Calculus II Study Guide To help you prepare for midterms and the final, this study guide is already updated for the whole semester…the idea is that if you go through the study guide on a regular basis and make sure you understand the concepts as we go along, and keep going over it to refresh your memory of key ideas throughout the course, you’ll be better able to pinpoint what topics you need help on, and studying for finals should be a lot easier. NOTE: To help you study, you should also look over exercises assigned for quizzes, honour homework, and assignments, as well as examples done in lectures and tutorials. (This may sound like a lot, but don’t panic…if you regularly attend class and do your homework, then you’ve got this all done already). Also, don’t forget any independent-study material from assignments, as that isn’t included in the study guide below. Good luck. Hope this helps!

7.1: Integration by Parts First of all, realize that this section is actually pretty easy. Really, the main thing here

is to understand the formula for integration by parts: duvuvdvu , and then the

tricky part is determining what to make u and what to make dv… in doing this, you do have the LIATE rule (from left to right, tells you which functions get first pick for u), but

common sense will tell you that you want the new integral duv to be easier to evaluate

than the original one, so this dictates your choice for the functions. Once you’ve selected which part is u and which is dv, the question becomes a lot easier…find du (differentiate) and v (integrate) since you’ll need these in the formula, and then just plug them into the formula. That’s it. The resulting integral must of course still be evaluated, but it should be a lot easier. As a word of caution, there are questions that will require you to do integration by parts more than once (or to use some other integration technique to evaluate the new integral), and there are questions where you come around “full circle” to return to the original integral, and those where you have a single function, so

dxdv …refer to the lecture notes for examples. Practice these types of questions…if you understand the concept, then integration by parts is fairly easy!!!

7.2: Trigonometric Integrals Yes, this section involves trig, but stop worrying…it’s pretty simple. This section is focused on integrating functions composed of the various trig functions, which means they’re quite easy to recognize. Most of the rules in this section apply to the “natural pairings” of 1. sin(x) and cos(x), 2. tan(x) and sec(x), or 3. cot(x) and csc(x) (i.e. powers of these functions multiplied together). [if you have to do an integral that involves trig functions that aren’t a “natural pairing”, you can always rewrite it in terms of sin(x) and

cos(x)]. Know the rules which apply for odd and for even powers when dealing with sin(x) and cos(x). If at least one of the powers is odd, split up the odd power so that 1 factor is left, while the remaining factors are converted using the trig identity sin2(x)+cos2(x)=1. Once this is done, a simple u-substitution should make integration easy. This technique applies regardless of what the other power is (it could be even, a fraction, or even negative (i.e. a term in the denominator)…just split up the odd power in the numerator as usual. If both powers are odd, split up the lower power, as this makes the arithmetic easier (although either one works). If both powers are even, use the double-angle and/or half-angle formulas (these are on the formula sheet) to transform the function into something you can integrate. For tan(x) and sec(x), things are a bit more complicated, but the identity 1+tan2(x)=sec2(x) is what’s useful here. If the power of sec(x) is even, leave a factor of sec2(x) and convert everything else using the trig identity. Then a u-sub will work, with u=tan(x), since du=sec2(x) and you have this available. If the power of tan(x) is odd, split off a factor of tan(x)sec(x) and use the trig identity to convert everything else to sec(x). Again, a u-sub now works (this time, u=sec(x) since du=tan(x)sec(x) appears in the integral). For powers of cot(x) and csc(x), the trig identity to use is 1+cot2(x)=csc2(x) and the rules/techniques are the same as for tan(x) and sec(x).

7.3: Trigonometric Substitution The idea behind trigonometric substitution is as follows: faced with an expression of the

form 22 xa , 22 xa , or 22 ax , make the appropriate trigonometric substitution. All terms with x get converted to be in terms of , and similarly, dx gets converted to be in terms of d. Trig identities are used to simplify the expression, and the result should be something that you can integrate (although you might have to use techniques from 7.2 or u-sub, etc.) After you’ve finished integrating, you must of course return to the original variable x for an indefinite integral…to do this, you can either use trig identities again, or set up a triangle, and make use of your inverse trig knowledge from Calc I/Intro Calc…your choice. If you’re dealing with a definite integral, then you can convert your limits of integration along the way to be in terms of the new variable, and then you don’t have to do any conversion at the end…just substitute in the new limits. Remember that it is up to you to figure out which substitution applies for which expression (figure this out using the trig identities 1cossin 22 and

22 sectan1 , or by drawing a triangle). ….realistically, if you choose the wrong substitution, you’ll find that you won’t be able to simplify using any of the trig identities, so that should be a sign that you’ve made a mistake. This section is a bit tricky, so practice! (it’s not hard, but some of the questions are long…make sure you fully understand inverse trig from Calc I/Intro Calc, and 7.2 before working through this section).

7.4: Integration of Rational Functions by Partial Fractions This section is focused on integrating rational functions (a polynomial divided by another polynomial…fairly easy to recognize) by splitting them up into several terms which are easier to integrate. To do this in the simplest case, when all of the factors in the

denominator are linear and distinct, we break up our rational expression into terms of the

form bax

A

where bax is a factor of the denominator. E.g.

1334

632

x

B

x

A

xx

x .

The beauty of this is that the expression on the right is VERY easy to integrate. So, all that remains is to determine the unknowns A, B, etc.; to do this, we combine the terms on the right by finding a common denominator, or cross-multiplying. Setting the numerators equal gives us a single equation (in this case, it’s )3()1(63 xBxAx ). You now have a choice of 2 methods to solve for A and B: 1. Equate coefficients for like terms (e.g. x2, x, numbers, etc.) on both sides…this leads to a system of equations which you can solve using some basic algebra (elimination or substitution) OR 2. You can substitute in convenient values of x to result in a system of equations which is much easier to solve (for example, using 1x or 3x in the above example). Of course, once you’ve found A and B, you can integrate, and you’re done. There are a few more complicated cases which you should know how to do: how to deal with irreducible quadratic factors in the denominator, and how to deal with repeating factors. Irreducible quadratic factors means that you can’t factor a quadratic term at all (i.e. it would have a

negative discriminant); in this case, you break it up as cbxax

BAx

2 where the major

difference is that we now have a linear term in the numerator. E.g.

15)1)(5(

1422

xx

CBx

x

A

xxx

x and then you proceed as before. Finally, in the case of

repeated factors, you’ll need extra terms for each time the factor occurs. E.g.

323 )2()2(21)1()2(

13

x

D

x

C

x

B

x

A

xx

x and proceed as before.

7.5: Strategy for Integration There is NO new math in this section. However, this is a wonderful section since it allows you to practice determining which method of integration to use for which question (a talent which you’ll need!). When you get an integration question, you should check:

Can I simplify; is it of a form I know (e.g. inverse trig) for which I have a formula?

Can I use u-substitution? Is it a trigonometric integral? (this is easy to recognize…powers of various trig

functions) Will a trig substitution help? (again, easy to recognize if you’ve memorized the

expressions in 7.3…root signs with a sum or difference of squares) Will integration by parts work? Try again, perhaps combining several methods.

A few important points: 1. Never resort to methods such as integration by parts or trig substitution until you’re certain that simpler methods such as u-substitution don’t work. 2. If nothing seems to work, perhaps try u-substitution and check where it gets you…it might get you to a form where a method such as integration by parts suddenly becomes feasible.

7.7: Approximate Integration Basically, we’re approximating the area under the curve using the sum of vertical pieces (e.g. rectangles); this is very useful when we have numerical data, or when the given function is too hard to integrate. In Calc I/Intro Calc, you already saw the right endpoint, left endpoint, and midpoint rules. Here, we also introduce the trapezoid rule (which is actually the average of the right and left endpoint methods) and Simpson’s rule. We will give you the formula for Trapezoid and Simpson’s rule, but you are responsible for knowing all of the other approximation formulas (if you really understand the concept, then there’s really nothing to memorize…you’re just adding up rectangle areas). In all cases, you need to know what x is (same for all methods). In addition, you should be able to use the error formulas (which will be provided) for Midpoint, Trapezoid, and Simpson’s rule to estimate the error, or figure out the number of rectangles that must be used to achieve a desired accuracy. Please note that various notation for the error formulas is possible (e.g. both NES and SE denote the error for Simpson’s rule) We will

NOT test you on Monte Carlo approximation.

7.8: Improper Integrals There really isn’t a lot of new math here. Basically, we’re interested in evaluating two types of improper integrals:

1. Those with infinite limits. 2. Those with infinite integrands. For the first case, these are really easy to recognize…the limits are and/or …can’t miss ‘em. Since infinity is not a real number and we won’t be able to substitute the limits in after integrating, the whole point here is that we do the integration as usual (working with the indefinite integral), but we take the limit as we approach or . So, we integrate as we always do, but write limits when applying the Fundamental Theorem of Calculus, and finally evaluate the limit (note: you may use L’Hospital’s rule in some cases to help evaluate the limit if you are familiar with it from Calc I, but this is not required in this course! For some limits, it’s just helpful to remember that in a tug of war between an exponential and another function, the behaviour of the exponential will dominate) If both limits are infinite, then you must first split up the integral at any point along the way to form 2 integrals, each with one infinite limit. For the second type of improper integral, it’s a little harder to recognize, because it’s the integrand which is infinite…in other words, the function explodes somewhere along the interval of integration because we divide by zero….so, every time that you do a definite integral, you should check that the denominator of the function is never zero along the interval of integration or at the endpoints. If it is, then it’s improper so you again integral as usual, then take the limit approaching the “problem” point, and evaluating the limit. If the “problem” point doesn’t occur at an endpoint, then you must split the integral into 2 parts, approaching that “problem” point from the left and right.

8.1: Arc Length

This section is really easy…really! There’s one formula:

b

a

dxdx

dyL

2

1 . Know it

(actually, I’ll give it to you). Then all you have to do is plug things into it…way easy!

8.2: Area of a Surface of Revolution

Again, this is a really easy section! There’s one main formula:

b

a

dxdx

dyyS

2

12

which gives the surface area of revolution when we rotate about the x-axis (which is what we’ll focus on). Again, I’m giving you the formula, and again, you just plug things into it! Easy!

8.3: Applications to Physics and Engineering In this section, we only covered the part about dams. So, I’ll give you the formula

b

a

dyywydg )()( , and you have to apply it for the given problem. What we’re

doing is trying to figure out the force on the dam, which is pressure times area. The challenge is that the pressure depends on the depth, so we have to integrate (add up many small horizontal strips); the other challenge is that unless the dam is a rectangle, the area (or rather the width) of the strips changes with depth as well. Now, and g are constants, so all you have to do is

1. set up a coordinate system (it is strongly suggested (and in fact the question may tell you to do this) that you set y = 0 at the water line)

2. figure out what a and b are (i.e. where do we start and stop adding up the strips) 3. what d (depth) is as a function of y (d = y if you set y = 0 at the water line, so this

makes things easier!) 4. what w (width) is as a function of y (this is probably the trickiest part…for

rectangles, it’s totally easy since it’s a constant; for trapezoids and triangles, it’s based on similar triangles, or assuming the width is a linear function, and then solving a linear system of equations; for semi-circles, it’s based on Pythagorean Theorem).

Know how to work with dams that are rectangles, trapezoids, triangles, semi-circles, or other similar shapes.

8.4: Applications to Economics and Biology In this section, we only covered the part about cardiac output. So, I’ll give you the

formula

T

dttc

AF

0

)(

, and you have to apply it for the given problem. The quantity of

dye A is given, as is the total time, T. For c(t), you’re either given a function, in which case you’re just doing a simple definite integral, or you’re given numerical data, in which case you have to use Simpson’s rule.

8.5: Probability In this section, you’re given a probability density function (pdf), and you’re asked to calculate the probability of certain events occurring. Now, the main point to know here is how to compute the probabilities:

Probability of outcome lying between a and b = b

a

dxxf )(

In other words, you integrate the pdf between the 2 values you want X to lie between. In addition, you’ve also learned that the mean (average value) is given by

dxxfx )( (this is on the formula sheet)

Finally, understand the exponential distribution, and how to apply it to solve problems

relating to waiting time. The formula is

t

etf

1

)( for 0t . I will provide you

with the formula, but you must know that is the mean waiting time. Again, be able to find the probability of waiting a certain time.

9.1: Modelling with Differential Equations First of all, understand what a differential equation is (an equation which involves a function and its derivatives), and how you might set up such an equation to model a given physical phenomenon. Also, know what “order” means….i.e. the highest derivative in the equation. Another thing to know is that to verify that a given function satisfies a differential equation, all you have to do is substitute the given function and its derivatives, as appropriate, into the differential equation, and verify that LHS=RHS (left-hand-side of equation equals right-hand-side of equation). You should of course know how to solve basic differential equations that contain the derivative, but not the function (e.g. 73 2 xy ), as you have done this in Calc I (antiderivatives!). In addition, you may be given an initial condition which can be used to solve for the constant of integration.

9.2: Direction Fields and Euler’s Method Understand that a direction field can be used to give us a graphical solution to a differential equation by showing us the form of the solutions. So if you’re given a direction field and asked to draw a solution curve for a given initial condition, then just draw a curve that passes through that point, but is also always following the direction given by the arrows in the direction field. While I will NOT ask you to construct a direction field by hand, it is important that you understand how this is done, as you may be required to match direction fields to equations, and this requires an understanding of what a direction field is, and how it is generated. Basically, for a differential equation of

the form ),( yxfy (i.e. the derivative depends on some function of x and y), a direction field is generated by plotting an arrow representing slope ( y ) for given values of x and y. So, to draw a direction field, you would select values of x and y, find y there using the differential equation, and draw an arrow with that particular slope at the value of x and y you were working with. So, if you are asked to determine if a given direction field matches a given differential equation (which you should know how to do), look for things like the value of x and y where you expect the slope to be zero or infinite in the direction field. Also, is the slope expected to be always positive or negative? Also, if a given differential equation does not have a dependence on x (or on y), then the arrows in the direction field should not have such a dependence either. For Euler’s Method, you will be given the formula ),( 111 nnnn yxhFyy , so you just have to know how to

apply it. First, to find h, you’re either told what h is in the question, or you’re given the number of steps desired (given the starting x value and the x value where you want to find the approximation, you can find the length of the interval and then divide by the number of steps to get the stepsize h). Now, F is just the right-hand-side of the differential

equation (once you’ve isolated for dx

dy on the left side, of course). So, given a starting

point ),( 00 yx , sub it in to find y1 (x1 of course is just x0+h). Now that you have ),( 11 yx ,

you can sub it into the formula to find y2. And so on, until you reach the x value at which you were asked to estimate the solution y.

9.3: Separable Equations There are three important concepts in this section: knowing what a separable equation is, setting up a differential equation, and solving a differential equation. All of the differential equations in this section will be separable….this means that you can write it

as )()( yfxgdx

dy …in other words, the derivative is equal to some function of x times

some function of y. You should know how to recognize if a given differential equation is separable or not. The second important topic is setting up differential equations to model a physical phenomenon…we are setting up an equation for the rate of change of some quantity (as an example, let’s consider the amount of salt in a tank, which we call A).

Thus, the left-hand-side of the equation is just the derivative (dt

dA). On the right-hand-

side, we must write how our quantity is actually changing. A change occurs based on how much of the quantity is coming in minus how much is leaving (salt into tank – salt out of tank). Often, the amount coming in is easy to determine (rate of flow * concentration of salt), but it’s the amount that leaves that requires a bit more thought, as it of course depends on the amount currently present (i.e. A). Once you’re done, you should have a separable differential equation which is subject to some initial condition (based on the amount of salt initially in the tank)…this really isn’t that hard once you get some practice at it. Now, the third part of this section involves actually solving the separable differential equations for the quantity of interest (e.g. finding the amount of salt as a function of time). To solve a separable equation, move the derivative to one side and all the other terms to the other side, and then cross-multiply/divide such that everything

to do with a particular variable is all on one side (e.g. yxdx

dy 5 becomes dxxy

dy 5

where all y’s are now on the left, and x’s now on the right). Next, integrate both sides (so you’ll need to know your integration techniques from midterm 1), introducing a constant of integration on the right. Finally, solve for y if possible (assume that you should unless I tell you otherwise), and finally, solve for the constant of integration using the initial condition, if provided.

9.4: Exponential Growth and Decay Really, this section is just a continuation of 9.3, but it presents some specific examples of differential equations related to population growth and Newton’s law of Cooling. I would NOT ask you to memorize the solutions to these equations, as the point of this chapter is really for you to learn to set up and solve equations yourself. So, there’s nothing really new to learn in this section as compared to 9.3…just know how to set up a differential equation, and know how to solve a separable equation subject to the given conditions (so you can solve for any constants). That’s it.

10.1: Curves Defined by Parametric Equations The main point of this section was to be able to convert from a set of parametric equations to a Cartesian equation. So, you’re given )(tfx and )(tgy , and your goal is to obtain )(xfy . In other words, solve the x equation for t, and substitute this into the y equation; this eliminates the parameter, and you obtain an equation for y in terms of x. That’s it. You should also have a basic understanding of how to recognize the graph of parametric equations without having to convert to Cartesian (e.g. match graphs of

)(tx and )(ty with a graph of )(xy .

10.2: Calculus with Parametric Curves Here, we were interested in finding derivatives of a curve that is defined by parametric

equations. You should know that the formulas are

dt

dxdt

dy

dx

dy as long as 0

dt

dx for the

first derivative, and

dt

dxdx

dy

dt

d

dx

yd

2

2

for the second derivative. (KNOW the first formula,

but I’ll give you the one for the second derivative!) Once you’ve written down the formula, basically all you have to do is find derivatives with respect to t…super easy

(totally Calc I stuff)! The other concept in this section was arclength for parametric

curves. I will give you the formula

dtdt

dy

dt

dxL

22

, and all you have to do is to

find the necessary derivatives, and plug them in (usually set up only)…about as easy as it gets! 10.3: Polar Coordinates In this section, you should first have a basic understanding of what simple polar curves look like, and how to convert between polar and Cartesian coordinates. You should also have a basic understanding of how to plot points in polar coordinates. Next, know how to convert a polar curve into a Cartesian equation (you should also know how to convert from Cartesian to polar, but that’s easy…just replace x with cos() and y with sin(). Anyways, to help you convert to Cartesian, you should know the formulas )cos(rx ,

)sin(ry , 222 yxr , and x

y)tan( (memorize these, or better yet, just draw a

circle and you’ll be able to figure them out easily). So, given a polar equation which has only r and in it, the goal is to convert it to a Cartesian equation which will only have x and y in it. To do this, the best strategy is to first convert any sec(), csc(), and cot() to be in terms of sin(), cos(), and tan(), since these are the only trig functions the above formulas give you any information about. Next, we typically eliminate by using the

above formulas to write r

x)cos( , etc. At this point, you’ll only have x, y, and r in the

equation. To get rid of r, use 222 yxr ; this produces an equation with only x and y in it, which is all we wanted. The other concept in this section is to find derivatives of polar curves. To do this, we proceed as in 10.2 for parametric curves, using the variable

instead of t. In other words, we obtain the same formula, which is

d

dxd

dy

dx

dy . The

problem is we’re given an equation with only r and in it, so what’s x and y? Well, we again use the equations )cos(rx and )sin(ry where r is the given polar curve equation, and then we can find the necessary derivatives.

10.4: Areas and Lengths in Polar Coordinates Know how to compute the area of a polar curve, )(fr . I will give you the formula,

drAb

a

2

2

1 , so all you have to do is plug everything into it and integrate. If limits of

integration are given, then it’s easy; if not, and if you’re asked to find the area enclosed by one loop of the curve, then set 0r to find the limits of integration, and choose two adjacent angles which satisfy the criteria 0r . If you’re finding the region between two curves, then you have to find where the two curves intersect by setting their formula for r equal to one another and solving for . Once you have these (p.s. this might be a good

time to review trig values at special angles), then it just comes down to plugging in and integrating from that point on. It is often the case that you might end up with an even power of cos( ) or sin( ) to integrate, so just use half-angle formulas (which are on the formula sheet, and which you’re familiar with from 7.2 anyways). The other concept in

this section is again arclength:

b

a

dd

drrL

22 (this is the third arclength formula

you have, but they’re all labelled on the formula sheet, so don’t worry…just find the necessary derivatives and substitute everything in…I’ll typically only ask you to set it up…super easy!)

14.1: Functions of Several Variables This section was really meant to serve as an introduction to functions of several variables. You should know how to evaluate them for given values of the variables (i.e. substitute into the function), and how to find the domain. To find domain, you simply make sure that you include only those values of the variables for which the function is defined. To help you find the restrictions, make sure you’re not taking square roots of negative numbers, or “ln” of negative numbers or “ln” of 0, and make sure you’re never dividing by zero, etc. I will not ask you to graph/sketch any multivariate functions or their contour plots, although you should be able to match basic contour plots with 3-d plots of a function, and sketch a contour plot given a 3-d plot.

14.3: Partial Derivatives Know how to find partial derivatives with respect to a given variable; to do this, consider all other variables as constants, and find the derivative as usual. Make sure you recognize the notation for partial derivatives, so that you know what the question is asking…given

),( yxfz , then xf , x

f

, and x

z

all mean the same thing, namely, the partial derivative

of the function with respect to x. Also, know how to find higher order derivatives. For example, yxf means that you first take the partial derivative with respect to x, and then,

once you’re finished, differentiate the resulting function with respect to y. Remember that, according to Clairaut’s Theorem, ),(),( bafbaf yxxy for the functions that we’ll

deal with in this course.

14.5: The Chain Rule Know how to apply the chain rule for a function ),( yxf where x and y both depend on one or more other variables as well (also, you could be given a function f which depends on more than 2 variables). To figure out the resulting chain rule formula, it helps to draw the tree structure which shows which variables are functions of which other variables. If we want to e.g. find the partial derivative of f with respect to t, and it is x and y that depend on t, then we must go through those two branches (x and y) to get to t. Really, the only tricky part is setting up the formula using the tree structure…once you have that,

then it’s just a matter of finding a few partial derivatives, which is the same as 14.3. Again, not hard, just practice!

14.6: Directional Derivatives and the Gradient Vector Understand the meaning of a directional derivative (i.e. finding slope in a given direction). Know how to find the gradient vector (e.g. for a function of 2 variables, it’s components are just the partial derivatives in the x and y directions, respectively). Know how to compute directional derivatives (this means finding the gradient, possibly finding the direction vector and/or converting the given direction vector into a unit vector, and then taking the dot product). Finally, know that the gradient vector is perpendicular to the level curves of a function, and it represents the direction of the steepest slope. Thus, if you are asked to find the direction of greatest change of a function, just find the gradient, and this vector gives the direction. If you are asked for what the value of the maximum rate of change actually is, just find the magnitude of the gradient vector.

14.7: Maximum and Minimum Values This section is an extension of the work that we did in Calc I/Intro Calc to find max/min, and the concepts are totally the same. Know that local max/min occur at critical points, which you find by solving 0xf AND 0yf . Once you find the critical points, you

may be asked to classify them as a local max/local min/saddle point. If asked to classify them, you have the 2nd derivative test to help you. The formula

2),(),(),( yxfyxfyxfff

ffD xyyyxx

yyyx

xyxx is provided on the formula sheet, and then 1)

If 0D and 0xxf at your point, you’ve got a local min. 2) If 0D and 0xxf

at your point, you’ve got a local max. 3) If 0D , then you’ve got a saddle point. (By the way, if D=0, the test gives no info)

15.1: Double Integrals over Rectangles In this section, we are finding numerical approximations to double integrals. This is similar to what you did in Calc I/Intro Calc and 7.7, except that we add up the volumes of rectangular columns rather than the areas of rectangles…in other words, we multiply the area of the base of the rectangular columns (xy) by the height of the function at the specified sample points (either bottom-right, bottom-left, top-right, top-left endpoints, or midpoints). I will specify the number of partitions to use, and you must compute x and y accordingly, and then find the resulting approximation. You are responsible for knowing the “formulas”...really, it’s just adding up the volumes of several rectangular columns together. You should also know how to find the average value of a two-dimensional integral (find the integral, and divide by the area of the whole rectangular region you’re integrating over).

15.2: Iterated Integrals Here, know how to compute double integrals. You start with the inner integral and work your way out. When you integrate with respect to x, then you consider y to be a constant,

and vice versa. E.g. 3

0

2

1

2 dxdyyx : the inner integration is over y between the limits of 1

and 2 (so we regard x as a constant when we do this), and the outer integral is over x between the limits of 0 and 3 (at this point, y has been evaluated, so it’s actually just a number, and this becomes a single-variable integral). Fubini’s Theorem guarantees that when we integrate over a rectangular region, for the functions we’ll deal with in this course, the order of integration can be switched if desired.

15.3: Double Integrals Over General Regions The integration here is exactly the same as in 15.2...no real new math. The difference is that rather than having a rectangular region of integration where the bounds on the inner and outer integral are all numbers and Fubini’s theorem applies, the bounds on the inner integral may involve functions of the other variable (i.e. the one that appears in the outer integral). What we’re doing is integrating over a non-rectangular region; in this case, Fubini’s theorem doesn’t apply...if we decide to switch the order of integration (which we may need to do to make the integration easier, for instance), we have to figure out the appropriate new bounds for both the inner and outer integral. The hard part here is figuring out the bounds...to do this, draw a diagram and figure out if it would be easier to add a bunch of vertical strips (in other words, integrating with respect to y first), or horizontal strips (in which we integrate with respect to x first). For vertical strips, the inner integral’s bounds denote the function that defines the top and bottom of each strip; the outer integral’s bounds tell us the x-value at which we start and stop adding up the strips. For horizontal strips, the inner integral in terms of x tells us the function )(yfx that defines the left and right endpoint of each strip. The outer integral tells us the y value at which we start and stop adding up the strips. Once you’ve figured out the bounds (it’s tricky, so get lots of practice at this!), the rest is just like in 15.2 (sure, when you do FTC for the inner integral, you’ll be subbing in a function rather than a number, but the math is still exactly the same).

11.1: Sequences First of all (and perhaps most importantly), stop convincing yourself that this chapter is so hard…it’s not. Now, you should of course understand the concept of a sequence of numbers, and what it means to find the limit. To find the limit of a sequence, you take

nn

a

lim . If you get a numerical value, then the sequence is convergent, and the value you

got is the limit. If not (i.e. you get or the value oscillates and hence doesn’t exist), then the sequence is divergent. Easy as that. To help you evaluate these limits, you are of course expected to know basic limit rules from Calc I/Intro Calc (L’Hopital’s Rule is NOT required since it isn’t covered in Intro Calc).

11.2: Series Again, understand the basic concept of a series, which is just a sum of terms. You should

know and be able to recognize a geometric series

1

21

n

n araraar (although

you may have to manipulate it a bit to get it into this form). Next, be able to determine if a given geometric series converges or diverges. This depends on the value of r… 1|| r means convergent, while 1|| r means divergent, since the terms are getting bigger. If

it’s convergent, then you should know that the sum is given by

1

1

1n

n

r

aar (actually,

this is on the formula sheet). Also in this section, know the Divergence Test (If nn

a

lim

does not exist or if 0lim n

na , then the series

1nna is divergent, since we’ll be adding up

infinitely-many non-zero terms, so the sum will explode). Note that it’s not called the DIVERGENCE test for nothing…it can only be used to prove divergence, not convergence. If 0lim

nn

a , this tells you that the test is inconclusive.

11.6: Absolute Convergence and the Ratio and Root Tests There are two different tests in this section: the ratio test, and the root test, but both have the same criteria/conclusions…if the result of the test is a number greater than 1, the series diverges; if the result of the test is less than 1, the series converges; if the result of the test is 1, the test is inconclusive. Both tests are actually testing for absolute convergence (hence, don’t forget the absolute values), but if a series converges absolutely, then it is convergent, which is all we’re interested in anyways. Know how to

apply the ratio test, which involves finding n

n

n a

a 1lim

. To find 1na , take na and change

all occurrences of n to n+1. Then divide by na , simplify (lots of terms usually cancel

out), and then take the limit as n . For the root test, we’re taking nn

na

lim , so this

test works best when all of the terms have an nth power.

11.8: Power Series A power series is a representation of a function as a sum of power functions. In this section, we are interested only in finding when a given power series converges. To do this, we apply the ratio or root test. This means that this section is EXACTLY the same as 11.6, except that there is the variable x as part of the series. In section 11.6, when we applied the ratio or root test, we got a number as a result, and that number was either less, greater, or equal to 1. Now, since we also have a variable as part of the answer, the value of the variable can affect whether or not the series converges. There are three possibilities that may occur when we apply the ratio or root test for a power series: 1. We get an answer of 0. Well, 0 is obviously less than 1 for all x, so that means that the power series converges for all x…in other words, its radius of convergence is . 2. We

get an answer of . Well, that’s obviously greater than 1 for all x, so this is always divergent, and the radius of convergence is then 0. 3. Finally, it’s possible that we get some answer involving x centred around some point e.g. 54 x . In this case, this

converges when 154 x , which means that 4

15 x ; so here, the radius of

convergence is 4

1 (i.e. x can move up to a distance of

4

1 in either direction from the point

at which the power series is centred, which is in this case 5, and the series will still converge).

11.10 (incl 11.11): Taylor and Maclaurin Series To try to represent a given function as a power series, we use the Taylor series formula.

You should know that, within the radius of convergence,

0

)(

)(!

)()(

n

nn

axn

afxf is

the Taylor series formula (while this will be on the formula sheet, it’s something that you should know/memorize anyways for your future studies). The Maclaurin Series is the exact same thing, but for the special case 0a (so, for a Taylor series question, I would give you a value of ‘a’, but if I tell you to find the Maclaurin series, then you should know that this means 0a ). To find the Taylor/Maclaurin series, plug in the appropriate value of ‘a’ into the formula. The only tricky part is to figure out )()( af n , which is the nth derivative of f, evaluated at “a”…to find this, find the first few derivatives of the function—there should be some sort of clear pattern (don’t multiply out the numbers…it makes the pattern harder to find…e.g. 2*2*2 is easier to work with than 8 here, because you can tell that there’s an extra factor of 2 each time you differentiate for instance); use this to establish a formula for the nth derivative, and then evaluate at “a”. Finally, plug everything into the above formula. Not as difficult as you think, just practice! You can also use the first few terms of a Taylor/Maclaurin series to approximate a function, which can be useful in evaluating difficult integrals or replace a function with its series representation to evaluate tricky limits. Finally, the only other thing in this section is to be able to find an estimate of the error (remainder) obtained when you approximate a function by only a finite number of terms in the Taylor series. The formula is

1

)!1()(

n

n axn

MxR , which gives an upper bound on this error, and it will be

provided on the formula sheet. Here, Mxf n )()1( for dax . Thus, to find an

upper bound on the error, you need to substitute in the given value of n (i.e. the order of the approximation) and a maximum value for how far x may get from the centre of the series a (this is ax ) on the given interval. To find M, find the form for the n+1

derivative of the function, and determine how big it can get on the given interval for x. After substituting everything in, you can solve for )(xRn . Alternatively, you may be

given a desired accuracy and instead be asked to solve for n.

NOTE: For Chapter 11, watch your notation. You must demonstrate that you understand the theorems. The tests for convergence/divergence all have different criteria, so if you perform a test and get a numerical answer, and then just say it converges or diverges, you will NOT receive full marks…you must explain why it converges or diverges based on your test result. E.g. For Divergence Test, if you get 5 when you take

nn

a

lim , then you must state that 5 is not equal to 0, hence it diverges, etc.