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Integrating Factors Found by Inspection Tags: integrating factor exact differential common differential This section will use the following four exact differentials that occurs frequently. 1. 2. 3. 4. The Determination of Integrating Factor Tags: exact equation partial differentiation integrating factor function of x alone function of y alone From the differential equation Rule 1 If , a function of x alone, then is the integrating factor. Rule 2 If , a function of y alone, then is the integrating factor.

Integrating Factors Found by Inspection

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Page 1: Integrating Factors Found by Inspection

Integrating Factors Found by InspectionTags: integrating factorexact differentialcommon differential

This section will use the following four exact differentials that occurs frequently.

1. 

2. 

3. 

4.

The Determination of Integrating FactorTags: exact equationpartial differentiationintegrating factorfunction of x alonefunction of y alone

From the differential equation 

 

Rule 1

If    ,   a function of x alone, then       is the integrating factor.

 

Rule 2

If    ,   a function of y alone, then       is the integrating factor.

 

Note that the above criteria is of no use if the equation does not have an integrating factor that is a function of x or y alone. 

Page 2: Integrating Factors Found by Inspection

Steps

1. Take the coefficient of dx as M and the coefficient of dy as N.2. Evaluate ∂M/∂y and ∂N/∂x.3. Take the difference ∂M/∂y - ∂N/∂x.4. Divide the result of Step 3 by N. If the quotient is a function of x alone, use the integrating

factor defined in Rule 1 above and proceed to Step 6. If the quotient is not a function of x alone, proceed to Step 5.

5. Divide the result of Step 3 by M. If the quotient is a function of y alone, use the integrating factor defined in Rule 2 above and proceed to Step 6. If the quotient is not a function of y alone, look for another method of solving the equation.

6. Multiply both sides of the given equation by the integrating factor u, the new equation which is uM dx + uN dy = 0 should be exact.

7. Solve the result of Step 6 by exact equation or by inspection.

Problem 01 | Determination of Integrating FactorTags: partial differentiationintegrating factorfunction of x alone

Problem 01

 

Solution 01

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Page 3: Integrating Factors Found by Inspection

 

       → a function of x alone 

 

Integrating factor

 

Thus,

           answer

Page 4: Integrating Factors Found by Inspection

Problem 02 | Determination of Integrating FactorTags: partial differentiationintegrating factorfunction of x alone

Problem 02

 

Solution 02

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       → a function of x alone 

 

Integrating factor

 

Thus,

Page 5: Integrating Factors Found by Inspection

           answer

Problem 03 | Determination of Integrating FactorTags: partial differentiationintegrating factorfunction of y alone

Problem 03

 

Solution 03

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Page 6: Integrating Factors Found by Inspection

       → neither a function of x alone nor y alone

       → a function of y alone 

 

Integrating factor

 

Thus,

           answer

Problem 04

 

Solution 04

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Page 7: Integrating Factors Found by Inspection

 

 

       → neither a function of x alone nor y alone

       → a function y alone 

 

Integrating factor

 

Thus,

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           answer

Substitution Suggested by the Equation | Bernoulli's EquationTags: substitutionlinear differential equationintegrating factorBernoulli's equationsuggested substitution

Substitution Suggested by the EquationExample 1

 

The quantity (2x - y) appears twice in the equation. Let

 

Substitute,

then continue solving.

 

Example 2

 

The quantity (-sin y dy) is the exact derivative of cos y. Let

 

Substitute,

then continue solving.

 

Page 9: Integrating Factors Found by Inspection

Bernoulli's EquationBernoulli's equation is in the form 

 

If x is the dependent variable, Bernoulli's equation can be recognized in the form .

 

If n = 1, the variables are separable.If n = 0, the equation is linear.If n ≠ 1, Bernoulli's equation. 

Steps in solving Bernoulli's equation

ndefinite IntegralsIndefinite Integrals

If F(x) is a function whose derivative F'(x) = f(x) on certain interval of the x-axis, then F(x) is called the anti-derivative of indefinite integral f(x). When we integrate the differential of a function we get that function plus an arbitrary constant. In symbols we write 

 

where the symbol  , called the integral sign, specifies the operation of integration upon f(x) dx; that is, we are to find a function whose derivative is f(x) or whose differential is f(x) dx. The dx tells us that the variable of integration is x.

- See more at: http://www.mathalino.com/reviewer/integral-calculus/indefinite-integrals#sthash.J5q1Fvlb.dpuf

1. Write the equation into the form    . 

2. Identify    ,    ,   and    . 

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3. Write the quantity       and let    . 

4. Determine the integrating factor    . 

5. The solution is defined by    . 

6. Bring the result back to the original variable.

7. 1 - 3 Examples | Indefinite Integrals8. Tags: 9. power formula10. integration

11. Evaluate the following integrals:

12.[13.] Example 1: 

13.[14.] Example 2: 

14.[15.] Example 3:  

15.[16.] Solution to Example 1:16.[17.] HideClick here to show or hide the solution

17.[18.]

18.[19.]

19.[20.]

20.[21.]

21.[22.]

22.[23.]

23.[24.]           answer24.[25.]  25.[26.] Solution to Example 2:26.[27.] HideClick here to show or hide the solution

27.[28.]  

28.[29.] This is the form  . If we let  , then   is raised to a power 4 and

is multiplied by the differential   of the function corresponding to  ,

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the integral can be evaluated as follows:

          answer 

29.[30.] It should be pointed out that no integral can be evaluated directly unless it contains, in addition to the expression identified with  , the exact differential of the function corresponding to  .

30.[31.]  31.[32.] Solution to Example 3:32.[33.] HideClick here to show or hide the solution

33.[34.]

34.[35.]

35.[36.]

36.[37.]            answer

4 - 6 Examples | Indefinite IntegralsTags: power formulaintegration

Evaluate the following:

Example 4: 

Example 5: 

Example 6:  

Solution to Example 4

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This is not of the form   because of the missing constant factor 3 in the integrand.

Identifying  ,  , then the differential  . We must then insert 3 in the integrand and to compensate for it, we place the reciprocal 1/3 before the integral sign. This in effect

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multiplying by one does not affect the value of the function. 

Let  , then  

           answer

 

Solution to Example 5

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Letu = 2x3 + 2x + 1du = (6x2 + 2) dx = 2(3x2 + 1) dxn = -2/3 

           answer

 

Solution to Example 6

Page 13: Integrating Factors Found by Inspection

HideClick here to show or hide the solution

If we let   and  , then  . But there is no   in the given integrand. It is easy to insert -4 in the integrand and offset this by placing -1/4 before the integral sign but nothing

can be done about the missing factor  . We therefore expand   and integrate term by term. 

           answer

Definite IntegralThe definite integral of f(x) is the difference between two values of the integral of f(x) for two distinct values of the variable x. If the integral of f(x) dx = F(x) + C, the definite integral is denoted by the symbol 

 The quantity F(b) - F(a) is called the definite integral of f(x) between the limits a and b or simply the definite integral from a to b. It is called the definite integral because the result involves neither x nor the constant C and therefore has a definite value. The numbers a and b are called the limits of integration, a being the lower limit and b the upper limit. 

General Properties of Definite Integral

1. The sign of the integral changes if the limits are interchanged. 

 

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2. The interval of integration may be broken up into any number of subintervals, and integrate over each interval separately. 

 

3. The definite integral of a given integrand is independent of the variable of integration. Hence, it makes no difference what letter is used for the variable of integration. 

Inverse Trigonometric Functions | Fundamental Integration FormulasIn applying the formula (Example: Formula 1 below), it is important to note that the numerator du is the differential of the variable quantity u which appears squared inside the square root symbol. We mentally put the quantity under the radical into the form of the square of the constant minus the square of the variable. 

1. 

2. 

3.  

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