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The Antiderivative Safa Faidi

The Antiderivative

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The Antiderivative. Safa Faidi. The definition of an Antiderivative. A function F is called the antiderivative of f on an interval I if F’(x) = f (x) for all x in the interval I Example: F(x ) = x 2 , g(x ) = 2x - PowerPoint PPT Presentation

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Page 1: The Antiderivative

The Antiderivative

Safa Faidi

Page 2: The Antiderivative

The definition of an Antiderivative

•A function F is called the antiderivative of f on an interval I if F’(x) =f(x) for all x in the interval I

•Example:

F(x) = x2, g(x) = 2x •If we look here, we can see that F’(x)

=2x= g(x), so with the definition mentioned above we can see that F(x) is the antiderivative of g(x)

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The rules of the Antiderivative

•When dealing with the antiderivative there are rules that have to be followed. The first basic rule is:

•If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is

F(x) +C Note that the C is a constant that can be any

number, even zero

Page 4: The Antiderivative

Table of Formulas for the Antiderivative: Given that F & G are antiderivative of f and g respectively and a, c are constants

Function antiderivative

af(x) aF(x)+Cf(x)+g(x) F(x)+G(x)Xn (n≠-1) Xn+1/n+1

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Trigonometric Functions

•*where f(x) represents the function and F(x) represents the antiderivative*

• f(x) = cos x then F(x)= sin x+C• f(x) =sin x then F(x)= - cos x + C • f(x)= sec2x then F(x)= tan x + C• f(x)= sec x tan x then F(x)= sec x + C

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Examples • Find the Antiderivative F(x) for each of the following functions

• f(x)= 2 ( x2+3x) F(x) = 2(x3/3 + + 3/2 x2) + C  • f(x)= 3 sin x + 2x5+√x f(x)= 3 sin x + 2x5+ x 1/2

F(x)= -3 cos x + 1/3 x6 +2/3 x 3/2 + C  • f(x)= cos x + 7sec2x F(x)= sin x + 7tan x + C  

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  Find f if f’’(x)= 20x3+x+√x

f’(x)=20/4 x4 + x2/2 +2/3 x3/2 + C

f’(x)=5x4 + x2/2 + 2/3 x3/2 + Cf(x)= x5 + x3/6 + 4/15 x5/2 + Cx + D   Where C, D stands for constants

Here, we simply need to work backwards from the second derivative, to the first, then to the actual functionYou can always try to check by differentiating your final answer f(x)

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The Antiderivative and the definite Integral

Dina Saleh

Page 9: The Antiderivative

Anti-derivatives and definite integrals

•An integral which is evaluated over an interval. A definite integral is

written as follows.

Page 10: The Antiderivative

Definite integrals are used to find the area between the graph of a function and the x-

axis. And there are many other applications.

Formally, a definite integral is the limit of a Riemann sum as the norm of the partition

Approaches zero.

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Theorem 11 . if f is continuous on [a , b] except maybe at a

finite number of points at which it has a jump discontinuity, then f will be integrated on [a , b].

2. if f is decreasing or increasing function on [a , b] then it could be integrated on

[a , b]

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3. if we could integrate f on [a , b] then f is bounded on [a , b] which is a necessary condition for it to be integrated over [a , b]

4. a ∫a f(x) dx = 0

5. if f is to be integrated on (a , b)

then we consider a∫b f(x) dx = - f(x) dx

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Theorem 2

1. c dx = c(b - a). where c is any constant

2. [ f(x) +g(x)]dx = f(x) dx + g(x) dx

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3. c f(x) dx = c f(x) dx. where c is a constant

4. [f(x) - g(x)] dx = f(x) dx - g(x) dx

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Theorem 3

if f is integrable on [a , b ] then f(x) dx = lim n →∞ b-a / n , n∑k=1 f[ a + (b-a)k/n ]

 

Page 16: The Antiderivative

Example :4∫1 ( 2x) dx

2x is a continuous function , therefore it could be integrated on the intervals [1,4] , and so

4∫1 2x dx = lim n →∞ 4-1/n n∑k=1 2[1+ (4-1)k/n]

= lim n →∞ 3/n .2 n∑k=1 (1+ 3k/n )

=6 lim n →∞ 1/n (n∑k=1 1+ 3/n n∑k=1 k )

= 6 lim n →∞1/n [( n + 3/n (n(n+1)/2) ]=6 lim n →∞ [1+ 3/2 (n2+n/n2]=6[1+3/2] = 6 . 5/2 = 15

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The Fundamental Theorem of Calculus (1)

Shaima Al Bikri

Page 18: The Antiderivative

The upper area is positive while the lower area is negative.

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Count the boxes in the red area it almost equal to g (6) =4

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Count the boxes in the purple area it almost equal to g (10) =2.5

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To find the net area = g(6) - g(10) ?

we subtract them because is down on the opposite side

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h(x+g) – h(x) =

Proof If x and x + g are

in (a, b)?

And so, for g 0