Calculus For Dummies®

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Generally speaking, when a quadratic factor is raised to the nth power, add n partial fractions. For example: The number of distinct quadratic factors in the denominator tells you how many partial fractions you get. So in this example, two factors in the denominator yield two partial fractions. Next, combine similar terms (using x as the variable by which you judge similarity). This is just algebra: For each squared linear factor in the denominator, add two partial fractions in the following form:

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Did you know we don’t learn calculus the way Newton and Leibniz discovered it? They used intuitive ideas of “fluxions” and “infinitesimals” which were replaced with limits because “Sure, it works in practice. But does it work in theory?”. Because this equation works for all values of x, you now take what appears to be a questionable step, breaking this equation into three separate equations as follows: Many calculus examples are based on physics. That’s great, but it can be hard to relate: honestly, how often do you know the equation for velocity for an object? Less than once a week, if that. It makes more sense to think about these problems in terms of division: area equals base times height, so the height of the mean value rectangle equals its area divided by its base.

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if I had to design a mechanism for the express purpose of destroying a child’s natural curiosity and love of pattern-making, I couldn’t possibly do as good a job as is currently being done — I simply wouldn’t have the imagination to come up with the kind of senseless, soul-crushing ideas that constitute contemporary mathematics education.” If you remember that, you can easily remember that the integral on the right is just like the one on the left, except with the u and v reversed.

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First, you’ve got to split up the integrand into a u and a dv so that it fits the formula. For this problem, choose ln(x) to be your u. Then, everything else is the dv, namely Here’s an example. What’s the average speed of a car between t = 9 seconds and t = 16 seconds whose speed in feet per second is given by the function,So the speed is 10 + 5Δt m/s, and Sam thinks about that Δt value ... he wants Δt to be so small it won't matter ... so he imagines it shrinking towards zero and he gets: Calculus for Beginners and Artists Chapter 0: Why Study Calculus? Chapter 1: Numbers Chapter 2: Using a Spreadsheet Chapter 3: Linear Functions Chapter 4: Quadratics and Derivatives of Functions Chapter 5: Rational Functions and the Calculation of Considering learning Calculus, READ THE REVIEW ) (Also, I recommend reading this 2 times, first read the book understanding the basics and concept, then second time you would actually read the book.) -you'll see why that in a minute Because -2 is in the left-most region on the number line below, and because the second derivative at -2 equals negative 240, that region gets a negative sign in the figure below, and so on for the other three regions. We’ve forgotten that math is about ideas, not robotically manipulating the formulas that express them. Ok bub, what’s your great idea?

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Imagine teaching art like this: Kids, no fingerpainting in kindergarten. Instead, let’s study paint chemistry, the physics of light, and the anatomy of the eye. After 12 years of this, if the kids (now teenagers) don’t hate art already, they may begin to start coloring on their own. After all, they have the “rigorous, testable” fundamentals to start appreciating art. Right? When you start out with a linear factor, using partial fractions leaves you with an integral in the following form: It’s because the little band width is slanted instead of horizontal (in which case it would be just dx). The fact that it’s slanted makes it work like the hypotenuse of a little right triangle. The fancy-looking expression for the width of the band comes from working out the length of this hypotenuse with the Pythagorean Theorem. That should make you feel a lot better! Use the chain rule when the argument of the function you’re differentiating is more than a plain old x. A positive sign on this sign graph tells you that the function is concave up in that interval; a negative sign means concave down. The function has an inflection point (usually) at any x-value where the signs switch from positive to negative or vice versa.

We’re looking at the sweetness of sugar from the level of brain-chemistry, instead of recognizing it as Nature’s way of saying “This has lots of energy. Eat it.” Most math teachers have at least a shred of mercy in their hearts, so they don’t tend to give you problems that include this most difficult case. When you start out with a quadratic factor of the form (ax2 + bx + C), using partial fractions results in the following integral: The mere thought of having to take a required calculus course is enough to make legions of students break out in a cold sweat. Others who have no intention of ever studying the subject have this notion that calculus is impossibly difficult unless you happen to be a direct descendant of Einstein. Well, the good news is that you can master calculus. It's not nearly as tough as its mystique would lead you to think. Much of calculus is really just very advanced algebra, geometry, and trig. It builds upon and is a logical extension of those subjects. If you can do algebra, geometry, and trig, you can do calculus.

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Next, you differentiate u to get your du, and you integrate dv to get your v. Finally, you plug everything into the formula and you’re home free. Unfortunately, x2 + 3 has no root in the real numbers, so you need a different approach. First, get rid of the parentheses on the right side of the equation: Variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration. For each quadratic factor in the denominator that’s raised to the third power, add three partial fractions in the following form: You can locate a function's concavity (where a function is concave up or down) and inflection points (where the concavity switches from positive to negative or vice versa) in a few simple steps. The following method shows you how to find the intervals of concavity and the inflection points ofTo start out, see how far you can get by plugging in the roots of equations. Begin by getting a common denominator on the right side of the equation: This time, you add 0 to the integral, which doesn’t change its value. At this point, you can split the integral in two: