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Antiderivatives

In calculus, an antiderivative, primeval or unclear integral of a function f is a task F whose imitative is equal to f, i.e., F? = f. The process of solving for antiderivatives is called antidifferentiation and its differing occupation is called differentiation, which is the development of result a derived. The antiderivative, or primeval, of a intricate-valued function g is a function whose intricate derivative is g. Antiderivatives are connected to definite integrals through the basic theorem of calculus the exact integral of a task over an interval is different to the difference between the values of an antiderivative evaluate at the endpoints of the distance.

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Antiderivatives Conditions

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• The basic of a satisfactory, state for a function f to have an antiderivative is that f has the middle value property. That is, if [a, b] is a subinterval of the sphere of f and d is any actual number between f (a) and f (b), then f(c) = d for some c between a and b.

• The set of discontinuities of f should be a meager set. They can make some significance values of f having an anti-derivative, which has to locate as it’s location of discontinuity.

• If f has an antiderivative, is bound on close finite subinterval of the domain and has a set of discontinuities of Lévesque measure 0, then an antiderivative may be found by integration.

                  The method of finding antiderivatives is called antidifferentiation or integration:

                                                                                         d/dx [F(x)] = f(x)

                                                                        `=>`   `int`  f(x) dx = F(x) + C.

 The expression F(x) + C is the general antiderivative of ƒ. If ƒ and g are defined on the same interval, then the general antiderivative of the sum of ƒ and g equals the sum of the general antiderivatives of ƒ and g.

Antiderivative Formulas

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`int` xn dx = `(x^(n + 1))/(n + 1)` + C  (for n ? - 1)

`int` `1/x` dx = ln |x| + C

`int` ex dx = ex + C

`int` sin x dx = - cos x + C

`int` cos x dx = sin x + C

`int` sec2 x dx = tan x + C

`int` cosec2 x dx = - cot x + C

`int` sec x tan x dx = sec x + C

`int` cosec x cot x dx = - cosec x + C

`int` tan x dx = ln |sec x| + C

`int` cot x dx = - ln |cosec x| + C

`int` sec x dx = ln |sec x + tan x| + C

`int` cosec x dx = ln |cosec x - cot x| + C

Finding Antiderivative

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Below are the examples on antiderivatives -

Example 1: Find antiderivative for the function, f(x) = 3x2 + 7x.

Solution:

      Step 1: Given function

                                         f(x) = 3x2 + 7x

                               `int` f(x) dx = `int` (3x2 + 7x) dx

      Step 2: Separate the integral function

                             `int`(3x2 + 7x) dx = `int` 3x2 dx + `int` 7x dx

      Step 3: Integrate each function with respect to ' x',

                             `int`(3x2 + 7x) dx = `(3x^3)/3` + `(7x^2)/2` + C

                                                 = x3+ `(7x^2)/2` + C

Example 2: Find antiderivative for the function, f(x) = `2/x^7`

Solution:

      Step 1: Given function

                                         f(x) = `2/x^7 `

                               `int` f(x) dx = `int` `2/x^7` dx

      Step 2: Integrate the function `2/x^7` with respect to ' x ',

                              `int` `2/x^7` dx = `int` 2x-7 dx

                                             = 2`((x^(-7 + 1))/(-7 + 1))`

                                             = 2 `((x^-6)/(-6))` + C

                                             = - `1/(3x^6)` + C

Example 3: Find antiderivative of y = 2cos 8x

Solution:

      Step 1: Given function

                                         y = 2cos 8x

                               `int` y dx = `int` 2cos 8x dx

      Step 2: Integrate the given function y = 2cos 8x with respect to ' x',

                              `int`2cos 8x dx = 2sin (8x) `1/8`

                                                = `(sin (8x))/4`


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