Why is the integral of the function related to its area?
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.
1050238
Here's a good statement:\n"Every continuous curve in the plane has zero area".\nIt looks like it should be true. But it is FALSE. In fact,there is a continuous curve that fills a whole square! This curve 'intersects itself' infinitely often, so we might then conjecture\n"Every arc (a one-to-one curve) has zero area".\nBut this is FALSE also! \nFinding an example of an arc with strictly positive area is not easy, but it can be done (the usual examples require the use of Cantor sets).\n\nDepending on the level, you might be able to say:\n"Every continuous function is differentiable except at finitely many points".\nThis is also FALSE. There are continuous functions that are differentiable NOWHERE.\nOr you might try:\n"Every differentiable function is monotone on some interval".\nAgain, this is FALSE.\nOr another:\n"A function with derivative zero almost everywhere is constant".\nThe 'Devil's staircase' is a counterexample.
783613
Area = length x width
480203
Area of a circle = pi*r^2\n\nr=radius os the circle
6015
intgral(from - infinity to infinity) of e^(-x^2)= sqrt(pi)\n\nThis is the area under the bell curve. It comes up very often in statistics.
759209
This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
826403
hmmm taking me way back here.  when you break the area under a curve into rectangles of differential width,  either the left top corner will lie on the curve or the right top corner will.  \n\nas you decrease the differential width the area will become more exact until bada bing you are doing calculus.  i think.
208438
There is a delightful article by the late Dr. John Kenny on that, back a few years ago, looking at the formulae for surface area and volumes of spheres; I'll see if I can find it for you--it's in my files SOMEWHERE--and give you the reference, and the formulae he generated. It turns out that they're not precisely the same as what you've got--but not terribly far from it.\n\nAs I said, I'll see if I can find it, and I'll get you the reference and the formulae.\n\nAbout 9pm CST...\n\nOK, here is the formulae, done the best I can on this thing...\n\n\nFor a sphere of dimension 2*k, radius R\n\nVolume: (pi^k)*(R^(2*k))/k!\n\nFor a sphere of dimension 2*k + 1 and radius R\n\nVolume:\n(2^(2*k+1))*k!*(pi^k)* (R^(2*k+1))/(2*k+1)!\n\n\nTo get the surface areas for each of these, all you do is differentiate with respect to R...\n\nAs I said, it was a little more complex than you thought--the surface areas come out a little bit differently for odd and even numbered dimensions.  I do not recall why, frankly.
241144
to calculate the area of a 2-dimensional space you always multiply length x width, or in this case height\n\nC
768242
it means that as a (linear) transformation from the plane to the plane it reverses the orientation.\nthe way we calculate the area, is really a signed area, in the sense that it takes into consideration the orientation of the 2 vectors.\nit is a bit subtle and that is why most teachers do not really explain it.\n'
184296
Differentiation of a function mapping R into R considers the limiting case of a secant to the curve to get the unique tangent line to the curve at each point. Ie., if we have a graph y=f(x), then the slope of a secant line between two points (x,f(x)) and (x+h, f(x+h)) on the curve is [f(x+h) - f(x)]/[x+h - x] = [f(x+h) - f(x)]/h. The tangent line to the curve at the point (x, f(x)) is hypothesized to have the slope given by the limit of the secant's slope as we rotate the secant so that the second intersection point moves closer to (x,f(x)). Algebraically, we want to study the limit of the above quotient as h approaches 0. The usefulness of the tangent at each point is that such a line gives a linear approximation to whatever f(x) stands for at each point. The tangent line also tells you when there is a maximum or minimum, as the line will be horizontal there (for certain well-behaved curves).\nThe first half of your study of calculus will involve justifying this concept and making concrete and mathematically precise what we mean when we say we are going to take the limit of an expression.\nThe second half of your calculus course will concetrate on integration, which studies the problem of calculating the area of arbitrarily shaped regions (instead of just the simple geometrical shapes you've studied before). Given an arbitrary curve y=f(x) or a bunch of curves enclosing a region, you will be able to calculate the area contained in the region. The motivation for the integral also involves limiting behavior. In the case of the Riemann integral, the hypothesis is that you can cover the area between a curve and the x-axis by a finite amount of rectangles, each with a finite area, and as the width of the rectangles get smaller, the difference between their total area and the true area under the curve goes to 0. The rigorous proof of this concept will be given by your textbook or teacher.\nYou will also study the Fundamental Theorem of Calculus, which connects the behavior of integrals to the behavior of derivatives in such a fundamental way that integrals are also known as antiderivatives. This theorem makes evaluating integrals a lot easier.\nAs you study further in higher-dimensional calculus and manifold theory, you will learn more general ways of thinking about derivatives and integrals, in that you will simply view them as forms over fields. We know all about linear algebra; we have to use derivatives to reduce the study of nonlinear forms to our familiar territory of linear algebra. This view is very useful in theoretical physics and differential geometry.\n\nAs some examples, let's study a curve we know well, the parabola y=x^2. To spice this up a little, let the parabola be the path of a bullet fired at some (horizontal) muzzle velocity v. Then from physics we know y(t) = -4.5*t^2 + 1 and horizontally x(t)=vt, so the path of the bullet is given by y(x) = (-4.5/v^2)x^2 + 1. We want to know how hard the bullet will hit the ground, so we need to know the instantaneous velocity of the bullet when y=0, which is when x=v/Sqrt[4.5] and thus t=Sqrt[4.5]. The velocity consists of the speed and direction of the bullet. First we get the direction, which is the slope of y(x) when y=0. Let -4.5/v^2 = k. Thus we want to study the quotient [f(x+h) - f(x)]/h = [ k(x+h)^2 + 1 - kx^2 - 1]/h = k*[x^2 + 2xh + h^2 - x^2]/h = k*[2xh + h^2]/h = k*[2x + h] as h approaches zero, which is just 2kx, so we have 2*(-4.5/v^2)*(v/Sqrt[4.5]) = -2*Sqrt[4.5]/v so the final slope decreases in absolute magnitude as muzzle velocity increases, as one would intuitively expect. The speed is a little more tedious, but if you've followed the above, you can do it yourself in no time.\nIntegration at this level requires knowledge of series and limits; if you're interested, pick up a good calculus text.
297111
Think of the first as the area of a flat round plate sweeping out the inside of the figure you get from the rotation.  What will be its area? πr² where the r is the y value of √(9 + X²).  That changing area will sweep out the volume from whatever the limits are (you don't say), so it's the integral of π(√(9 + X²))², or π(9 + X²).\n\nIn the second case, the thing sweeping out the volume is the surface area of a cylinder as its radius grows.  Thats the circumfrence of the base times the height.  The circumfrence is 2πr where r is the X value and the height is the difference between Y = x and Y = X/2.  So it's the integral from 0 to 2 of:\n\n   2πX(X-X/2).\n\nThinking of an area sweeping thru the volume helps.
787409
that would be if it was an indefinite integral\nbut in the case of definite one, it would be the difference of the values of the function at the limits of the integral.
594683
To find area you multiply the length times the width. It's pretty easy once you know
354577
The function f(x)=Sigma of((f^n)(c)/n!)*((x-c)^n) is what you start with.  You can replace the Sigma(sum) with an integral.  Derivatives are the basically the opposite of Integrals.  If you derive the integral, you end up back at the start: ((f^n)(c)/n!)*((x-c)^n).  It might be easier to use a graph to see what happens. \n\nBut basically, intervals are merely used to look at a part of whatever is on the graph.  It doesn't matter what you do to that part, you are still looking at that part.  That's why the intervals don't change.\n\nDoes that help you or confuse you?
40514
When you are doing integrations, all you are doing is computing an area under a curve. Sometimes the formula for the area will be obvious to you, from something you already know or recognize. Othertimes, you have to transform the problem into one you know already. \n\nThis is the method called integration by substitution. \n\nFor example, suppose you want to integrate the function sqrt(1 - x^2) with respect to x, between some limits. Most times, you will not remember or recognize this integral. However, you might remember that sqrt(1-(sin a)^2) is cos a, which you know how to integrate, and therefore if you let x=sin a, you have transformed the integral into one you recognize and can easily solve. \n\nThere is a HUGE amount of material written on this on the internet and in books, but this should get you started. All you are doing is transforming the problem into one that you can solve easier.
1019108
Do you have limits between which to calculate the area? If not you can solve for the values where f(x) = g(x)\nyou should get two values since g(x) is a second degree function.\nThe 2 values are your limits for integration.\nIntegrate each function with respect to dx between the limits and take the difference of the two. Since area is always positive the value should be the absolute value of the difference.\n\nI think that is enough hint. I won't do it for you. But here is what you should start with:\n\nsolve for x where f(x) = g(x), the two values for x are the two limits for your integration, for example x1 and x2.\nthen integrate\n  x1\n| ∫( f(x) - g(x) ) dx |\n  x2