To: Canuck Dave who wrote (156710 ) 5/5/2009 2:55:46 PM From: E. Charters Read Replies (1) | Respond to of 312733 calculus.org I have a super short calculus course. If all the points on your x-y curve fit the formula, and there are no undefined points, then there is an area under that curve which equates to that under a straight line drawn thru some y height of the graph and drawn between beginning and end points of your integration. This is called Green's theorem and it makes perfect sense. To every continuous function there is a simple average function that equates to it. Integration is the opposite of derivation. Derivation is the tangency to any point on the curve, which is point at which the limit of the function is reached. The limit of the function, is demonstrated by two factors. When you have lost all patience trying to figure it out, and where the value of the function is when delta x and delta y are so close to zero god can't see the difference anymore. For Integration you just flip derivation to the enemy side. If that is too hard thing of a bunch of skyscrapers that reach the height of the function at some X point. As the skyscrapers get narrower and narrower and closer and closer together eventually they disappear, but there are an infinite number of them. So infinitely narrow infinite number of skyscrapers times their height equals the area a windwasher would wash of all the skyscrapers is the area under the curve. As some have figured out, the amount the skyscrapers poke over the curve and lose under the curve with their square tops should even out positive and negative, if they are small enough. An exercise in visual calculus is to determine if this is true. In engineering we can ignore it. There is no really good method of taking limits. It is enough for 99% of injunirs to know the concept and use Leibniz and Heavyside. The rules work. When in doubt use tables. All the integrations are in the back of the math book. Differential calculus is conceptual and ridiculously easy. If you take integrals you can equate them if parts are unknown. If you take differentials, and realize that derivative is the slope which is the rate of change or the rise over the run, the amount y changes with x at any point, or the slope of a point, or the tangent to the line of the function, then you should be able to equate that in some way to other values. the rate of change of distance with time is velocity. The rate of change of velocity with time is acceleration. If these are derivatives of the curves of each relation of one variable with another, then the opposite must be the integral. So the integral of velocity (the opposite of the derivative of a distance over time) -- with respect to time is distance. So what is a foot-second? As a final test, the only question is, is calculus confusing and hard? Students get top marks for "yes". You get one mark for your name in the top right hand corner, and half marks for "I don't know". A bonus question is, "How many times will the average engineer use calculus in this career?" The correct answer is, "the limit of this equation y=e(-x) as x --> infinity." EC<:-}