What is the difference between differential and integral calculus. any practical example?
353544
http://mathforum.org/library/drmath/sets/high_calculus.html\n\nask here...:)
266181
dy/dx = x/y\n\nso y dy=x dx\n\nintegrate both sides u get....\n\n(y^2)/2 = (x^2)/2 + c\n\nwhere c is a constant.
442091
The integral of a function can be defined as the limit of those rectangles as they get smaller and smaller, but I think what you are asking is--why is this connected to the anti-derivative?  Why is it that you can take the anti-derivative of a function, plug in at the endpoints, and you get the area under the curve?\n\nThis is a hard question to explain; it's generally called the "Fundamental Theorem of Calculus", and most textbooks prove it in detail.  You can also check out the wikipedia article below, although it's certainly not easy to understand.\n\nIf I could sum it up in a nutshell I would say that the reason why it works is that taking area under a curve is like adding up lots of little pieces, but taking the derivative (slope) is like subtraction...little tiny pieces at a time.  Just as addition and subtraction are opposites, differentiation (taking slopes) and integration (taking areas) are opposites.\n\nIt also helps to play around with the two functions f(x)=1, and f(x)=x.  Sometimes experimenting with these simple functions can help give you a feel for why this crazy connection really works.
99672
honestly....good luck\n\ntry www.mathworld.com\n\ndo a search for differential equations, i think that would be your best bet...but that can get confusing...good luck
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
786528
Integration is one of the two major operations of calculus (the other being the inverse, differentiation. There are indefinite and definite types of integration. When the integral of a function is expressed symbolically (as another function), this is the indefinite integral (e.g., the ind. int. of y=2x is y=x^2 + a constant). When the integral is numerically evaluated over some range of x, this is the definite integral. Much of time-domain simulation consists of successive evaluation and accumulation of the definite integral of some function, or "time stepping". Typically, each time step starts with evaluation of the 2nd derivative of something, let's say position (whose 2nd deriv. is acceleration), accumulation of the 1st deriv. (velocity) as the previous value + time-step size * 2nd deriv., then accumulation of the 0th deriv. (position) as the previous value + time-step size * 1st deriv. Of course it's a bit more complex than that, and many integration formulas exist to minimize errors.\nAn "integration" function could be something like what I've described for time-stepping, or it could be a much more sophsticated function that accepts a function definition and gives you the indefinite integral in symbolic form. So you need more information about what kind of integration function is desired.
982897
the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
219346
The first link below is a problem discussing improper integrals (which classifies all integrals with an infinite limit of integration). Just scroll to the bottom of the page--the top of the page is a different problem altogether. Basically, the tip is to replace the "troublesome" limit values with a constant (like k) and then evaluate the limit as k approaches that value. In this case, k becomes one of the limits of integration and you evaluate the integral as k approaches infinity.\n\nI recommend The Complete Idiot's Guide to Calculus (follow the second link in the Sources list below) when brushing off calculus rust, and not only because I am the author of the book. :) It's a light-hearted but content-heavy review of calculus, beginning with advanced algebra review and ending with function approximation by infinite series approximation.\n\nAlso, spend some time with "Tutorials for the Calculus Phobe," free Flash-based calculus tutorials on my website, Calculus-Help.com. They're fun but help you conceptualize calculus basics. The third and final link in the sources houses the Phobe tutorials.\n\nBest of luck!
794869
Here's a good website that discuss implicit differentiation with some example problems.\n\nHope this helps
45131
Calculus was confusing enough in English - I can't imagine taking it in Japanese.  You might want to try to get hold of a textbook or something in English.\n\nIt's an integral.  More specifically an indefinite integral.  It's kind of like the inverse of the derivative function except for that there is no way to get the constant back (hence the "C").  Basically the indefinite integral is the set of all functions whose derivative is equal to the thing between the integral sign and the dx (or dy, dq, whatever the variable is... I'm blanking on general name of the thing).  The integral symbol tells you it's an integral (obviously), and the "x" in dx tells you what variable you're integrating with respect to (that will make more sense when you learn to integrate more complicated expressions).  So the indefinite integral of x^2 is 1/3x^3 + C with respect to x where C is some constant because the derivative of 1/3x^3 + C is x^2 (remember that the derivative of a constant is zero).\n\n***A Note on Definite and Indefinite Integrals***\nA couple of the answers below appear to be describing the definite integral.  The definite integral can be described as the sum of the areas of rectangles, and can be used to find the area under a curve (among other things).  The indefinite integral (or antiderivative), however, which is the one refered to in the question, describes a function.  Although the function described by the indefinite integral is extremely useful in calculating the area under a curve, it does not directly describe this area.  In addition to being used to caculate the definite integral, the indifinite integral can also be used, for example, to find an equation for the position of an object given an equation for its velocity.\n\nAnother afterthough: I may have gotten some of the terminology wrong as my high school math department had a strange affinity for inventing their own terms.
929142
In electronics, which is my specialty, determining the dynamic response of a circuit (with respect to time or with respect to frequency) often requires the use of differential and integral calculus.  In structural engineering, designing buildings or bridges to make them strong and earthquake resistant also requires calculus.  Certainly physics and mechanics rely heavily upon calculus.  I'm not sure about chemistry or biology.
261953
Calculus \n  \nDuring the plague years Newton laid the foundation for elementary differential and integral CALCULUS, several years before its independent discovery by the German philosopher and mathematician LEIBNIZ. The "method of fluxions," as he termed it, was based on his crucial insight that the integration of a function (or finding the area under its curve) is merely the inverse procedure to differentiating it (or finding the slope of the curve at any point). Taking differentiation as the basic operation, Newton produced simple analytical methods that unified a host of disparate techniques previously developed on a piecemeal basis to deal with such problems as finding areas, tangents, the lengths of curves, and their maximal and minima. Even though Newton could not fully justify his methods--rigorous logical foundations for the calculus were not developed until the 19th century--he receives the credit for developing a powerful tool of problem solving and analysis in pure mathematics and physics. Isaac Barrow, a Fellow of Trinity College and Lucasian Professor of Mathematics in the University, was so impressed by Newton's achievement that when he resigned his chair in 1669 to devote himself to theology, he recommended that the 27-year-old Newton take his place.
770990
I don't understand your statement of the problem.  It sounds like it might be an exercise in integral calculus, in which you integrate the expression at hand, evaluate the resulting function at the two limits, and take the difference.
71756
Here's a classic:\n\nUsing the following partial deriviatives:\n\nInt(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx\n\nwhere G(x): the primitive function of g(x)\n     and f'(x): the derivative of f(x)\n\nNow watch the following 'proof':\n\n    Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx\n           {just take f(x) = 1/x^2 and g(x) = 2x}\n\n    this yields:\n\n    Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx    \n\n    substracting Int(2/x)dx on both sides yields:\n\n    0 = 1
303750
i'm an accountant and i've never had to use calculus...\n\nthat's more mathematics and physics i think...
402001
Watch it ….! \n\ndelta\n\n x =t^2  ---> t= sqrt(x)\ndx = 2t dt .............(not dx = dt /2t )\n-->\ndt = dx/(2 Sqrt(x)\n\nand this Integral can not be done using Analytical Method; Also, the (method of Integration by Part) can not apply in such integral because either of the cos(t) and sqrt(t) have infinitive derivatives.
763570
in calculus, integration is just an extension of sum. for discrete values, we use the summation. for continuous values, we use integration.\n\nindefinite integration is just integrating indefinitely, ie., without any bounds. this usually results in a polynomial or trigonometric equation etc. \n\ndefinite integration is done between two specific points, be it on the x axis or y axis. this results in a constant value, rather than a polynomial or other equations.\nthe applications of definite integration is to find the area under curves, etc..