Antiderivatives Help Calculus

In this article we are going to discuss about the antiderivatives help calculus concept. The process of short form the calculus consequent to over screening the significant concept in antiderivatives help calculus are referred as review calculus. This article helps to improve the knowledge for antiderivatives help calculus and below the content are helping toll for the exam. Go forward of the test using review to this article. Antiderivatives help calculus rules also show below.


Properties of antiderivatives help calculus:-


Let `f(x) ` is continuous on `[a, b].` If `G(x) ` is permanent on` [a, b]` and `G'(x) =f(x)` for all` x in (a, b),` then `G` is called an antiderivative of `f.` We know how to put up antiderivatives by integrating. The function `F(x) =int ^x_a f (t) dt` is an antiderivatives for` f` because it can be shown that` F(x)` constructed in this way is continuous on `[a, b] ` and `F, (x) =f(x) ` for all `x in` `(a, b).`

Let` F(x)` is any antiderivative for` f(x).`

• For any constant `C, F(x) + C` is an antiderivative for `f(x).`


Since `(d)/(dx)[F(x)]=f(x),`

`(d)/ (dx) [F(x) + C]`

`= (d)/ (dx) [F(x)] + (d)/ (dx) [C]`

`=f(x) +0`


Hence `F(x) + C ` are an antiderivative for `f(x)` .

• Every antiderivative of` f(x) ` can be written in the form

`F(x) + C`

for some `C.` That is, each two antiderivatives of f vary by at most a constant.


Let `F(x)` and `G(x) ` are antiderivatives of `f(x)` . Then `F'(x) =G'(x) =f(x), ` so `F(x)` and G(x) differ by at the majority an invariable.

The development of finding antiderivatives is called antidifferentiation or integration:

`(d)/(dx)[F(x)] = f(x) lArr rArr int f(x)dx=F(x)+ C .`

`(d)/ (dx) [g(x)] = g' (x) lArr rArr int g' (x) dx = g(x) + C.`


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Properties of antiderivatives help calculus:-


• `(d)/ (dx) [int f(x) dx] = f(x).`


Let `int f(x) dx =F(x) ` , where` F(x)` is an antiderivative of `f` . Then

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

` int [alpha f(x) + beta g(x)] dx = alpha int f(x) dx + beta int g(x)dx .`


We need only show that alpha int `f(x) dx + beta int g(x) dx ` is an antiderivative of int` [alpha f(x) + beta g(x)] dx:`

`(d)/ (dx) [alpha int f(x) dx + beta int g(x) dx]`

`= alpha (d)/ (dx) [int f(x) dx] + beta (d)/ (dx) [int g(x) dx]`

`=alpha f(x) +beta g(x)`


1. Every antiderivative of `x^2 has the form x^3/3 + C`

since `(d)/ (dx) [(x^3)/ (3)] =x^2.`

2. `(d)/ (dx) [intx^5 dx] = x^5.`


Key Concepts of antiderivatives help calculus:-


If `G(x) ` is continuous on `[a, b]` and `G'(x) =f(x) ` for all x in `(a, b),` then `G` is called an antiderivative of `f` .

We know how to make antiderivatives by integrating. The function `F(x) = int ^a_x f (t) ` dt is an antiderivative for `f.` In detail every antiderivative of f(x) can be written in the form `F(x) + C` , for some `C`.

`(d)/ (dx) [F(x)] =f(x) lArr rArr int f(x) dx = F(x) + C.`

`(d)/ (dx) [g(x)] =g'(x) lArr rArr int g'(x) dx = g(x) + C.`