Is integration the direct opposite to differentiation?
150385
f(x)=Int(f'(y), 0, x)-f(0) => yes, they're opposite operation, but if x has a given value, e.g. x=2, 3...no, they're not, b/c in that case, the RHS of the previous egality is a real number, not a function
696606
Unless you are talking about math then yes it is the opposite.
854317
I think you're slightly confused. ∑ is sigma, and refers to summation. The integration symbol ∫ is called the "long s." And what it refers to is integration. Although the simplest method of evaluating an integral is usually antidifferentiation (especially when exact solutions are desired), this is not the only method, and there are many functions which are integrable, but whose antiderivative cannot be written in terms of elementary functions. e^(x²) is one such function. In these cases, other methods, such as numerical integration or taylor series expansion, must be used to evaluate the integral.\n\nAlso, as mentioned by the previous poster, if the ∫ symbol appears without limits it refers to the antiderivative in general (which can still be defined using the integral, even if it cannot be expressed in terms of elementary functions).
840388
This is differentiation of the composite function \n\nJ(x) = I(a(x),b(x))f(t)dt\n\nwhere \nI(a,b) is the integral from a to b, \na(x) = tan(x), b(x) = x^2 \nand f(t) = sqrt(1+t^4).\n\nApply the Chain Rule and then the Fund. Thm of Calc., which implies that dI/da = -f(a) and dI/db = f(b). These should be written with curly d's because they are partial derivatives of I (as a fctn of the two integration limits). And work out da/dx and db/dx. So \n\ndJ/dx = (dI/da) (da/dx) +  (dI/db) (db/dx)\n = -f(a(x)) (da(x)/dx) + f(b(x)) (db(x)/dx)\n\nNow just put in your particular functions f, a and b.
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
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/
969305
This initial value problem is surprisingly simple, since niether y nor any derivatives other than the second appear in the equation, so all that is necessary is to integrate twice:\n∫∫d²y/dx² dx dx = ∫∫6x dx dx\n∫dy/dx dx = ∫3x² + C dx\ny=x³+Cx+K\n\nNow we need merely find the constants C and K. We know that y'(0)=3(0)² + C = 0, so C=0. Further, we know that y(0) = K = 1, so K=1.\n\nThus y=x³+1\n\nb: There is exactly one such function. All antiderivatives of a function are the same up to addition of an arbitrary constant, and as we showed when solving the problem, there is only one consistent set of values for the two constants of integration introduced in this problem.
486469
There are many reasons why one would "want" to.\nIf you mean: 'as a layman; what is the practical point of this' then (basically) differentiation is a way of finding a tangent and integration is a way of finding "the area under the graph".\nAs a Mathematician, my motives are somewhat different; I consider Differentiation and Integration as interesting functions on function spaces. For example, Integration of Real-valued functions of on Reals is really of interest as a "measure" of various sets. The topic of Measure Theory is a large and interesting one.\n\nBut for your purposes I guess it suffices to say "because you get useful information about the function".
937153
B is false: Let a be any number other than zero. If a>0, 2(2a)-a > 0, and f'(2a) = 2. Conversely, if a<0, 2(2a)-a<0, and f'(2a) = -2. Thus, any value of a other than zero serves as a counterexample.\n\nA is also false, although I'm not sure your teacher knows this. She may be thinking of something like f(x)=[a, b]∫g(t) dt, which is alawys a constant function regardless of g(t), and thus its derivative is zero. However, if the free variable appears in the limits of integration, the function may be nonconstant: for instance, let f(x) = [1, x]∫1/t dt, which is equal to ln x for positive values of x, and its derivative is certainly not zero. So the correct answer here is d: both A and B are false.
719380
Let y = x^(1/2)\n\nso y^2 = x,\n\ndx = 2y dy\n\nThe problem becomes:\n\nint [1 / (y + y^3)] 2y dy\n\nOr:\n\nint 2 [1 / (1 + y^2)] dy\n\nA recognizable integral. Integrate and substitute back x^1/2 = y
971343
neither\nit is -1/2 e^cos 2x + C\n\nto confirm differentiate the result and find itr\n\nif you need to integate let cos 2x = y\nthen -2 sin 2x dx = dy\nsin 2x dx = -dy/2\n\nput the value we need to integrate - e^y dy/2 \nresult = - e^y/2 + C\n       = -1/2e^(cos 2x) + C
914813
Your correct, it is equal to f(x).\n\nthe integral and the differential cancel eachother out.!
149854
babymegan101 is wrong because the derivative of xsin(ax^2+b)+C is sin(ax^2+b) + 2ax^2*cos(ax^2+b). (And we expect the derivative to be sin(ax^2 + b)). I've tried every trick in a physicist's toolbox to solve this integral but I have not succeeded. Therefore I propose that there isn't a definitive answer to the question, and that it can only be solved by approximation using numerical methods.
736484
For all the rules of differentiation and integration, check sources
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.
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
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards