why if we get  integral  frome one function it gives us that  functions  defransial ..???
97560
Are you referring to a fluid mechanics problem, or?\n\nCan you restate the question?
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
294124
This integral does not have closed form using elementary functions. It is defined to be the logarithmic integral function, and you can find out more about it here: http://mathworld.wolfram.com/LogarithmicIntegral.html .
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.
141606
There is an application to convert an AC current to DC.\nAssume that f is a sigmoid function, then integration of\nf^2 over one period gives a constant number which is considered as flattened sigmoid and is analogous with converting AC to DC.
245944
The integral of a function over a definite interval.
695595
Each question is a function of two variables... and it has two derivatives, relative to each variable.\n\nDerivative of F relative to Z1   is   uZ1^(u-1)*Z2^v\nDerivative of F relative to Z2   is   Z1^u*vZ2^(v-1)\n\nDerivative of MRTS relative to Z1 is  -uZ2/(vZ1^2)\n\nDerivative of MRTS relative to Z2 is  u/(vZ1)
274471
i made a little comment for you on your integral answer:)
796566
A function relates each of its inputs to exactly one output. That is for each value of x in the domain, there must be one and only one y.\n1) 2) are OK for each x in R there is exactly one y (and vice versa)\n3) Not OK.  Take y=-2 and y=2. Both have x=4\n4) OK. There are x-values that lead to the same y, but that is allowed.\n6) Not OK. Take x=2, y>4. There are an infinite number of y-values.
793695
' the integral of 1/e^u du = integral of e^{-u}du\n=-e^{-u} + C
213905
Think of this as a linear approximation thing. The Jacobian *matrix* is essentially the linear part of the approximation. Now use the fact that matrix multiplication is one-to-one when the matrix is invertible and not otherwise. We test invertibility of a square matrix by looking at its determinant. So, finally, the determinant of the Jacobian matrix is the Jacobian.\n\nOne thing you should be careful of, though. The above paragraph only works locally. It is possible for the Jacobian to never be 0, yet the global function to fail to be one-to-one.
276904
The Laplace transform *doesn't* exist for every function. For example, f(x)=exp(x^2).
996184
You know that the integral from negative infinity to infinity (or in your case, since it's in a box, from 0 to a, which is the width of the box) of the absolute value of the wavefunction squared is equal to one (via statistics).  So once you do this integral, you should have a function of one variable, a.  Then you just have to solve for a.  Easy money.  I'm going to assume you are referring to the one dimensional wavefunction (although you would do it the same way in 2 or 3D, just doing separable integrals).\n\npsi(x)=A*sin(kx) is the wavefunction (whatever it is, it doesn't matter) \n\n1=int from 0 to a ((psi(x))^2 *dx)\n1=int from 0 to a (A^2 * sin^2(kx) *dx)\n1=A^2 * int from 0 to a (sin^2(kx) *dx)\n1=A^2*a/2\nSolve for A\nA=sqrt(2/a)\n\n:P
263227
This integral does not have an indefinite closed form of elementary functions. It can be evaluated numerically by either integrating the series representation of e^(-x^2) or using integral approximation methods like Simpson's rule.\nIt also has an interesting value if evaluated over its entire domain; by changing coordinates to polar coordinates, one can easily show that the integral over the entire real line is equivalent to Sqrt[Pi].\nThis fact is used in statistics to normalize the Gaussian distribution to have area 1 under the curve. The integral of your expression on the interval [0,t] is then known as the error function. See the source for details.
71115
You need some theory background first. Integration is most rigorously defined as a sum of an infinite number of rectangles that start on the x-axis and stretch up or down to the function (called a Riemann sum). The rectangles are infinitely thin, but each one still has area length * width, and integration is merely the sum of the areas of all those rectangles.\n\nImagine a rectangle is attached to the x-axis, and the x-axis acts as an axle, so you can spin the rectangle around that axle. The spinning rectangle will look like a cylinder whose radius is the height of the rectangle and whose width is the width of the rectangle (the amount touching the axis). \n\nThe area of a cylinder is pi * r^2 * w. As I stated, the radius is the height of the rectangle, which stretches from the x-axis (y=0) to the function (y=f(x)). Subtract those y-values to get the radius: f(x)-0=f(x). So, r=f(x).\n\npi * r^2 * w = pi * [f(x)]^2 * w\n\nThe width of the cylinder is how much is touching the x-axis and is usually represented by delta-x.  However, when an infinite number of rectangles are used, you use dx instead.\n\npi * [f(x)]^2 * dx\n\nSo, just like the definite integral from a to b of f(x)dx gives you the sum of the areas of the infinite rectangles beneath f(x), the definite integral from a to b of pi * [f(x)]^2 * dx is equal to the volume of the cylinders created when those rectangles are rotated.
1013129
Please explain what is ~f.  I have curiosity, because neither inverse multiplicative, complex conjugate or inverse function could do that result...\n\nIt should be interesting!! Please post what is ~f !
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated