An integral and an anti-derivative are the same thing. Integration means the process of finding the integral, just as anti-differentiation means the process of finding the anti-derivative.
there is no diffference, i think...
The anti-derivative of X2 plus X is the same as the anti-derivative of X2 plus the anti-derivative of X. The anti derivative of X2 is X3/3 plus an integration constant C1 The anti derivative of X is X2/2 plus an integration constant C2 So the anti-derivative of X2+X is (X3/3)+(X2/2)+C1+C2 The constants can be combined and the fraction can combined by using a common denominator leaving (2X3/6)+(3X2/6)+C X2/6 can be factored out leaving (X2/6)(2X+3)+C Hope that helps
In simple language, derivative is rate of change of something and integral represents the area of a curve whose equation is known.
In Calculus, integration is the process of finding the area under the curve of a function, usually between two boundaries. For example, the area under the curve of the graph y=x between 0 and 1 (the two boundaries) is equal to the area of the triangle formed by the x axis, the graph and a vertical line at x=1. Since this triangle covers half the area of a square of length 1 unit, the integral of y=x from 0 to 1 is 1/2. For more complex curves such as y=x^2, integration is easier and more accurate by finding the anti-derivative, or integral, of y=x^2. Finding the anti-derivative, as the name suggests, is the reverse process of finding a function's derivative. So, the anti-derivative of x^2 is the function whose derivative is x^2. I'm assuming you are familiar with differentiation (the process of finding a derivative of a function) if you are doing problems with integration. So, the anti-derivative of x^2 is (1/3)x^3. The purpose of finding the anti-derivative is to use the Fundamental Theorem of Calculus, which states that the area under the curve of a function from a to b (the two boundaries) is equal to the difference between the values of the function's anti-derivative at b and a. So, plugging in 0 and 1 again for x gives: Area under curve of x^2 from 0 to 1 = (1/3)(1)^3 - (1/3)(0)^3 = (1/3) - 0 = 1/3 Notice the reverse order of the boundaries, since subtraction is not commutative (order matters).
Well if you are talking about calculus, integration is the anti-derivative. So as my teacher explained to us, instead of going down, you will go up. For example if you have the F(x) = 2x, the F'(x) = 2. F'(x) is the derivative here, so you will do the anti of a derivative. So with the same F(x) = 2x the integral, is SF(x) = 1/3x^3. The Integral will find you the area under the curve.
The indefinite integral is the anti-derivative - so the question is, "What function has this given function as a derivative". And if you add a constant to a function, the derivative of the function doesn't change. Thus, for example, if the derivative is y' = 2x, the original function might be y = x squared. However, any function of the form y = x squared + c (for any constant c) also has the SAME derivative (2x in this case). Therefore, to completely specify all possible solutions, this constant should be added.
We say function F is an anti derivative, or indefinite integral of f if F' = f. Also, if f has an anti-derivative and is integrable on interval [a, b], then the definite integral of f from a to b is equal to F(b) - F(a) Thirdly, Let F(x) be the definite integral of integrable function f from a to x for all x in [a, b] of f, then F is an anti-derivative of f on [a,b] The definition of indefinite integral as anti-derivative, and the relation of definite integral with anti-derivative, we can conclude that integration and differentiation can be considered as two opposite operations.
The anti derivative of negative sine is cosine.
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