Decomposition of Fractions

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# Decomposition of Fractions - PowerPoint PPT Presentation

Decomposition of Fractions. Integration in calculus is how we find the area between a curve and the x axis.

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### Decomposition of Fractions

Integration in calculus is how we find the area between a curve and the x axis.

Examples: vibration, distortion under weight, or one of many types of "fluid flow" -- be it heat flow, air flow (over a wing), or water flow (over a ship\'s hull, through a pipe, or perhaps even groundwater flow regarding a contaminant) All these things can be either directly solved by integration (for simple systems), or some type of numerical integration (for complex systems).

Decomposition of fractions makes the process of integration easier.

Goal: To split this fraction into two or more parts

Step 1: Determine if the numerator is linear or quadratic or if the degree of the numerator is greater than the denominator.

Step 2: IF the degree of the numerator is greater than the degree of the denominator you must divide the numerator by the denominator.

Step 3: Once division is done, if needed, FACTOR the denominator.

Step 2: If an improper fraction (the degree of the numerator is greater than the degree of the denominator) must divide numerator by the denominator FIRST.

Step 4: Linear Factors: For EACH factor of the form (px + q)m , the partial fraction decomposition must include the following sum of m fractions.

Step 5: Quadratic Factors: For EACH factor of the form (ax2 + bx + c)n , the partial fraction decomposition must include the following sum of n fractions.

Step 1: Numerator is linear.

Step 2: Factor denominator.

Step 3: Clear the denominator by multiplying both sides by (x-3)(x+2).

Step 2: The expression is proper so no need to divide.

Step 1: x can be any number we want. So let’s pick an x so that it’s easy to solve. Like -2 and 3.

Step 2: To solve for A let x = 3

(3) + 7 =A (3 + 2) + B(3 -3)

10 = 5A

A = 2

Step 3: To solve for B let x = -2

(-2) + 7 =A (-2 + 2) + B(-2 -3)

5= -5 B

B = -1

We must divide

Step 1: IMPROPER, so must divide!

We must now factor denominator.

Step 2: Factor denominator.

x(x2 + 2x + 1)

Or x(x+ 1)2

Step 3: That squared term means we have to include one partial fraction for each power up to 2.

Notice how the linear factors had to be included like step 3 stated.

Step 4: Solve for A,B,C by clearing the denominator and letting x = 0, -1.

Clear denominator

To Solve for C let

x = -1

Step 5: Solve for C by letting x = -1.

Step 6: Solve for A by letting x = 0.

Step 7: We’ve used all the convent x’s so let x equal anything. So letting A = 6 and C = 9,

let’s pick x = 1