the integral within .01.
This week we have added one new topic,
on p. 582 of 8.3, the Limit Comparison Test. This allows us to
determine whether or not one series converges from the behavior
of a sequence when the limit of the ratio of corresponding terms
is any finite nonzero constant. For instance, consider the series
Sum_(n from 10 to oo) (n+7)/sqrt(n^5 - 27 n^2 + 2700) = Sum a_n
In the limit, this acts like series Sum n/n^(2.5) = Sum n^(-1.5)
which does converge. So take b_n to be n^(-1.5); Sum b_n converges
by the p-test (as p = 1.5 > 1). Since the limit of a_n/b_n is 1
which is finite and nonzero, the series Sum a_n must also converge.
Try #22,24,26 on p. 585 using this technique. Don't worry about
the material in this section on estimating the remainder.
#22 diverges by LCT with Sum (1/n) (the limit of the ratio
is either 1/2 or 2, depending which one you put on top).
The latter series diverges since it is the harmonic series
(or by p-test for series).
#24 converges by LCT with Sum(1/10^n); the limit of the
ratio of corresponding terms is 1 (by the Squeeze Thm).
The series Sum(1/10^n) is convergent since it is a geometric
series with r = 1/10 < 1.
#26 converges since 7/3 - 1 = 4/3 > 1 (fill in the details).
Try 21,23,25 which have the answers in back. More later.
Bring your questions regarding 8.1 -- 8.3 to class for Monday.
Also for Chapter 5, including error estimates, etc.
Oh, about formulas, rather than listing some here, as I suggested
in class, make up your own list of what you think are the key
formulas and put them on a piece of paper. You'll need both sides.
Some key theorems include the squeeze theorem and the monotone convergence
theorem.
Here are a couple more to try.
Find the antiderivative of sec^4.
int sec^4(x) dx = int (u^2 + 1) du (u = tan(x)) ...
What is I = int_(0 to 1) dx/(1 + x^2)^(3/2) ?
Try u = (1 + x^2)^(1/2). Then du = (1 + x^2)^(-1/2) dx
so I = int u^(-2) du
p. 438: #3,6,7,13. Int sin^3(t)dt/(1 - cos^2(t)) (easy!)
I = int sin^3(t)dt/(1 + cos^2(t)). The integrand is a rational
function in sin(t) (after re-expressing the denominator).
Now divide, integrate the easy part, put the denom back as it was,
and make the substitution u=cos(t).
Try the following problem for section 5.9:
(See Example 1 in this section.)
(1) Use the trapezoid and midpoint rules
with n = 3 to approximate the integral
integral_(1 to 2) x^(-1) dx
(Of course, we know the exact value of the
integral here is ln(2) so you can check to
see if you are getting something reasonable.)
(2) Repeat this using both trapezoid and midpt
rules but with n = 4.
(3) Use the error formulas for trapezoid and
midpoint rules to find an upper bound
on the errors for all four cases (two
different rule with two different choices
for n.
Here is the solution to the first part of (1):
Let f(x) = x^(-1). Then by the formula on p. 413, for n = 3,
Int_(1 to 2) f(x) dx is approximately equal to T_3, where
T_3 = (1/6) [f(x_0) + 2f(x_1) + 2f(x_2) + f(x_3)] with
x_0 = a = 1, x_1 = 4/3, x_2 = 5/3, and x_3 = 2. Hence,
T_3 = (1/6) [1 + 2(3/4) + 2(3/5) + 1/2] =
= (1/6)*[1 + 1.5 + 1.2 + .5]
= .7
Note that the sum starts at 0 and goes to n (in this case, 3),
so there should be n + 1 terms (in this case, 4) in the sum.
Since Delta x = 1/3, we multiply by half of that which is 1/6.
The first and last terms in the sum have coeff = 1; all others
have coeff = 2.
If you look at the corresponding formula for
the midpoint rule (p. 412), you will see that in this case,
the sum has only n terms, all coefficients = 1, and the factor
which multiplies the sum is Delta x. The summands for the
midpoint rule, however, rather than being determined by f(x_i),
are given by f(x_i-bar) where x_i-bar = (1/2)[x_(i-1) + x_i].
That is, x_i-bar is the midpoint of the interval [x_(i-1),x_i].
So the first summand will be f(7/6) since (1/2)[1 + 4/3] = 7/6.
Hence, for the Midpoint Rule the approximation is
(1/3) [ 6/7 + ... ] ... you fill in the rest!
Here are some other parts of the above (for the trapezoid rule):
The error estimate for the trapezoid rule is
E_T <= K(b-a)^3/12*n^2 = 2/12*9 = 1/54, about 0.02,
where f''(x) = 2*x^(-3) which has max of K = 2
on [1,2] since f(x) = x^(-1).
The actual value of the integral, ln(2), is
0.693... so the true error is about .007.
For section 5.7, try the following:
#4,5,7,12(b) (skip part (a) of #12),15,17,25,29.
For section 5.9, try #17(part c only)
Here are some answers for the practice problems, including
what the substitutions are, not just the final answer.
5.7: #4 Let I_m = int sin^3(mx) dx and take u = cos(mx).
Then I_m = (-1/m) int (1 - u^2) du
= (-1/m)[cos(mx) - (1/3)cos^3(mx)] + C.
#5 uses the half-angle formula cos^2(6t) = (1/2)(1 + cos(12t));
the value of the integral is then pi.
#7 Let u = sec x. Then int tan^3(x)sec(x) dx = int (u^2 - 1) du
#15 2/(x^2 + 3x -4) = A/(x+4) + B/(x-1)
#17 If (x-9)/(x+5)(x-2) = A/(x+5) + B/(x-2), then
x-9 = A(x-2) + B(x+5) = (A+B)x -2A +5B, so that
1 = A + B and -9 = -2A + 5B
Adding twice the first equation to the second gives
-7 = 7B so B = -1; hence, A = 1 - B = 2.
Thus, the integral of the rational function is
2 ln |x+5| - ln |x-2| + C.
#25 Use long-division on the integrand.
#29 Let u = sqrt(x), so u^2 = x and dx = 2udu. Hence, the integral
I = int_(from 9 to 16) sqrt(x)/(x-4) dx = 2*int_(3 to 4) u^2/(u^2-4) du.
The last integral can be simplified by long division and we get
I = 2 + 8*I_1,
where I_1 = int_(3 to 4) du/(u+2)(u-2). This integral can be done
by partial fractions to give I_1 = (-1/4)ln |u+2| + (1/4)ln |u-2|,
evaluated from 3 to 4, and the resulting value of I is 2 + ln(25/9).
Another practice problem for 5.9:
(a) Find T_4 when the integral is int_(0 to 2*Pi) sin(x) dx.
(b) What is the actual error in your estimate?
(c) According to the error formula for the Trapezoid Rule,
how large must n be to ensure that T_n approximates
the above integral with error less than 10^(-4)?
(d) Repeat this problem using M_2 instead.
Here is the answer to the above:
T_4 = (1/8)(0 + 2 + 0 - 2 + 0) = 0. Similarly, M_2 = 0.
The actual integral is identically zero.
The function sin(x) has second derivative -sin(x), and
|sin(x)| is largest at pi/2 (and -pi/2), where it has
value K = 1. Hence, the error bound is
E_T <= (2 pi)^3/12*16, equal to about 0.323
but in fact in this particular case the actual error is zero.
To solve the equation (2 pi)^3/12*n^2 < 10^(-4),
thereby finding the least value of n which guarantees
an error for the trapezoid rule of at most 10^(-4),
invert both sides (and the sense of the inequality),
and then do a skoshi bit of algebra:
n^2 > (1/12)[(2 pi)^3] * 10^4
so, since n is an integer, n >= 228. (This particular
part of the problem requires a calculator, unlike our
normal practice.)
For homework due Fri. Feb. 9, 5.7: #2,6,8,10,18,26,30;
Answers to some: #2 (8/15) #6 (pi/16) #8 (one way: take u = tan x)
#18 PF gives (-2 ln(2) + 3 ln(3)) - (-2 ln(1) + 3 ln(2)) = ln(27/32)
#26 long division #30 Let u = sqrt(x+2). The integral becomes
I = int 2udu/(u-2)(u+1). Now use PF.
5.9: (a) Find T_3 when the integral is int_(1 to 4) x^2 dx.
(b) What is the actual error in your estimate?
(c) According to the error formula for the Trapezoid Rule,
how large must n be to ensure that T_n approximates
the above integral with error less than 10^(-4)?
(d) Repeat this problem using M_3 instead.
Here is the answer to the last problem:
T_3 = (1/2)[1 + 2*4 + 2*9 + 16] = 43/2
The true value of the integral is (1/3)x^3 (eval from 4 to 1)
which is (1/3)[4^3 - 1] = 21 so the error is |43/2 - 21| = 1/2.
The error bound guarantee is
E_T <= 2*27/12*9 = 1/2.
Finding the least n so that the bound is < 10^(-4), we must solve
2*27/12*n^2 < 10^(-4) so n > 100*sqrt(54/12) so n >= 213 (calculator!)
Sections 8.1 through 8.3 bring out the
strong connections between improper integrals and series.
The integral comparison test and the test for divergence
both have corresponding versions for series. Basically,
improper integral is to integrand as series is to sequence.
So read sections 8.1 -- 8.3, skipping
Example 6 of 8.2 (p. 571) and also from the end of
Example 4 of 8.3 (bottom of p. 581
and pp. 582--585).
Here are some practice problems to try.
Try these: 8.1: #10,14,18,32,42; 8.2: #16,18,22,26,52;
8.3: #6,9,16,18.
Homework problems due for collection on Friday Feb. 2.
8.1: #12,30,34; 8.2: #14,20,24; 8.3: #10,20 .
Answers and hints to practice: 8.1: #32 use Thm. 2, p. 559;
#42 see p. 563 - increasing and the sequence is bounded
8.2: #22 try a consequence of Thm 8, p. 573; 26 (1 radian
is between 0 and pi/2 - now draw the graph of cos x)
8.3: #6 (the Integral Test applies and the corresp.integral
is divergent by the p-test p = 1/4 < 1); #16 converges;
#18 sum of convergent and divergent series is convergent (why?)
For Tuesday Jan. 23, here were a few problems to try.
Try 5.6: #10 and #22 [ ans. is 16/3 - (7/3)sqrt(5) ].
(Hint: for #22, try u = r^2 with IBP.)
Also try 5.10: #6 (divergent), #8 (covergent; ans. is 1/4), #12 div
Here are some problems for review. You don't need to write them up
but I may call on you to work some of them at the board.
Ch. 2 p. 176 #2,4,12,14,24,38
Ch. 3 p. 255 #12,28,40,50,52,54,60
Ch. 4 p. 336 #2,4,8,12,24,26,28,38
For Tuesday, we'll review the above and some new material
which reviews Chap. 5 and begins IBP (section 5.6):
rev #8, 12, 20 pp. 434--435, and #2,4,6 p. 298.
For homework due on Friday 1/26
5.6 (p. 398): 10, 28; 5.10 (p. 431): 8, 20, 42, 48
about homework Friday 1/26
5.6: #10 IBP(u = ln p) ans. is (1/6)p^6 ln p - (1/36)p^6 + C
#28 First subst. w for sqrt(x) to get
int_(1 to 4) e^(sqrt(x)) dx = 2 int_(1 to 2) w e^w dw
the latter integral can be determined by IBP(u = w)
ans. to the problem is 2 e^2
5.10: #8 int_(0 to t) x/(2 + x^2)^2 = (-1/2) [(2 + t^2)^(-1) - 1/2]
taking the limit as t --> oo, ans. is (-1/2)(-1/2) = 1/4
#20 ans. is 1/4; use IBP(u = ln x) and l'H rule
#42 integrand is larger than 2/x. But int_(1 to oo) 2/x dx
= 2 int_(1 to oo) 1/x dx which diverges by the p-test
#48 converges if and only if p > 1; ans. is 1/(p-1) (in class)
We showed in class that
int (from 0 to oo) x e^(-x) dx = 1
int (from 0 to oo) x^2 e^(-x) dx = 2
More generally, using IBP and l'Hospital's rule, we showed that for n > 1
int (from 0 to oo) x^n e^(-x) dx = n * int(from 0 to oo) x^(n-1) e^(-x) dx
so the integral on the left is equal to n! = n*(n-1)*(n-2)*...*3*2*1.
To check that you are on track, here are some answers to practice problems:
section 5.6: #6
-(1/2)t*cos(2t) + (1/4)sin(2t) + C
is the indefinite integral of t*sin(2t).
Check: (d/dt) (-1/2) t*cos(2t) = (-1/2) [cos(2t) + t*2*(-sin(2t))]
= (-1/2)cos(2t) + t*sin(2t)
and (d/dt) (1/4)sin(2t) = (1/2)cos(2t).
For #2 you should get theta sin theta + cos theta + C
Check: (d/dx) (x sin x + cos x) = sin x + x cos x - sin x = x cos x.
For #4 to find the integral int x e^(-x) dx,
take u = x and dv = e^(-x) dx.
Then du = dx and v = -e^(-x), so
int x e^(-x) dx = -x e^(-x) + int e^(-x) dx
= -x e^(-x) - e^(-x) = (-x-1)*e^(-x)
(and check it as usual)
Question: What is int(from 0 to pi/2) x cos x dx? Ans. pi/2 - 1.
Answers to some review problems:
Ch. 2 pp 176 et seq #4 ans. is 0, #12 ans. is 1/3
Ch. 3 pp 256 et seq. #12 y' = r x^(r-1) e^(sx) + s x^r e^(sx)
#40 y = -1 is an equation of the tangent line
#60(a) the line x - 4y = 1 can be rewritten y = (1/4)x - (1/4),
so has slope 1/4. At a point on the curve y = e^x the slope of
the tangent line is e^x so this line is parallel to the given line
if and only if e^x = 1/4 which holds if and only if x = ln(1/4).
Note ln(1/4) = -ln(4). Hence, an equation of the tangent line
to the curve at this point is y - 1/4 = (1/4)(x + ln(4) ).
Ch. 4 pp 336 et seq. #26 0, #28 +oo (pos. infinity)