8 techniques of integration. there are two situations in which it is impossible to find the exact...

8TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION

There are two situations in which it is

impossible to find the exact value of

a definite integral.

TECHNIQUES OF INTEGRATION

The first situation arises from the fact that,

in order to evaluate using the

Fundamental Theorem of Calculus (FTC),

we need to know an antiderivative of f.

af x dx

However, sometimes, it is difficult, or

even impossible, to find an antiderivative

(Section 7.5).

For example, it is impossible to evaluate the following integrals exactly:

21 1 3

0 11xe dx x dx

The second situation arises when the function

is determined from a scientific experiment

through instrument readings or collected data.

There may be no formula for the function (as we will see in Example 5).

In both cases, we need to find

approximate values of definite

integrals.

7.7Approximate Integration

In this section, we will learn:

How to find approximate values

of definite integrals.

APPROXIMATE INTEGRATION

We already know one method for

approximate integration.

Recall that the definite integral is defined as a limit of Riemann sums.

So, any Riemann sum could be used as an approximation to the integral.

If we divide [a, b] into n subintervals

of equal length ∆x = (b – a)/n, we have:

where xi* is any point in the i th subinterval

[xi -1, xi].

( ) ( *)nb

f x dx f x x

Ln APPROXIMATION

If xi* is chosen to be the left endpoint of

the interval, then xi* = xi -1 and we have:

The approximation Ln is called the left endpoint approximation.

( ) ( )nb

f x dx L f x x

Equation 1

If f(x) ≥ 0, the integral represents an area

and Equation 1 represents an approximation

of this area by the rectangles shown here.

Ln APPROXIMATION

If we choose xi* to be the right endpoint,

xi* = xi and we have:

The approximation Rn is called right endpoint approximation.

Equation 2

( ) ( )nb

f x dx R f x x

Rn APPROXIMATION

In Section 5.2, we also considered the case

where xi* is chosen to be the midpoint

of the subinterval [xi -1, xi].

Mn APPROXIMATION

The figure shows

the midpoint

approximation Mn.

Mn APPROXIMATION

Mn appears to be better

than either Ln or Rn.

THE MIDPOINT RULE

[ ( ) ( ) ... ( )]

f x dx M

x f x f x f x

11 12 ( ) midpoint of [ , ]i i i i ix x x x x

TRAPEZOIDAL RULE

Another approximation—called the

Trapezoidal Rule—results from averaging

the approximations in Equations 1 and 2,

as follows.

TRAPEZOIDAL RULE

0 1 1 2

1( ) ( ) ( )

( ( ) ( ))2

( ( ) ( )) ( ( ) ( ))2

... ( ( ) ( ))

( ) 2 ( ) 2 ( )2

... 2 ( ) ( )

i iai i

f x dx f x x f x x

xf x f x

xf x f x f x f x

f x f x

xf x f x f x

f x f x

THE TRAPEZOIDAL RULE

where ∆x = (b – a)/n and xi = a + i ∆x

( ) 2 ( ) 2 ( )2

... 2 ( ) ( )

f x dx T

xf x f x f x

f x f x

TRAPEZOIDAL RULE

The reason for the name can be seen

from the figure, which illustrates the case

f(x) ≥ 0.

TRAPEZOIDAL RULE

The area of the trapezoid that lies above

the i th subinterval is:

If we add the areas of all these trapezoids,we get the right side of the Trapezoidal Rule.

( ) ( )[ ( ) ( )]

2 2i i

f x f x xx f x f x

Approximate the integral

with n = 5, using:

a. Trapezoidal Rule

b. Midpoint Rule

Example 12

1(1/ )x dx

With n = 5, a = 1 and b = 2,

we have: ∆x = (2 – 1)/5 = 0.2

So, the Trapezoidal Rule gives:

1 0.2[ (1) 2 (1.2) 2 (1.4)

22 (1.6) 2 (1.8) (2)]

1 2 2 2 2 10.1

1 1.2 1.4 1.6 1.8 2

0.695635

dx T f f fx

Example 1 a

The approximation is illustrated

Example 1 a

The midpoints of the five subintervals

are: 1.1, 1.3, 1.5, 1.7, 1.9

Example 1 b

So, the Midpoint Rule gives:

1[ (1.1) (1.3) (1.5)

(1.7) (1.9)]

1 1 1 1 1 1

5 1.1 1.3 1.5 1.7 1.9

0.691908

dx x f f fx

Example 1 b

In Example 1, we deliberately chose

an integral whose value can be computed

explicitly so that we can see how accurate

the Trapezoidal and Midpoint Rules are.

By the FTC,

1ln ] ln 2 0.693147...dx x

APPROXIMATION ERROR

The error in using an approximation is

defined as the amount that needs to be

added to the approximation to make it

exact.

From the values in Example 1, we see that

the errors in the Trapezoidal and Midpoint

Rule approximations for n = 5 are:

ET ≈ – 0.002488

EM ≈ 0.001239

In general, we have:

E f x dx T

E f x dx M

The tables show the results of calculations

similar to those in Example 1. However, these are for n = 5, 10, and 20 and for

the left and right endpoint approximations and also the Trapezoidal and Midpoint Rules.

We can make several observations

from these tables.

In all the methods. we get more accurate

approximations when we increase n. However, very large values of n result in so many

arithmetic operations that we have to beware of accumulated round-off error.

OBSERVATION 1

OBSERVATION 2

The errors in the left and right endpoint

approximations are: Opposite in sign Appear to decrease by a factor of about 2

when we double the value of n

OBSERVATION 3

The Trapezoidal and Midpoint Rules

are much more accurate than the endpoint

approximations.

OBSERVATION 4

The errors in the Trapezoidal and Midpoint

Rules are: Opposite in sign Appear to decrease by a factor of about 4

when we double the value of n

OBSERVATION 5

The size of the error in the Midpoint Rule

is about half that in the Trapezoidal Rule.

MIDPOINT RULE VS. TRAPEZOIDAL RULE

The figure shows why we can usually expect

the Midpoint Rule to be more accurate than

the Trapezoidal Rule.

The area of a typical rectangle in

the Midpoint Rule is the same as the area

of the trapezoid ABCD whose upper side is

tangent to the graph at P.

The area of this trapezoid is closer to

the area under the graph than is the area

of that used in the Trapezoidal Rule.

The midpoint error (shaded red) is smaller

than the trapezoidal error (shaded blue).

OBSERVATIONS

These observations are corroborated

in the following error estimates—which

are proved in books on numerical

analysis.

OBSERVATIONS

Notice that Observation 4 corresponds

to the n2 in each denominator because:

(2n)2 = 4n2

That the estimates depend on the size of

the second derivative is not surprising if you

look at the figure.

f’’(x) measures how much the graph is curved.

Recall that f’’(x) measures how fast the slope of y = f(x) changes.

ERROR BOUNDS

Suppose | f’’(x) | ≤ K for a ≤ x ≤ b.

If ET and EM are the errors in

the Trapezoidal and Midpoint Rules,

then3 3

( ) ( )and

12 24T M

K b a K b aE E

Estimate 3

ERROR BOUNDS

Let’s apply this error estimate to the

Trapezoidal Rule approximation in Example 1.

If f(x) = 1/x, then f’(x) = -1/x2 and f’’(x) = 2/x3.

As 1 ≤ x ≤ 2, we have 1/x ≤ 1; so,

2 2''( ) 2

ERROR BOUNDS

So, taking K = 2, a = 1, b = 2, and n = 5

in the error estimate (3), we see:

2(2 1) 1

12(5) 150

0.006667

ERROR BOUNDS

Comparing this estimate with the actual error

of about 0.002488, we see that it can happen

that the actual error is substantially less than

the upper bound for the error given by (3).

ERROR ESTIMATES

How large should we take n in order to

guarantee that the Trapezoidal and Midpoint

Rule approximations for are

accurate to within 0.0001?

Example 2

1(1/ )x dx

ERROR ESTIMATES

We saw in the preceding calculation

that | f’’(x) | ≤ 2 for 1 ≤ x ≤ 2

So, we can take K = 2, a = 1, and b = 2 in (3).

Example 2

ERROR ESTIMATES

Accuracy to within 0.0001 means that

the size of the error should be less than

0.0001

Therefore, we choose n so that:3

2(1)0.0001

Example 2

ERROR ESTIMATES

Solving the inequality for n,

we get

Thus, n = 41 will ensure the desired accuracy.

12(0.0001)n

0.0006n

Example 2

ERROR ESTIMATES

It’s quite possible that a lower value

for n would suffice.

However, 41 is the smallest value for which the error-bound formula can guarantee us accuracy to within 0.0001

Example 2

ERROR ESTIMATES

For the same accuracy with the Midpoint Rule,

we choose n so that:

This gives:

2(1)0.0001

0.0012n

Example 2

ERROR ESTIMATES

a. Use the Midpoint Rule with n = 10 to

approximate the integral

b. Give an upper bound for the error involved

in this approximation.

0.xe dx

Example 3

ERROR ESTIMATES

As a = 0, b = 1, and n = 10, the Midpoint Rule

gives:21

0.0025 0.0225 0.0625 0.1225 0.2025

0.3025 0.4225 0.5625 0.7225 0.9025

[ (0.05) (0.15) ... (0.85) (0.95)]

1.460393

x f f f f

e e e e e

Example 3 a

ERROR ESTIMATES

The approximation is illustrated.

Example 3 a

ERROR ESTIMATES

As f(x) = ex2, we have:

f’(x) = 2xex2 and f’’(x) = (2 + 4x2)ex2

Also, since 0 ≤ x ≤ 1, we have x2 ≤ 1.

Hence, 0 ≤ f’’(x) = (2 + 4x2) ex2 ≤ 6e

Example 3 b

ERROR ESTIMATES

Taking K = 6e, a = 0, b = 1, and n = 10 in

the error estimate (3), we see that an upper

bound for the error is:3

24(10) 400

Example 3 b

ERROR ESTIMATES

Error estimates give upper bounds

for the error.

They are theoretical, worst-case scenarios.

The actual error in this case turns out to be about 0.0023

Another rule for approximate integration

results from using parabolas instead

of straight line segments to approximate

a curve.

As before, we divide [a, b] into n subintervals

of equal length h = ∆x = (b – a)/n.

However, this time, we assume n is an even

number.

Then, on each consecutive pair of intervals,

we approximate the curve y = f(x) ≥ 0 by

a parabola, as shown.

If yi = f(xi), then Pi(xi, yi) is the point on

the curve lying above xi.

A typical parabola passes through three consecutive points: Pi, Pi+1, Pi+2

To simplify our calculations, we first

consider the case where:

x0 = -h, x1 = 0, x2 = h

We know that the equation of

the parabola through P0, P1, and P2

is of the form

y = Ax2 + Bx + C

Therefore, the area under the parabola

from x = - h to x = h is: 2 2

( ) 2 ( )

(2 6 )3

Ax Bx C dx Ax C dx

However, as the parabola passes through

P0(- h, y0), P1(0, y1), and P2(h, y2), we have:

y0 = A(– h)2 + B(- h) + C = Ah2 – Bh + C

y1 = C

y2 = Ah2 + Bh + C

Therefore,

y0 + 4y1 + y2 = 2Ah2 + 6C

So, we can rewrite the area under

the parabola as:0 1 2( 4 )

hy y y

Now, by shifting this parabola

horizontally, we do not change

the area under it.

This means that the area under the parabola

through P0, P1, and P2 from x = x0 to x = x2

is still:0 1 2( 4 )

hy y y

Similarly, the area under the parabola

through P2, P3, and P4 from x = x2 to x = x4

is:2 3 4( 4 )

hy y y

Thus, if we compute the areas under all

the parabolas and add the results, we get:

0 1 2 2 3 4

0 1 2 3 4

( ) ( 4 ) ( 4 )3 3

... ( 4 )3

( 4 2 4 23

... 2 4 )

h hf x dx y y y y y y

hy y y

hy y y y y

Though we have derived this approximation

for the case in which f(x) ≥ 0, it is a reasonable

approximation for any continuous function f .

Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1

SIMPSON’S RULE

This is called Simpson’s Rule—after

the English the English mathematician

Thomas Simpson (1710–1761).

SIMPSON’S RULE

where n is even and ∆x = (b – a)/n.

0 1 2 3

[ ( ) 4 ( ) 2 ( ) 4 ( )3

... 2 ( ) 4 ( ) ( )]

f x dx S

xf x f x f x f x

f x f x f x

SIMPSON’S RULE

Use Simpson’s Rule

with n = 10 to approximate

1(1/ )x dx

Example 4

SIMPSON’S RULE

Putting f(x) = 1/x, n = 10, and ∆x = 0.1 in

Simpson’s Rule, we obtain:

1[ (1) 4 (1.1) 2 (1.2) 4 (1.3)

3... 2 (1.8) 4 (1.9) (2)]

1 4 2 4 2 4 2 40.1 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

2 4 13

1.8 1.9 20.693150

xdx S f f f fx

Example 4

SIMPSON’S RULE

In Example 4, notice that Simpson’s Rule

gives a much better approximation

(S10 ≈ 0.693150) to the true value of the

integral (ln 2 ≈ 0.693147) than does either:

Trapezoidal Rule (T10 ≈ 0.693771)

Midpoint Rule (M10 ≈ 0.692835)

SIMPSON’S RULE

It turns out that the approximations in

Simpson’s Rule are weighted averages of

those in the Trapezoidal and Midpoint Rules:

Recall that ET and EM usually have opposite signs and | EM | is about half the size of | ET |.

1 22 3 3n n nS T M

SIMPSON’S RULE

In many applications of calculus, we need to

evaluate an integral even if no explicit formula

is known for y as a function of x.

A function may be given graphically or as a table of values of collected data.

SIMPSON’S RULE

If there is evidence that the values are not

changing rapidly, then the Trapezoidal Rule

or Simpson’s Rule can still be used to find

an approximate value for .b

SIMPSON’S RULE

The figure shows data traffic on the link from

the U.S. to SWITCH, the Swiss academic and

research network, on February 10, 1998.

D(t) is the data throughput, measured in megabits per second (Mb/s).

Example 5

SIMPSON’S RULE

Use Simpson’s Rule to estimate the total

amount of data transmitted on the link up to

noon on that day.

Example 5

SIMPSON’S RULE

Since we want the units to be consistent

and D(t) is measured in Mb/s, we convert

the units for t from hours to seconds.

Example 5

SIMPSON’S RULE

If we let A(t) be the amount of data (in Mb)

transmitted by time t, where t is measured in

seconds, then A’(t) = D(t).

So, by the Net Change Theorem (Section 5.4), the total amount of data transmitted by noon (when t = 12 x 602 = 43,200) is:

43,200

0(43,200) ( )A D t dt

Example 5

SIMPSON’S RULE

We estimate the values of D(t) at hourly

intervals from the graph and compile them

Example 5

SIMPSON’S RULE

Then, we use Simpson’s Rule

with n = 12 and ∆t = 3600 to estimate

the integral, as follows.

Example 5

SIMPSON’S RULE

43,200

[ (0) 4 (3600) 2 (7200)3... 4 (39,600) (43,200)]

3600[3.2 4(2.7) 2(1.9) 4(1.7)

32(1.3) 4(1.0) 2(1.1) 4(1.3)

2(2.8) 4(5.7) 2(7.1) 4(7.7) 7.9] 143,880

A t dt

tD D D

Example 5

The total amount of data transmitted up to noon is 144,000 Mbs, or 144 gigabits.

SIMPSON’S RULE VS. MIDPOINT RULE

The table shows how Simpson’s Rule

compares with the Midpoint Rule for

the integral , whose true value

is about 0.69314718

1(1/ )x dx

SIMPSON’S RULE

This table shows how the error Es

in Simpson’s Rule decreases by

a factor of about 16 when n is doubled.

SIMPSON’S RULE

That is consistent with the appearance of n4

in the denominator of the following error

estimate for Simpson’s Rule.

It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules.

However, it uses the fourth derivative of f.

ERROR BOUND (SIMPSON’S RULE)

Suppose that | f (4)(x) | ≤ K for a ≤ x ≤ b.

If Es is the error involved in using

Simpson’s Rule, then 5

K b aE

Estimate 4

How large should we take n to guarantee

that the Simpson’s Rule approximation

for is accurate to within 0.0001?2

1(1/ )x dx

Example 6

If f(x) = 1/x, then f (4)(x) = 24/x5.

Since x ≥ 1, we have 1/x ≤ 1, and so

Thus, we can take K = 24 in (4).

24( ) 24f x

Example 6

So, for an error less than 0.0001, we should

choose n so that:

This gives

24(1)0.0001

180(0.0001)n

0.00075n

Example 6

Therefore, n = 8 (n must be even)

gives the desired accuracy.

Compare this with Example 2, where we obtained n = 41 for the Trapezoidal Rule and n = 29 for the Midpoint Rule.

Example 6

a. Use Simpson’s Rule with n = 10

to approximate the integral .

b. Estimate the error involved in this

approximation.

Example 7

If n =10, then ∆x = 0.1 and the rule gives:

0 0.01 0.04 0.09 0.16

0.25 0.36 0.49 0.64 0.81 1

[ (0) 4 (0.1) 2 (0.2) ...3

2 (0.8) 4 (0.9) (1)]

0.1[ 4 2 4 2

4 2 4 2 4 ]

1.462681

x xe f f f

e e e e e

e e e e e e

Example 7 a

The fourth derivative of f(x) = ex2 is:

f(4)(x) = (12 + 48x2 + 16x4)ex2

So, since 0 ≤ x ≤ 1, we have:

0 ≤ f(4)(x) ≤ (12 + 48 +16)e1 = 76e

Example 7 b

Putting K = 76e, a = 0, b = 1, and n = 10

in (4), we see that the error is at most:

Compare this with Example 3.

76 (1)0.000115

180(10)

Example 7 b

Thus, correct to three decimal places,

we have:

01.463xe dx

Example 7 b

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