In simple language, derivative is rate of change of something and integral represents the area of a curve whose equation is known.
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∙ 11y agothere is no diffference, i think...
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.
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
For positive x, this expression is equal to 1. The integral (anti-derivative) is therefore x + C (where C is the arbitrary integration constant). For negative x, this expression is equal to -1, and the integral is -x + C. Wolfram Alpha gives the integral as x times sgn(x), where sgn(x) is the "sign" function.
find anti derivative of f(x) 5x^4/3 + 8x^5/4
there is no diffference, i think...
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.
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 Derivative is the instantaneous rate of change of a function. An integral is the area under some curve between the intervals of a to b. An integral is like the reverse of the derivative, Derivatives bring functions down a power, integrals bring them up, in-fact indefinite integrals (ones that do not have specifications of the area between a to b) are called anti derivatives.
In all but very exceptional cases there is no difference.
According to Wolfram Alpha, input:integral csc x it is -log[cot(x) + csc(x)] + constant You can verify this by taking the derivative of the purported integral.
The derivative is the inverse of the integral. ∫ f'(x) dx = f(x) + C
For positive x, this expression is equal to 1. The integral (anti-derivative) is therefore x + C (where C is the arbitrary integration constant). For negative x, this expression is equal to -1, and the integral is -x + C. Wolfram Alpha gives the integral as x times sgn(x), where sgn(x) is the "sign" function.
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 anti derivative of negative sine is cosine.
The other name of an anti-derivative is an integral. An integral is the function which finds the area under a line. Let me give you an example. The integral of xn would be (xn+1/n+1)+C. So the integral of x2 would be (x3/3)+C. If you wanted to know what the area under the equation x2 when x=3 (in other words, all values of x between 0 and 3 on the x-axis), then you can do the equation ((3)3/3)+C and area would be 3 units squared. C is the constant of integration. For more info on C, you can go to http://en.wikipedia.org/wiki/Constant_of_integration or simply look it up
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).