Antiderivatives/Indefinite Integrals

A function F( x) is called an antiderivative of a function of f( x) if F′( x) = f( x) for all x in the domain of f. Note that the function F is not unique and that an infinite number of antiderivatives could exist for a given function. For example, F( x) = x³, G( x) = x³ + 5, and H( x) = x³ − 2 are all antiderivatives of f( x) = 3 x² because F′( x) = G′( x) = H′( x) = f( x) for all x in the domain of f. It is clear that these functions F, G, and H differ only by some constant value and that the derivative of that constant value is always zero. In other words, if F( x) and G( x) are antiderivatives of f( x) on some interval, then F′( x) = G′( x) and F( x) = G( x) + C for some constant C in the interval. Geometrically, this means that the graphs of F( x) and G( x) are identical except for their vertical position.
The notation used to represent all antiderivatives of a function f( x) is the indefinite integral symbol written , where . The function of f( x) is called the integrand, and C is reffered to as the constant of integration. The expression F( x) + C is called the indefinite integral of F with respect to the independent variable x. Using the previous example of F( x) = x³ and f( x) = 3 x², you find that .
The indefinite integral of a function is sometimes called the general antiderivative of the function as well.
Example 1: Find the indefinite integral of f( x) = cos x.


Example 2: Find the general antiderivative of f( x) = –8.
Because the derivative of F( x) = −8 x is F′( x) = −8, write

Integration Techniques

Many integration formulas can be derived directly from their corresponding derivative formulas, while other integration problems require more work. Some that require more work are substitution and change of variables, integration by parts, trigonometric integrals, and trigonometric substitutions.

Basic formulas

Most of the following basic formulas directly follow the differentiation rules.

1. 
   
2. 
   
3. 
   
4. 
   
5. 
   
6. 
   
7. 
   
8. 
   
9. 
   
10. 
    
11. 
    
12. 
    
13. 
    
14. 
    
15. 
    
16. 
    
17. 
    
18. 
    
19. 
    
20. 
    

Example 1: Evaluate
Using formula (4) from the preceding list, you find that .
Example 2: Evaluate .
Because using formula (4) from the preceding list yields


Example 3: Evaluate
Applying formulas (1), (2), (3), and (4), you find that


Example 4: Evaluate
Using formula (13), you find that
Example 5: Evaluate
Using formula (19) with a = 5, you find that

Substitution and change of variables

One of the integration techniques that is useful in evaluating indefinite integrals that do not seem to fit the basic formulas is substitution and change of variables. This technique is often compared to the chain rule for differentiation because they both apply to composite functions. In this method, the inside function of the composition is usually replaced by a single variable (often u). Note that the derivative or a constant multiple of the derivative of the inside function must be a factor of the integrand.
The purpose in using the substitution technique is to rewrite the integration problem in terms of the new variable so that one or more of the basic integration formulas can then be applied. Although this approach may seem like more work initially, it will eventually make the indefinite integral much easier to evaluate.
Note that for the final answer to make sense, it must be written in terms of the original variable of integration.
Example 6: Evaluate
Because the inside function of the composition is x³ + 1, substitute with




Example 7:
Because the inside function of the composition is 5 x, substitute with


Example 8: Evaluate
Because the inside function of the composition is 9 – x², substitute with

Integration by parts

Another integration technique to consider in evaluating indefinite integrals that do not fit the basic formulas is integration by parts. You may consider this method when the integrand is a single transcendental function or a product of an algebraic function and a transcendental function. The basic formula for integration by parts is


where u and v are differential functions of the variable of integration.
A general rule of thumb to follow is to first choose dv as the most complicated part of the integrand that can be easily integrated to find v. The u function will be the remaining part of the integrand that will be differentiated to find du. The goal of this technique is to find an integral, ∫ v du, which is easier to evaluate than the original integral.
Example 9: Evaluate ∫ x sech² x dx.


Example 10: Evaluate ∫ x⁴ In x dx.


Example 11: Evaluate ∫ arctan x dx.

Trigonometric integrals

Integrals involving powers of the trigonometric functions must often be manipulated to get them into a form in which the basic integration formulas can be applied. It is extremely important for you to be familiar with the basic trigonometric identities, because you often used these to rewrite the integrand in a more workable form. As in integration by parts, the goal is to find an integral that is easier to evaluate than the original integral.
Example 12: Evaluate ∫ cos³ x sin⁴ dx


Example 13: Evaluate ∫ sec⁶ x dx


Example 14: Evaluate ∫ sin⁴ x dx

Trigonometric substitutions

If an integrand contains a radical expression of the form a specific trigonometric substitution may be helpful in evaluating the indefinite integral. Some general rules to follow are

1. If the integrand contains
   
   
   
2. If the integrand contains
   
   
   
3. If the integrand contains
   
   
   

Right triangles may be used in each of the three preceding cases to determine the expression for any of the six trigonometric functions that appear in the evaluation of the indefinite integral.
Example 15: Evaluate
Because the radical has the form

Figure 1 Diagram for Example 15.

Example 16: Evaluate
Because the radical has the form

Figure 2 Diagram for Example 16.

Distance, Velocity, and Acceleration

The indefinite integral is commonly applied in problems involving distance, velocity, and acceleration, each of which is a function of time. In the discussion of the applications of the derivative, note that the derivative of a distance function represents instantaneous velocity and that the derivative of the velocity function represents instantaneous acceleration at a particular time. In considering the relationship between the derivative and the indefinite integral as inverse operations, note that the indefinite integral of the acceleration function represents the velocity function and that the indefinite integral of the velocity represents the distance function.
In case of a free-falling object, the acceleration due to gravity is –32 ft/sec². The significance of the negative is that the rate of change of the velocity with respect to time (acceleration), is negative because the velocity is decreasing as the time increases. Using the fact that the velocity is the indefinite integral of the acceleration, you find that


Now, at t = 0, the initial velocity ( v₀) is


hence, because the constant of integration for the velocity in this situation is equal to the initial velocity, write
Because the distance is the indefinite integral of the velocity, you find that


Now, at t = 0, the initial distance ( s₀) is


hence, because the constant of integration for the distance in this situation is equal to the initial distance, write
Example 1: A ball is thrown downward from a height of 512 feet with a velocity of 64 feet per second. How long will it take for the ball to reach the ground?
From the given conditions, you find that


The distance is zero when the ball reaches the ground or


hence, the ball will reach the ground 4 seconds after it is thrown.
Example 2: In the previous example, what will the velocity of the ball be when it hits the ground?
Because v( t) = –32( t) – 64 and it takes 4 seconds for the ball to reach the ground, you find that


hence, the ball will hit the ground with a velocity of –192 ft/sec. The significance of the negative velocity is that the rate of change of the distance with respect to time (velocity) is negative because the distance is decreasing as the time increases.
Example 3: A missile is accelerating at a rate of 4 t m/sec² from a position at rest in a silo 35 m below ground level. How high above the ground will it be after 6 seconds?
From the given conditions, you find that a( t) = 4 t m/sec², v₀ = 0 m/sec because it begins at rest, and s₀ = –35 m because the missile is below ground level; hence,


After 6 seconds, you find that
hence, the missile will be 109 m above the ground after 6 seconds.

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By Muhammad Atif Saeed on 03:41. Filed under Calculus , feature , Mathandstat , Recent . Follow any responses to the RSS 2.0. Leave a response