whats the bar of an integral called?
142152
integration sign.\nWhen reading a formula, you simply read it "integral of" if there are no boundaries, or "integral from a to b" if there are boundaries a (lower) and b (upper)
110772
0.58 (with a horizontal bar right above the 58)
941714
If you break it into real and imaginary parts, you get integrals of\ncos(x^2) and sin(x^2) from -infty to infty. These integals are finite. The anti-derivatives are called Fresnel integrals and show up in studies of diffraction. The infinite integrals are sqrt(pi/2) each.
420145
The antideriative is the same as an integral.  For this, we can use the fact that sin^2(t) + cos^2(t) = 1.  Then,\n\n[int] cos^2(t)/sin(t) dt = [int] (1 - sin^2(t))/sin(t) dt\n= [int] (1/sin(t)) - sin(t) dt\n= [int] csc(t) - sin(t) dt\n= ln(tan(t/2)) + cos(t) + C
859250
integral = x*tan(x)+ln(sec(x))  (use integration by parts with du = sec^2(x) and v = x).\n\ng'(x) = f(x), so a, c, e and g are true (b is true if f is continuous)\n\nSteve
292177
i call it a roll bar..but for convertables its called a light bar....
171024
In "C", it is called the "floor" function. A quick search on wikipedia\n\nhttp://en.wikipedia.org/wiki/Floor_function\n\nshows that mathematicians favour the same terminology.
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?
189548
Integrable Function.The idea of integration is twofold.\n\nWe may seek the function which has a given function as its differential coefficient, or we may generalize the question of finding the area of a curve. The first inquiry leads directly to the indefinite integral, the second directly to the definite integral. Following the second method we define the definite integral of the function f(s) through the interval between a and b to be the limit of the sum ~f(Xr) (x~ - X~_i)\n\nwhen the interval is divided into ultimately indefinitely small partial intervals by points 5i, XI, . . . X,,_i. Here x, denotes any point in the rth partial interval, Sf is put for a, and x~for b. It can be shown that the limit in question is finite and independent of the mode of division into partial intervals, and of the choice of the points such as x~, provided (I) the function is defined for all points of the interval, and does not tend to become infinite at any of them; (2) for any one mode of division of the interval into ultimatey indefinitely small partial intervals, the sum of the products of the oscillation of the function in each partial interval and the difference of the end-values of that partial interval has limit zero when n is increased indefinitely. &#92;Vhen these conditions are satisfied the function is said to be integrable in the interval. The numbers a and b which limit the interval are usually called the lower and upper limits. We shall call them the nearer and further end-values. The above definition of integration was introduced by Riemann in his memoir on trigonometric series (1854). A still more general definition has been given by Lebesgue. As the more general definition cannot be made intelligible without the introduction of some rather recondite notions belonging to the theory of aggregates, we shall, in what follows, adhere to Riemanns definition.\n\nWe have the following theorems:\n\nr. Any continuous function is integrable.\n\n2. Any function with restricted oscillation is integrable.\n\n3. A discontinuous function is integrable if it does not tend to become infinite, and if the points at which the oscillation of the (unction e~fceeds a given number e, however small, can be enclosed in partial intervals the sum of whose breadths can be diminished indefinitely.\n\nThese partial intervals must bea set chosen out of some complete set obtained by the process used in the definition of integration.\n\n4. The sum or product of two integrable functions is integrable.\n\nAs regards integrable functions we have the following theorems:\n\n1. If S and I are the superior and inferior limits (or greatest and least values) of f(x) in the interval between a and b, f f(x)dx is intermediate between S(ba) and I(ba).\n\n2. The integral is a continuous function of each of the end-values.\n\n3. If the further end-value b is variable, and if f f(x)dx=F(x), then if f(s) is continuous at b, F(s) is differentiable at b, and F(h) =f(b).\n\n4. In case f(s) is continuotis throughout the interval F(s) is continuous and differentiable throughout the interval, and F(x) =f(x) throughout the interval.\n\n5. In case f(s) is continuous throughout the interval between a and b, ff(x)4x =f(b) f(a).\n\n6. In casef(x) is discontinuous at one or more points of the interval between a and b, in which it is integrable, f f(x)dx is a function of x, of which the four derivates at any point of the interval are equal to the limits of indefiniteness off(s) at the point.\n\n7. It may be that there exist functions which are differentiable throughout an interval in which their differential coefficients are not integrable; if, however, F(x) is a function whose differential coefficient, F(-x), is integrable in an interval, then F(s) = fF(x)dx+const.,\n\nwhere a is a fixed point, and x a variable point, of the interval. Similarly, if any one of the four derivates of a function is integrable in an interval, all are integrable, and the integral of either differs from the orig
761317
We have to find derivative of, y = the integral where lower limit is e^x and upper limit is 0 of sin^3 t dt.\n\ny = int(e^x)(0) [sin^3 t] dt     [I would write lower limit in first bracket and upper limit in bracket after that, so dont confuse]\n\nNow, the derivative of the integrand f(t)=sin^3 t is f'(t)=3 (sin^2 t) *cos t\ndy /dt= d {int(e^x)(0) [sin^3 t] dt }/dt\n=>dy /dt= d {f(t)}/dt (at 0) - d {f(t)}/dt (at e^x)\n=>dy /dt= f'(t) (at 0) - f(t) (at e^x)\n=>dy /dt= [3 (sin^2 t) *cos t] (at 0) - [3 (sin^2 t) *cos t ] (at e^x)\n=>dy /dt= [3 (sin^2 0) *cos 0] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= [3 (0) *1] - [3 (sin^2 (e^x)) *cos (e^x) ] \n=>dy /dt= -[3 (sin^2 (e^x)) *cos (e^x) ]\n\nI hope that answers your question.\n\nAll the bset
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.
74089
http://integrals.wolfram.com/
352249
This integral does exist (it can be evaluated for any endpoints you choose). Since it cannot be represented as a finite combination of elementary functions and operations, it is given its own name, the error function. Many integrals like this have their own names and are quite useful in applied mathematics. See http://mathworld.wolfram.com/Erf.html .\nTo say that it does not exist is unrigorous.
1143501
you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
242020
i'll do it by 2 ways \nthe first one if u know what integration by parts is:\n(int=integral)\nint[udv]=uv-int[vdu]\nlet u=lnx =>du=dx/x\n& let dv=xdx => v=(x^2)/2\nso we'll have\nint[xlnx]=((x^2)/2) lnx -1/2int[xdx]\n=((x^2)/2)lnx -((x^2)/4)+ Constante\nthe second method is\ntry derivating (x^2)lnx u'll get [(x^2)lnx]'=2xlnx +x\nthen integrate int{[(x^2)lnx]dx}=(x^2)lnx\n&int[x]=(x^2)/2\nso u'll get (x^2)lnx=2int[xlnxdx]+(x^2)/2\nso int[xlnxdx]=((x^2)/2)lnx-(x^2)/4+ C
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!
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]