Block #9: Integration by Partial Fractions, Approximate ...apsc171/Notes/notes09.pdf · Partial Fractions - 1 Partial Fractions This is the last of the techniques of integration (that

Block #9: Integration by Partial Fractions, ApproximateIntegration

Goals:

• Partial Fractions• Approximate Integration

1

Partial Fractions - 1

Partial Fractions

This is the last of the techniques of integration (that is, techniques foranti-differentiation) covered in this course. The method of PartialFractions is purely algebraic. It consists of a series of algorithmicsteps that simplify the integrand so that an anti-derivative can beeasily found.For example, consider the integral∫

x + 5

x2 + x− 2dx.

List the methods of integration you might try to evaluate this integral,and why they might or might not work.


The method of partial fractions is used only for expressions of theform

P (x)

Q(x)(rational function),

where P and Q are polynomials.A proper rational function is one for which the degree of P is strictlyless than the degree of Q. To be able to use the method of PartialFractions, we must first make sure that the integrand is a properrational function; this is our first step.


Step 1.

TurnP (x)

Q(x)into an expression involving a proper rational function

and a polynomial:P (x)

Q(x)= S(x) +

R(x)

Q(x).

Separate the rational functionx2 − 4x + 3

x− 5into a polynomial and a

proper rational function.


Problem. Find

∫x2 + 1

x2 − 1dx.


The next step (after we have turned the integral into one involving aproper rational) consists of four cases, distinguished by what happenswhen you factor the denominator.Step 2. CASE I:The denominator is the product of distinct linear factors. Say

Q(x) = (a1x + b1)(a2x + b2) · · · (akx + bk).

The goal is to look for numbers A1, · · · , Ak so thatR(x)

Q(x)=

A1

a1x+ b1+

A2

a2x+ b2+ . . .+

Ak

akx+ bk.

“Partial Fractions”

Partial Fractions - Examples 1 - 1

Partial Fractions - Examples 1

Problem. Find

∫x

x2 + 3x + 2dx.

Partial Fractions - Examples 1 - 2 ∫x

x2 + 3x + 2dx.


Problem. Find

∫3x2 − 2

(x− 1)(x− 2)(x + 1)dx.

Partial Fractions - Examples 1 - 4 ∫3x2 − 2

(x− 1)(x− 2)(x + 1)dx.

Partial Fractions - Examples 1 - 5∫3x2 − 2

(x− 1)(x− 2)(x + 1)dx.


Problem. Find the volume of the “hat-shaped” solid you get whenyou take the region between the graph of the function 1/(x2 +3x+2)and the x-axis, bounded by the y-axis and the line x = 1, and rotateit around the vertical axis.



Step 2. CASE II:The polynomial Q(x) is a product of linear factors, some of whichare repeated. Rule: If (ax + b) occurs to the power r, then insteadof one term, put down the following r terms:

A1

ax + b+

A2

(ax + b)2+ . . . +

Ar

(ax + b)r.

Problem. Find

∫dx

x2(x− 1)2.

Partial Fractions - Examples 1 - 9∫dx

x2(x− 1)2.


Why do we have to learn these rules for setting theseproblems up?It is natural at this point to wonder why we have to set up thequestion precisely as we did. Why could we not reduce the numberof variables by letting

1

x2(x− 1)2=

B

x2+

C

x− 1+

D

(x− 1)2?

If we did this we would get the equations

C = 0

B − C + D = 0

2B = 0

B = 1

You can see immediately that this system of equations does not havea solution. That is, there do not exist numbers B, C, and D thatsatisfy all four equations at once.


Problem. Suppose we want to integrate

∫1

(x− 1)(x + 2)2dx. How

should we set up the partial fractions?

(A.)A

x− 1+

B

x + 2(B.)

A

x− 1+

B

x + 2+

C

x + 2

(C.)A

x− 1+

B

(x + 2)2(D.)

A

x− 1+

B

x + 2+

C

(x + 2)2

(E.)A

x− 1+

B

(x− 1)2+

C

x + 2+

D

(x + 2)2


Problem. Find

∫1

(x− 1)(x + 2)2dx.

Partial Fractions - Examples 1 - 13∫1

(x− 1)(x + 2)2dx.

Partial Fractions - Examples 1 - 14 ∫1

(x− 1)(x + 2)2dx.


Step 2. CASES III AND IV:The polynomial Q(x) contains quadratic factors. These cases willbe included only if lecture time remains.

Academic Note: The method of partial fractions is included hereas an integration technique. However, it will also re-appear in a dif-ferential course next year for many of you (MTHE 225 being the mostcommon). In that course, partial fractions will be used to simplifyan integration-related transform called the Laplace transform whichis frequently used in analyzing engineering control systems.

Approximate Integration - Introduction - 1

Approximate Integration

Approximate Integration is very important when we cannot find aformula for the anti-derivative of a given function. For example, thefunctions in the integrals∫

e−x2dx and

∫ √1 + x3 dx

do not have a formula for their anti-derivatives.


Problem. If we don’t have a formula for the anti-derivative of

∫e−x2

dx,

what does that tell us about the definite integral∫ 2

0e−x2

dx?

A. The integral’s value is undefined because e−x2will be undefined.

B. The integral’s value is undefined because there is no formula for

the anti-derivative of e−x2.

C. The integral’s value is defined, but we can’t calculate it exactly

because there is no formula for the anti-derivative of e−x2.

D. The integral’s value is infinite because e−x2grows infinitely large

on the interval.


Approximating a Definite Integral

In engineering, it is not uncommon to want to find the definite integral

of a function,

∫ b

af (x) dx, but not be able to use the Fundamental

Theorem of Calculus (i.e. an anti-derivative, F (x) =∫f (x) dx) to

help us.Two common reasons are:

• we have a function like f (x) = ex2

for which no formula for theanti-derivative exists.

• we don’t have a formula for f (x), but rather just a set measureddata values (xi, f (xi)).


Sketch graphically how we could estimate the integral

∫ b

af (x) dx

in both these cases, using a Riemann Sum.

Approximate Integration - LEFT and RIGHT - 1

Approximating a Definite Integral, Given a Function

Suppose we wanted to know how close we could get to calculating∫ 1

0

x

x2 + 1dx by using a simple Riemann sum. For this example, we

note that we can use the Fundamental Theorem of Calculus to getan exact value: ∫ 1

0

x

x2 + 1dx =

1

2ln 2 ≈ 0.34657 . . . .

Here are some values for f (x) =x

x2 + 1between 0 and 1:

x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

In the next few examples we will see that, even using the same set ofnumbers, some ways of combining those numbers give more accurateresults than others.


x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

Problem. Calculate the Riemann sum using two intervals and theirmidpoints. (This is the “Midpoint Rule”.)

0

0.25

0.50

0 0.25 0.50 0.75 1.00

x

y

b

b

bb b


x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

Problem. Calculate the Riemann sum using four intervals and theirleft-end points. (This is the “left end-point approximation”, or LEFT(4).)

0

0.25

0.50

0 0.25 0.50 0.75 1.00

x

y

b

b

bb b


x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

Problem. Calculate the Riemann sum using four intervals and theirright-end points. (This is the “right end-point approximation”, orRIGHT(4).)

0

0.25

0.50

0 0.25 0.50 0.75 1.00

x

y

b

b

bb b


The general formula for the left end-point approximation with n in-tervals, LEFT(n), is:

LEFT(n) =

The general formula for the right end-point approximation with nintervals, RIGHT(n) is:

RIGHT(n) =

The Trapezoidal and Simpson’s Rules - 1

The Trapezoidal Rule

The average of the left end-point and right end-point approximationsis known as the “Trapezoidal Rule”. The Trapezoidal Rule is bestunderstood with a picture of the area interpretation on each interval:From the picture, what is the area of the trapezoid in the ith subin-terval?


How does this trapezoidal area relate to the average of the two rec-tangle areas from the left and right end-point rules?

We can write the Trapezoidal Rule in terms of the formula:∆x

2

[f (x0) + 2f (x1) + 2f (x2) + · · · + 2f (xn−1) + f (xn)

].

You should memorize this formula.


x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

Problem. Calculate the Riemann sum using four intervals and theTrapezoidal rule, TRAP(4).

0

0.25

0.50

0 0.25 0.50 0.75 1.00

x

y

b

b

bb b

Simpson’s Rule - 1

Simpson’s Rule

While better than LEFT or RIGHT, the Trapezoidal rule can stillbe improved further. Simpson’s Rule is an even better method,in that it provides a more accurate integral estimate for the sameamount of data or labour.


Simpson’s rule is based on approximating the area under the inte-grand using pieces of parabolas:

x

y

a b

x

y

a b

b

bb

b

b

b b

x

y

a b

b

bb

b

b

b b


The formula for Simpson’s Rule is: (n must be even)

∆x

3

[f (x0) + 4f (x1) + 2f (x2) + 4f (x3) + 2f (x4) + · · ·

· · · + 2f (xn−2) + 4f (xn−1) + f (xn)

].

You should memorize this formula.


Problem. Apply Simpson’s Rule to the

∫ 1

0

x

x2 + 1dx example.

x 0 0.25 0.5 0.75 1

f (x) 0 0.23529 . . . 0.4 0.48 0.5

0

0.25

0.50

0 0.25 0.50 0.75 1.00

x

y

b

b

bb b


Comment on the accuracy and the effort required to compute Simp-son’s Rule compared to the other methods.

Block #9: Integration by Partial Fractions, Approximate ...apsc171/Notes/notes09.pdf · Partial Fractions - 1 Partial Fractions This is the last of the techniques of integration (that

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