decomposition of fractions. integration in calculus is how we find the area between a curve and the...
TRANSCRIPT
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.
6
72
xx
x
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.
6
122
2
xx
xxxxx
xxxx
23
234
2
62062Step 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 3: Once division is done, if needed, FACTOR the denominator.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.
mm
qpx
A
qpx
A
qpx
A
qpx
A
)(.....
)()()( 33
221
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.
nnn
cbxax
CxB
cbxax
CxB
cbxax
CxB
cbxax
CxB
)(...
)()()( 23233
2222
211
6
72
xx
x
Step 1: Numerator is linear.Step 2: The expression is proper so no need to divide.
Step 3: Factor denominator.
)2)(3(
7
xx
x
23)2)(3(
7
x
B
x
A
xx
x
Step 4: Clear the denominator by multiplying both sides by (x-3)(x+2)
23
)2)(3()2)(3(
7)2)(3(
x
B
x
Axx
xx
xxx
)3()2(7 xBxAx
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.
)3()2(7 xBxAx
Step 2: To solve for A let x = 3
(3) + 7 =A (3 + 2) + B(3 -3)10 = 5A
A = 2Step 3: To solve for B let x = -2
(-2) + 7 =A (-2 + 2) + B(-2 -3)5= -5 BB = -1
2
1
3
2
)2)(3(
7
xxxx
x
86
82
xx
x
xxx
xxxx
23
234
2
62062
Step 1: IMPROPER, so must divide!
xxx
xxx
23
2
2
6205
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.
22
2
)1_(1)1(
6205
x
C
x
B
x
A
xx
xx
Step 4: Solve for A,B,C by clearing the denominator and letting x = 0, -1.
22
2
)1(1)1(
6205
x
C
x
B
x
A
xx
xx
x
CxxBxxAxx )1()1(6205 22
Step 5: Solve for C by letting x = -1.)1()11)(1()11(6)1(20)1(5 22 CBA
)1(6205 C
C19 9C
Step 6: Solve for A by letting x = 0.
2)1(
9
1
xx
B
x
Ax
)0()10)(0()10(6)0(20)0(5 2 CBA
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)1(9)11)(1()11(66)1(20)1(5 22 B
92)4(631 B
1B
A16
2)1(
9
1
6
xx
B
xx
2)1(
9
1
16
xxx
x
xxx
xx
23
2
2
53