Calculus ii workbook for dummies pdf

It covers intermediate calculus topics in plain English, featuring in-depth coverage of integration, including substitution, integration techniques and when to use them, approximate integration and improper integrals.

This is a collection of my Calculus II midterm exam problems. The solutions are written by me using methods taught during lecture. For further explanation as to the why behind the methods, please see CalcCoach. There you will find my lecture notes, lecture videos, and premium problem solution videos explaining in detail the thought process involved in solving different problems.

If your goal is to gain a good understanding of the topics typically found in a Calculus II class, then the combination of this workbook and the other three components found on CalcCoach. By breaking down differentiation and integration into digestible concepts, this guide helps you build a stronger foundation with a solid understanding of the big ideas at work.

This user-friendly math book leads you step-by-step through each concept, operation, and solution, explaining the "how" and "why" in plain English instead of math-speak. Through relevant instruction and practical examples, you'll soon learn that real-life calculus isn't nearly the monster it's made out to be.

Calculus is a required course for many college majors, and for students without a strong math foundation, it can be a real barrier to graduation. Breaking that barrier down means recognizing calculus for what it is—simply a tool for studying the ways in which variables interact. It's the logical extension of the algebra, geometry, and trigonometry you've already taken, and Calculus For Dummies, 2nd Edition proves that if you can master those classes, you can tackle calculus and win.

Includes foundations in algebra, trigonometry, and pre-calculus concepts Explores sequences, series, and graphing common functions Instructs you how to approximate area with integration Features things to remember, things to forget, and things you can't get away with Stop fearing calculus, and learn to embrace the challenge. With this comprehensive study guide, you'll gain the skills and confidence that make all the difference.

Calculus For Dummies, 2nd Edition provides a roadmap for success, and the backup you need to get there. Plus, an online component provides you with a collection of calculus problems presented in multiple-choice format to further help you test your skills as you go. Gives you a chance to practice and reinforce the skills you learn in your calculus course Helps you refine your understanding of calculus Practice problems with answer explanations that detail every step of every problem The practice problems in Calculus Practice Problems For Dummies range in areas of difficulty and style, providing you with the practice help you need to score high at exam time.

This chapter gets down to the nitty-gritty and presents several techniques for calculating the answers to limits problems. Easy Limits A few limit problems are very easy. Okay, so are you ready? Limits to memorize You should memorize the following limits.

If you fail to memorize the limits in the last three bullets, you could waste a lot of time trying to figure them out. The limit is simply the func- tion value. Beware of discontinuities. In that case, if you get a number after plugging in, that number is not the limit; the limit might equal some other number or it might not exist. See Chapter 7 for a description of piecewise functions. What happens when plugging in gives you a non-zero number over zero?

This is the main focus of this section. These are the interesting limit problems, the ones that likely have infinitesimal holes, and the ones that are important for differ- ential calculus — you see more of them in Chapter 9. Note on calculators and other technology. With every passing year, there are more and more powerful calculators and more and more resources on the Internet that can do calculus for you.

A calculator like the TI-Nspire or any other calculator with CAS — Computer Algebra System can actually do that limit problem and all sorts of much more difficult calculus problems and give you the exact answer. The same is true of websites like Wolfram Alpha www. Many do not allow the use of CAS calculators and com- parable technologies because they basically do all the calculus work for you. Method one The first calculator method is to test the limit function with two numbers: one slightly less than the arrow-number and one slightly more than it.

The result, 9. Since the result, The answer is 10 almost certainly. Method two 2 The second calculator method is to produce a table of values. Hit the Table button to produce the table. Now scroll up until you can see a couple numbers less than 5, and you should see a table of values something like the one in Table These calculator techniques are useful for a number of reasons. Your calculator can give you the answers to limit problems that are impossible to do alge- braically.

Also, for problems that you do solve on paper, you can use your calculator to check your answers. This gives you a numerical grasp on the problem, which enhances your algebraic understand- ing of it.

If you then look at the graph of the function on your calculator, you have a third, graphical or visual way of thinking about the problem. Many calculus problems can be done algebraically, graphically, and numerically.

When possible, use two or three of the approaches. Each approach gives you a different perspective on a problem and enhances your grasp of the relevant concepts.

Gnarly functions may stump your calculator. By the way, even when the non-CAS-calculator methods work, these calcu- lators can do some quirky things from time to time.

This can result in answers that get further from the limit answer, even as you input numbers closer and closer to the arrow-number. Try plugging 5 into x — you should always try substitution first.

You get 0 — no good, on to plan B. Now substitution will work. And note that the limit as x approaches 5 is 10, which is the height of the hole at 5, Conjugate multiplication — no, this has nothing to do with procreation Try this method for fraction functions that contain square roots. Try this one: Evaluate lim. Try substitution. Plug in 4: that gives you 0 — time for plan B. Definition of conjugate: The conjugate of a two-term expression is just the same expression with subtraction switched to addition or vice versa.

The product of conjugates always equals the first term squared minus the second term squared. Now do the rationalizing. Now substitution works. Plug in 0: That gives you 0 — no good. The best way to understand the sandwich or squeeze method is by looking at a graph. Look at functions f, g, and h in Figure g is sandwiched between f and h.

The limit of both f and h as x approaches 2 is 3. So, 3 has to be the limit of g as well. Figure The sandwich method for solv- ing a limit. Functions f and h are the bread, and g is the salami. Plug 0 into x. Try the algebraic methods or any other tricks you have up your sleeve. Knock yourself out. Plan C. Try your calculator. It definitely looks like the limit of g is zero as x approaches zero from the left and the right.

Table gives some of the values from the calculator table. Table Table of Values for x g x 0 Error 0. Get it? To do this, make a limit sand- wich. Fooled you — bet you thought Step 3 was the last step. Because the range of the sine function is from negative 1 to positive 1, whenever you multiply a number by the sine of anything, the result either stays the same distance from zero or gets closer to zero. Thus, will never get above x or below. Figure shows that they do.

The function is thus continuous everywhere. Figures and and discussed in the sec- tion about making a limit sandwich. If we now alter it or connect to 0, 0 from the left or the right?

The function is now drive through the origin, are you on one of the continuous everywhere; in other words, it has up legs of the road or one of the down legs? But at 0, 0 , it seems to contradict Neither seems possible because no matter how the basic idea of continuity that says you can close you are to the origin, you have an infinite trace the function without taking your pencil number of legs and an infinite number of turns off the paper.

There is no last turn before you reach 0, 0. Now, keeping the line vertical, slowly between you and 0, 0 is infinitely long! It winds up and down with such increas- you pass over 0, 0. There are no gaps in the ing frequency as you get closer and closer to function, so at every instance, the vertical line 0, 0 that the length of your drive is actually infi- crosses the function somewhere. On this long and function.

How can winding road. And because you reconcile all this? I wish I knew. This is confirmed by con- sidering what happens when you plug bigger and bigger numbers into 1 : x the outputs get smaller and smaller and approach zero. Determining the limit of a function as x approaches infinity or negative infinity is the same as finding the height of the horizontal asymptote.

That quo- tient gives you the answer to the limit problem and the height of the 3 asymptote. A horizontal asymptote occurs at this same value. If you have any doubts that the limit equals 0. All you see is a column of 0. Try substitution — always a good idea. No good. On to plan B. Now do the conjugate multiplication. Remember those problems from algebra? One train leaves the station at 3 p. Two hours later another train leaves going east at 50 mph. You can handle such a problem with algebra because the speeds or rates are unchanging.

Think about putting man on the moon. Apollo 11 took off from a moving launch pad the earth is both rotating on its axis and revolving around the sun. On top of all that, the rocket had to hit a moving target, the moon.

All of these things were changing, and their rates of change were changing. Much of modern economic theory, for example, relies on differentiation. In economics, everything is in constant flux. Prices go up and down, supply and demand fluctuate, and inflation is constantly changing. These things are con- stantly changing, and the ways they affect each other are constantly chang- ing. You need calculus for this. Differential calculus is one of the most practical and powerful inventions in the history of mathematics.

The derivative is just a fancy calculus term for a simple idea you know from algebra: slope. Slope, as you know, is the fancy algebra term for steepness. And steepness is the fancy word for. These are big words for a simple idea: Finding the steepness or slope of a line or curve. Throw some of these terms around to impress your friends. By the way, the root of the words differ- ential and differentiation is difference — I explain the connection at the end of this chapter in the section on the difference quotient.

A steepness of 1 means that as the stickman walks one 2 foot to the right, he goes up 1 foot; where the steepness is 3, he goes up 3 2 feet as he walks 1 foot to the right. This is shown more precisely in Figure Negative slope: To remember that going down to the right or up to the left is a negative slope, picture an uppercase N, as shown in Figure Figure This N line has a Negative slope.

How steep is a flat, horizontal road? Not steep at all, of course. Zero steepness. So, a horizontal line has a slope of zero. Like where the stick man is at the top of the hill in Figure There are more. Variety is not the spice of the wall trying to figure out things like why some 2 author uses one symbol one time and a different mathematics.

When mathematicians decide symbol another time, and what exactly does the on a way of expressing an idea, they stick d or f mean anyway, and so on and so on, or 2 to it — except, that is, with calculus. Hold on to your hat. I all mean exactly the same thing: or df or strongly recommend the second option. I realize that no calculation is necessary here — you go up 2 as you go over 1, so the slope is automatically 2.

But bear with me because you need to know what follows. Now, take any two points on the line, say, 1, 5 and 6, 15 , and figure the rise and the run. You rise up 10 from 1, 5 to 6, 15 because 5 plus 10 is 15 or you could say that 15 minus 5 is And you run across 5 from 1, 5 to 6, 15 because 1 plus 5 is 6 or in other words, 6 minus 1 is 5. Table shows six points on the line and the unchanging slope of 2. That was a joke. So why did I start the chapter with slope?

Because slope is in some respects the easier of the two concepts, and slope is the idea you return to again and again in this book and any other calculus textbook as you look at the graphs of dozens and dozens of functions. A slope is, in a sense, a picture of a rate; the rate comes first, the picture of it comes second. Just like you can have a function before you see its graph, you can have a rate before you see it as a slope. Calculus on the playground Imagine Laurel and Hardy on a teeter-totter — check out Figure Figure Laurel and Hardy — blithely unaware of the calculus implications.

Assuming Hardy weighs twice as much as Laurel, Hardy has to sit twice as close to the center as Laurel for them to balance. And for every inch that Hardy goes down, Laurel goes up two inches. So Laurel moves twice as much as Hardy. A derivative is a rate. You can see that if Hardy goes down 10 inches then dH is 10, and because dL equals 2 times dH, dL is 20 — so Laurel goes up 20 inches. But a rate can be anything per anything. Again, a derivative just tells you how much one thing changes compared to another.

It tells you that for each mile you go the time changes 1 of an hour. We just saw miles per hour and hours per mile. Rates can be constant or changing. In either case, every rate is a derivative, and every derivative is a rate. The rate-slope connection Rates and slopes have a simple connection. All of the previous rate examples can be graphed on an x-y coordinate system, where each rate appears as a slope.

Consider the Laurel and Hardy example again. Laurel moves twice as much as Hardy. The line goes up 2 inches for each inch it goes to the right, and its slope is thus 2 , or 2. One last comment. Well, you can think of dL as the run rise and dH as the run. That ties everything together quite nicely. The Derivative of a Curve The sections so far in this chapter have involved linear functions — straight lines with unchanging slopes. Calculus is the mathematics of change, so now is a good time to move on to parabolas, curves with changing slopes.

You can see from the graph that at the point 2, 1 , the slope is 1; at 4, 4 , the slope is 2; at 6, 9 , the slope is 3, and so on. Unlike the unchanging slope of a line, the slope of a parabola depends on where you are; it depends on the x-coordinate of wherever you are on the parabola. Table shows some points on the parabola 2 and the steepness at those points. Beginning with the original function, x , take the power and put it in 2 4 front of the coefficient.

Reduce the power by 1. In this example, the 2 becomes a 1. So the derivative is 1 x 1 or just 1 x. The Difference Quotient Sound the trumpets! You come now to what is perhaps the cornerstone of dif- ferential calculus: the difference quotient, the bridge between limits and the derivative.

Okay, so here goes. I keep repeating — have you noticed? You learned how to find the slope of a line in algebra. In Figure , I gave you the slope of the parabola at several points, and then I showed you the short-cut method for finding the derivative — but I left out the important math in the middle. That math involves limits, and it takes us to the threshold of calculus.

For a line, this is easy. You just pick any two points on the line and plug them in. You can see the line drawn tangent to the curve at 2, 4. Figure shows the tangent line again and a secant line intersecting the parabola at 2, 4 and at 10, Definition of secant line: A secant line is a line that intersects a curve at two points.

Now add one more point at 6, 36 and draw another secant using that point and 2, 4 again. Now, imagine what would happen if you grabbed the point at 6, 36 and slid it down the parabola toward 2, 4 , dragging the secant line along with it. Can you see that as the point gets closer and closer to 2, 4 , the secant line gets closer and closer to the tangent line, and that the slope of this secant thus gets closer and closer to the slope of the tangent?

So, you can get the slope of the tangent if you take the limit of the slopes of this moving secant. When the point slides to 2.

Sure looks like the slope is headed toward 4. As with all limit problems, the variable in this problem, x2 , approaches but never actually gets to the arrow-number 2 in this case. Herein lies the beauty of the limit process. A fraction is a quotient, right? You may run across other, equivalent ways. Figure is the ultimate figure for.

Chapter 9: Differentiation Orientation Have I confused you with these two figures? They both show the same thing. Both figures are visual representations of. Figure shows this general definition graphically. Note that Figure is virtually identical to Figure except that xs replace the 2s in Figure and that the moving point in Figure slides down toward any old point x, f x instead of toward the specific point 2, f 2.

Plug any number into x, and you get the slope of the parabola at that x-value. Figure sort of sum- marizes in a simplified way all the difficult preceding ideas about the differ- ence quotient. Like Figures , , and , Figure contains a basic slope stair-step, a secant line, and a tangent line. The slope of the tangent line is. If, for example, the y-coordinate tells you distance traveled in miles , and the x-coordinate tells you elapsed time in hours , you get the familiar rate of miles per hour.

That slope is the average rate over the interval from x1 to x2. When you take the limit and get the slope of the tangent line, you get the instan- taneous rate at the point x1 , y1. Again, if y is in miles and x is in hours, you get the instantaneous speed at the single point in time, x1.

Because the slope of the tangent line is the derivative, this gives us another definition of the derivative. To Be or Not to Be? By now you certainly know that the derivative of a function at a given point is the slope of the tangent line at that point. These types of discontinuity are dis- cussed and illustrated in Chapter 7. Continuity is, therefore, a necessary condition for differentiability. Dig that logician-speak. See function f in Figure See function g in Figure Inflection points are explained in Chapter You also now know the mathematical foundation of the derivative and its technical definition involving the limit of the difference quotient.

Some of this material is unavoidably dry. If you have trouble staying awake while slogging through these rules, look back to the last chapter and take a peek at the next three chapters to see why you should care about mastering these differentiation rules. You may want to order up a latte with an extra shot. Learning these first half dozen or so rules is a snap. If you get tired of this easy stuff, however, I promise plenty of challenges in the next section. The constant rule This is simple.

End of story. To find its derivative, take the power, 5, bring it in front of the x, and then reduce the power by 1 in this example, the power becomes a 4.

To repeat, bring the power in front, then reduce the power by 1. Instead of all that, just use the power rule: Bring the 2 in front, reduce the power by 1, which leaves you with a power of 1 that you can drop because a power of 1 does nothing. But the difference quotient is included in every calculus book and course because it gives you a fuller, richer understanding of calculus and its foun- dations — think of it as a mathematical character builder.

Or because math teachers are sadists. You be the judge. Make sure you remember how to do the derivative of the last function in the above list. The slope m of this line is 1, so the derivative equals 1. Or you can just memorize that the derivative of x is 1.

But if you forget both of these ideas, you can always use the power rule. Rewrite functions so you can use the power rule. Makes no difference. A coefficient has no effect on the process of differentiation. You just ignore it and differentiate according to the appropriate rule.

The coef- ficient stays where it is until the final step when you simplify your answer by multiplying by the coefficient. Solution: You know by the power rule that the derivative of x 3 is 3x 2, so the derivative of 4 x 3 is 4 3x 2.

The 4 just sits there doing nothing. Then, as a final step, you simplify: 4 3x 2 equals 12x 2. By the way, most people just bring the 3 to the front, like this: , which gives you the same result. Constants in problems, like c and k also behave like ordinary numbers. Be sure to treat them like regular numbers. Solution: Just use the power rule for each of the first four terms and the con- stant rule for the final term. You still differentiate each term separately. The addition and subtraction signs are unaffected by the differentiation.

A third option is to use the following mnemonic trick. Write these three down, and below them write their cofunctions: csc, csc, cot. Put a negative sign on the csc in the middle. Look at the top row. The bottom row works the same way except that both derivatives are negative. Differentiating exponential and logarithmic functions Caution: Memorization ahead. This is a special function: ex and its multiples, like 5ex, are the only functions that are their own derivatives.

Think about what this means. See Chapter 4 if you want to brush up on logs. Okay, ready for a challenge? The following rules, especially the chain rule, can be tough. The trick is knowing the order of the terms in the numerator. And is it more natural to begin at the top or the bottom of a fraction? The top, of course. So the quotient rule begins with the derivative of the top. If you remember that, the rest of the numerator is almost automatic. Note that the product rule begins with the derivative of the first function you read as you read the product of two functions from left to right.

In the same way, the quo- tient rule begins with the derivative of the first function you read as you read the quotient of two functions from top to bottom. All four of these functions can be written in terms of sine and cosine, right? See Chapter 6. The other three functions are no harder. Give them a try. To make sure you ignore the inside, tempo- rarily replace the inside function with the word stuff.

All chain rule problems follow this basic idea. You do the derivative rule for the outside function, ignoring the inside stuff, then multiply that by the derivative of the stuff.

Differentiate the inside stuff. Now put the real stuff and its derivative back where they belong. The outside function is the sine function, so you start there, taking the derivative of sine and ignoring the inside stuff, x 2. The derivative of sin x is cos x, so for this problem, you begin with cos stuff 2. Multiply the derivative of the outside function by the derivative of the stuff.

Parentheses are your friend. For chain rule problems, rewrite a composite function with a set of parentheses around each inside function, and rewrite trig functions like sin2 x with the power outside a set of parentheses: sin x 2. Okay, now that you know the order of the functions, you can differentiate from outside in: 1. The outermost function is stuff 3 and its derivative is given by the power rule.

Use the chain rule again. The derivative of sin x is cos x, so the derivative of sin stuff begins with cos stuff. Plug those things back in. This can be simplified a bit.

It may have occurred to you that you can save some time by not switching to the word stuff and then switching back. Make sure you. The argument of this natural logarithm function is x 3. This rule tells you to put the argument dx x of the function in the denominator under the number 1. You then finish the problem by x multiplying that by the derivative of x 3 which is 3x 2.

Final answer after sim- plifying: 3. In the example above, ln x 3 , you first use the natural log rule, then, as a separate step, you use the power rule to differentiate x 3. At no point in any chain rule problem do you use both rules at the same time.

For example, with ln x 3 , you do not use the natural log rule and the power rule at the same time to come up with 1 2.

The rate of movement of the noise- the chain rule is based on a very simple idea. So, how fast is riding a bike. If the biker goes four times as fast dL is the noisemaker moving compared to Hardy?

The noisemaker is the walker, then the biker goes 4 times 2, or 8 times moving 3 times as fast as Laurel, and Laurel is as fast as the walker, right? Here it is in symbols note that this is the on a teeter-totter? Recall that for every inch same as the formal definition of the chain rule Hardy goes down, Laurel goes up 2 inches. Differentiate 4x 2 sin x 3. This problem has a new twist — it involves the chain rule and the product rule. How should you begin? Where do I begin? Your last computation tells you the first thing to do.

Say you plug the number 5 into the xs in 4x 2 sin x 3. Because your last computation is multiplication, your first step in differentiating is to use the product rule.

Remember the product rule? In such cases, y is written explicitly as a function of x. This means that the equation is solved for y; in other words, y is by itself on one side of the equation. For such a problem, you need implicit differentiation.

Remember using the chain rule to differentiate something like sin x 3 with the stuff technique? With implicit differentiation, a y works like the word stuff. But the concept is exactly the same, and you treat y just like the stuff. Here goes. Differentiate each term on both sides of the equation. For the second term, you use the regular power rule.

And for the third term, you use the regular sine rule. Divide for the final answer. So, if you want to evaluate the derivative to get the slope at a particular point, you need to have values for both x and y to plug into the derivative.

Either way is fine. Take your pick. Now, you could multiply the whole thing out and then differentiate, but that would be a huge pain. Or you could use the product rule a few times, but that would also be too tedious and time-consuming.

The better way is to use loga- rithmic differentiation: 1. Take the natural log of both sides. Now use the property for the log of a product, which you remember of course if not, see Chapter 4. Differentiate both sides. The right side of the first line gets multiplied by the second line. Figure The graphs of inverse functions, f x and g x. As with any pair of inverse functions, if the point 10, 4 is on one function, 4, 10 is on its inverse.

And, because of the symmetry of the graphs, you can see that the slopes at those points are reciprocals: At 10, 4 the slope is 1 and at 4, 10 the slope is 3 3. The algebraic explanation is a bit trickier, however. The point 4, 10 on g can be written as 4, g 4. Okay, so maybe it was a lot trickier. Scaling the Heights of Higher Order Derivatives Finding a second, third, fourth, or higher derivative is incredibly simple. The second derivative of a function is just the derivative of its first derivative.

This site is like a library, Use search box in the widget to get ebook that you want. Pre-Calculus Workbook For Dummies is the perfect tool for anyone who wants or needs more review before jumping into a calculus class.

You'll get guidance and practical exercises designed to help you acquire the skills needed to excel in pre-calculus and conquer the next contender-calculus. The theorem and how it applies to special right triangles are set out here:.

In pre-calculus you deal with inequalities and you use interval notation to express the solution set to an inequality. The following formulas show how to format solution sets in interval notation.