LESSON X - Mathematics & Statistics
LESSON 7 INTEGRATION BY PARTS
Let u and v be differentiable functions of x and consider the differential [pic]. Since the function [pic] is a function of x, then by the definition of differential, we have that [pic]. By the Product Rule, we have that [pic] =
[pic]. Then [pic] = [pic] =
[pic] = [pic]. Thus, [pic]. Then we have that [pic]. Since [pic], then we obtain that [pic] = [pic] [pic].
The equation [pic] is the formula for evaluating an integral by the technique of integration called Integration by Parts. To use this formula on an integral, you must identify the function u and the differential dv.
Example Evaluate [pic] using Integration by Parts.
Let [pic] and [pic]
Then [pic] [pic]
Thus, [pic] = [pic] = [pic] = [pic] =
[pic] = [pic] =
[pic]
Answer: [pic]
COMMENT: It appears that the constant of integration C that was involved in integrating the differential dv in order to obtain the function v was not needed for the example above. Let’s see if this is true in general.
Example Evaluate [pic] using integration by parts.
Let [pic] and [pic]
Then [pic] [pic]
Thus, [pic] = [pic] = [pic] =
[pic] =
[pic] =
[pic] =
[pic] =
[pic] =
[pic]
Answer: [pic]
COMMENT: Thus, when we integrate the differential dv in order to obtain the function v, we do not need to include a constant of integration. This will simplify our work.
When applying the technique of Integration by Parts, it is helpful to know the following integral formulas, where k is a constant:
1. [pic]
2. [pic]
3. [pic]
4. [pic]
5. [pic]
6. [pic]
7. [pic]
8. [pic]
9. [pic]
Examples Evaluate the following integrals.
1. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Answer: [pic] OR [pic]
2. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic]
Answer: [pic]
3. [pic]
There are two ways to apply the Integration by Parts technique to evaluate a definite integral. The first way is to apply the technique to the indefinite integral in order to obtain all the antiderivatives of the integrand. Then evaluate the definite integral using the Fundamental Theorem of Calculus with one on these antiderivatives. This way is preferable if you have to apply the Integration by Parts technique more than one time. The second way is to work with the definite integral and keep the limits of integration in the work.
Using the first method, consider the indefinite integral [pic].
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] =
[pic]
Now, that we have all the antiderivatives of the integrand [pic], we can apply the Fundamental Theorem of Calculus to evaluate the definite integral [pic]. Note that since the function [pic] is continuous on its domain of definition, which is the set of all real numbers, then it is continuous on the closed interval [pic]. Thus, the Fundamental Theorem of Calculus can be applied. Thus,
[pic] = [pic] =
[pic] =
NOTE: [pic]; [pic]
[pic]; [pic]
[pic] =
[pic] = [pic] =
[pic] = [pic]
Using the second method, consider the definite integral [pic].
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] =
[pic] =
[pic] =
[pic] = [pic] =
[pic] = [pic]
Answer: [pic]
4. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Answer: [pic]
5. [pic]
First, we will apply the Integration by Parts technique to evaluate the indefinite integral [pic]. Then we will choose one of these antiderivatives to apply the Fundamental Theorem of Calculus to find the value of the given definite integral.
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
Now, apply the Integration by Parts technique again in order to evaluate the indefinite integral [pic].
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Thus, we have that
[pic] = [pic] =
[pic] =
[pic] =
[pic]
Now, that we have all the antiderivatives of the integrand [pic], we can apply the Fundamental Theorem of Calculus to evaluate the definite integral [pic]. Note that since the function [pic] is continuous on its domain of definition, which is the set of all real numbers, then it is continuous on the closed interval [pic]. Thus, the Fundamental Theorem of Calculus can be applied. Thus,
[pic] = [pic] =
[pic] = [pic] =
[pic] = [pic] = [pic]
Answer: [pic]
NOTE: There is a faster way to apply the Integration by Parts technique to
evaluate the indefinite integral [pic]. We will discuss it later in this lesson.
6. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
Now, apply the Integration by Parts technique again in order to evaluate the indefinite integral [pic].
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
Thus, we have that
[pic] = [pic] =
[pic] =
[pic]
Thus, we have that
[pic] = [pic]
Notice that the indefinite integral [pic], which we are evaluating, appears on both sides of the equation above. Thus, we could solve for [pic].
Adding [pic] to both sides of the equation, we obtain that
[pic] = [pic]. Since [pic],
where C is a constant, then we have that [pic] =
[pic]. Now, multiplying both sides of this equation
by [pic], we obtain that [pic] = [pic] =
[pic] = [pic], where
[pic].
Answer: [pic]
7. [pic]
First, we will apply the Integration by Parts technique to evaluate the indefinite integral [pic]. Then we will choose one of these antiderivatives to apply the Fundamental Theorem of Calculus to find the value of the given definite integral.
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] = [pic]
Now, that we have all the antiderivatives of the integrand [pic], we can apply the Fundamental Theorem of Calculus to evaluate the definite integral [pic]. Note that since the function [pic] is continuous on its domain of definition, which is the set of all positive real numbers, then it is continuous on the closed interval [pic]. Thus, the Fundamental Theorem of Calculus can be applied. Thus,
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] = [pic]
Or, you could have applied the Integration by Parts technique to evaluate the definite integral [pic] in the following manner.
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] = [pic]
Answer: [pic]
8. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
The indefinite integral [pic] can be evaluated by simple substitution. Let [pic]. Then [pic]. Then [pic] =
[pic] = [pic] = [pic] = [pic] =
[pic]
Thus, we have that
[pic] = [pic] =
[pic] = [pic] =
[pic]
Answer: [pic]
9. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] =
[pic]
Answer: [pic]
10. [pic]
[pic] = [pic] = [pic]
The indefinite integral [pic] can be evaluated by simple substitution.
Let [pic]. Then [pic].
[pic] = [pic] = [pic] = [pic] = [pic]
Answer: [pic]
11. [pic], where n is a rational number such that [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] =
[pic]
Answer: [pic]
12. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
The indefinite integral [pic] can be evaluated by rewriting the integrand in the following manner.
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Thus, we have that
[pic] = [pic] =
[pic] = [pic]
Answer: [pic]
13. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] = [pic]
The indefinite integral [pic] can be evaluated by rewriting the integrand in the following manner.
[pic] = [pic] = [pic] =
[pic] = [pic]
Thus, we have that
[pic] = [pic]
[pic] =
[pic] =
[pic]
Answer: [pic]
14. [pic]
Let [pic] and [pic]
Then [pic] [pic]
NOTE: [pic] = [pic] = [pic] = [pic]
[pic] = [pic] = [pic] = [pic]
The indefinite integral [pic] can be evaluated by rewriting the integrand.
First, carry out the long division of [pic] by [pic]:
[pic]
Thus, [pic]. Then [pic] =
[pic] = [pic]
Thus, we have that
[pic] = [pic] =
[pic] =
[pic] =
[pic]
Answer: [pic]
Let return to Example 5, where we applied the Integration by Parts technique two times on the indefinite integral [pic] to obtain that [pic] =
[pic]. Since the expression [pic] can be differentiated down to 0, we could set up the following table to help us carry out the repeated Integration by Parts:
D I
[pic] [pic] To obtain the value of the indefinite integral,
[pic] [pic] + add the product of [pic] and [pic],
2 [pic] [pic] subtract the product of [pic] and [pic],
0 [pic] + add the product of 2 and [pic],
and add the constant of integration.
Thus, we have that [pic] = [pic] =
[pic].
NOTE: You are NOT allowed to use this table method for one application of the Integration by Parts technique.
15. [pic]
Since the expression [pic] can be differentiated down to 0, we could set up the following table to help us carry out the repeated Integration by Parts:
D I
[pic] [pic] To obtain the value of the indefinite integral,
[pic] [pic] + add the product of [pic] and [pic],
[pic] [pic] [pic] subtract the product of [pic] and [pic],
6 [pic] + add the product of [pic] and [pic],
0 [pic] [pic] subtract the product of 6 and [pic],
and add the constant of integration.
Thus, we have that [pic] =
[pic] =
[pic]
Answer: [pic]
16. [pic]
Since the expression [pic] can be differentiated down to 0, we could set up the following table to help us carry out the repeated Integration by Parts:
D I
[pic] [pic] To obtain the value of the indefinite integral,
[pic] [pic] + add the product of [pic] and [pic],
[pic] [pic] [pic] subtract the product of [pic] and [pic],
[pic] [pic] + add the product of [pic] and [pic],
24 [pic] [pic] subtract the product of [pic] and [pic],
0 [pic] + add the product of 24 and [pic],
and add the constant of integration.
Thus, we have that [pic] =
[pic] =
[pic]
Answer: [pic]
17. [pic]
Since [pic] and we have the x in the integral, then this indefinite integral can be evaluated by simple substitution.
Let [pic]. Then [pic].
[pic] = [pic] = [pic] = [pic] = [pic]
Answer: [pic]
18. [pic]
We will use the Integration by Parts technique on this integral. Note that you can not integrate [pic]. However, the expression [pic] can be integrated by simple substitution. Thus,
Let [pic] and [pic]
Then [pic] [pic]
To find [pic]: Let [pic]. Then [pic].
Then [pic] = [pic] = [pic] = [pic] = [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic]
Answer: [pic]
19. [pic]
We will use the Integration by Parts technique on this integral. Note that you can not integrate [pic]. However, the expression [pic] can be integrated by simple substitution. Thus,
Let [pic] and [pic]
Then [pic] [pic]
To find [pic]: Let [pic]. Then [pic].
Then [pic] = [pic] = [pic] = [pic] = [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] =
[pic]
Answer: [pic]
20. [pic]
Let [pic] and [pic]
Then [pic] [pic]
[pic] = [pic] = [pic] =
[pic] =
NOTE: [pic] [pic]
[pic] =
[pic] =
[pic] =
[pic] =
[pic]
Thus, we have that
[pic] = [pic]
Notice that the indefinite integral [pic], which we are evaluating, appears on both sides of the equation above. Thus, we could solve for [pic].
Adding [pic] to both sides of the equation, we obtain that
[pic] = [pic].
Since [pic], where C is a constant, then we have that [pic] =
[pic]. Now, dividing both sides of this equation by 2, we obtain that
[pic] = [pic] =
[pic] =
[pic], where [pic].
Answer: [pic]
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