Chapter 7 Techniques of...

Chapter 7

Techniques of Integration

7.1 Integration by Parts

Every differentiation rule has a corresponding integration rule. For instance, the Substi-tution Rule for integration corresponds to the Chain Rule for differentiation. The rulethat corresponds to the Product Rule for differentiation is called the rule for integrationby parts .

The Product Rule state that if f and g are differentiable functions, then

d

dx

[

f(x)g(x)]

= f(x)g′(x) + g(x)f ′(x)

In the notation for indefinite integrals this equation becomes

∫

[

f(x)g′(x) + g(x)f ′(x)]

dx = f(x)g(x)

or∫

f(x)g′(x) dx+

∫

g(x)f ′(x) dx = f(x)g(x)

We can rearrange this equation as

∫

f(x)g′(x) dx = f(x)g(x)−∫

g(x)f ′(x) dx

This formula is called the formula for integration by parts.

Let u = f(x) and v = g(x). Then the differential are du = f ′(x)dx and dv = g′(x)dx,so, by the Substitution Rule, the formula for integration by parts becomes

∫

u dv = uv −∫

v du

151

MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 152

Find

∫

x cosx dx.

Example 7.1.

Solution

Note. Our aim in using integration by parts is to obtain a simpler integral than theone we start with. Thus, in Example 7.1 we start with

∫

x cos x dx and expressed it interms of the simpler integral

∫

sin xdx. If we had chosen u = cosx and dv = xdx, thendu = − sin xdx and v = x2/2, so integration by parts gives

∫

x cos x dx = (cosx)x2

2+

1

2

∫

x2 sin x dx

Although this is true,∫

x2 sin x dx is a more difficult integral than the one we startedwith.

In general, when deciding on a choice for u and dv, we usually try to choose u =f(x) to be a function that becomes simpler when differentiated (or at least not morecomplicated) as long as dv = g′(x)dx can be readily integrated to give v.

Find

∫

cosx ln(sin x) dx.

Example 7.2.

Solution


Find

∫

x2 sin 2x dx.

Example 7.3.

Solution

Find

∫

ex cosx dx.

Example 7.4.

Solution


If we combine the formula for integration by parts with Part 2 of the FundamentalTheorem of Calculus, we can evaluate definite integrals by parts. Assuming f ′ and g′

are continuous, and using the Fundamental Theorem, we obtain

∫ b

a

f(x)g′(x) dx = f(x)g(x)

]b

a

−∫ b

a

g(x)f ′(x) dx

That is, if u = f(x) and v = g(x), then

∫ b

a

u dv = uv]b

a−∫ b

a

v du.

Find

∫

1

0

tan−1 x dx.

Example 7.5.

Solution


Exercise 7.1

Evaluate the integral.

1.

∫

xe2x dx 2.

∫

x2 ln x dx

3.

∫

x sin 4x dx 4.

∫

x2e−3x dx

5.

∫

x2 cos 3x dx 6.

∫

ex sin 4x dx

7.

∫

(lnx)2 dx 8.

∫

cos x cos 2x dx

9.

∫

x sec2 x dx 10.

∫

cosx ln(sin x) dx

11.

∫

cos(lnx) dx 12.

∫

cos−1 x dx

13.

∫

sin√x dx 14.

∫

1

0

x sin 2x dx

15.

∫

1

0

xe−x dx 16.

∫

2

1

ln x

x2dx

17.

∫

4

1

ln√x dx 18.

∫

2

1

x4(ln x)2 dx

Answer to Exercise 7.1

1. 1

2xe2x − 1

4e2x + C 2. 1

3x2 ln x− 1

9x3 + C 3. −1

4x cos 4x+ 1

16sin 4x+ C

4. −1

3x2e−3x − 2

9xe−3x − 2

27e−3x + C 5. 1

3x2 cos 3x+ 2

9x cos 3x− 2

27sin 3x+C

6. 1

17ex sin 4x− 4

17ex cos 4x+ C 7. x(ln x)2 − 2x ln x+ 2x+ C

8. 2

3sin 2x cosx− 1

3cos 2x sin x+ C 9. x tan x+ ln | cosx|+ C

10. sin x ln(sin x)− sin x+ C 11. 1

2x[sin(ln x) + cos(ln x)] + C

12. x cos−1 x−√1− x2 + C 13. −2

√x cos

√x+ 2 sin

√x+ C 14. 1

4sin 2− 1

2cos 2

15. 1− 2

e16. 1

2− 1

2ln 2 17. 2 ln 4− 3

218. 3

2(ln 2)2 − 64

25ln 2 + 62

125

7.2 Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigono-metric functions. Our first aim is to evaluate integrals of the form

∫

sinm x cosn x dx,

where m and n are positive integers.


Case I: m or n Is an Odd Positive Integer

If m is odd, first isolate one factor of sin x. (You’ll need this for du). Then, replace anyfactors of sin2 x with 1− cos2 x and make the substitution u = cosx.

Likewise, if n is odd, first isolate one factor of cosx. (You’ll need this for du). Then,replace any factors of cos2 x with 1− sin2 x and make the substitution u = sin x.

Evaluate

∫

cos3 xdx.

Example 7.6.

Solution

Find

∫

cos5 x sin4 x dx.

Example 7.7.

Solution


Find

∫

cos2/3 x sin5 x dx.

Example 7.8.

Solution

Case II: m and n Are Both Even Positive Integers

In this case, we can use the half-angle formulas for sine and cosine to reduce the powerof in the integrand.

Half-angle formulas:

sin2 θ = 1

2(1− cos 2θ) and cos2 θ = 1

2(1 + cos 2θ)

Find

∫

sin4 x cos2 x dx.

Example 7.9.

Solution


Our next aim is to devise a strategy for evaluating integrals of the form

∫

tanm x secn x dx,

where m and n are integers.

Case I: m Is an Odd Positive Integer

First, isolate one factor of sec x tan x. (You’ll need this for du). Then, replace any factorsof tan2 x with sec2 x− 1 and make the substitution u = sec x.

Find

∫

tan3 x sec3/2 x dx.

Example 7.10.

Solution

Case II: n Is an Even Positive Integer

First, isolate one factor of sec2 x. (You’ll need this for du). Then, replace any remainingfactors of sec2 x with 1 + tan2 x and make the substitution u = tanx.

Find

∫

tan6 x sec4 x dx.

Example 7.11.

Solution


For other cases, the guidelines are not as clear-cut. We may need to use identities,integration by part, and occasionally a little ingenuity. We will sometimes need to beable to integrate tan x by using the formula

∫

tanx dx = ln | sec x| + C

We will also need the indefinite integral of secant:

∫

sec x dx = ln | sec x+ tanx|+ C

Find

∫

sec3 x dx.

Example 7.12.

Solution

Integrals of the form

∫

cotm x cscn x dx can be found by similar methods because of

the identity1 + cot2 x = csc2 x.

Finally, we can make use of another set of trigonometric identities: To evaluate theintegrals

(a)

∫

sinmx cos nx dx,

(b)

∫

sinmx sin nx dx, or

(c)

∫

cosmx cos nx dx,


we use the corresponding identity:

sinA cosB = 1

2

[

sin(A− B) + sin(A+B)]

sinA sinB = 1

2

[

cos(A−B)− cos(A+B)]

cosA cosB = 1

2

[

cos(A−B) + cos(A+B)]

Find

∫

cos 5x cos 4x dx.

Example 7.13.

Solution

Find

∫

sin 2x cos 3x dx.

Example 7.14.

Solution


Exercise 7.2


1.

∫

cosx sin4 x dx 2.

∫

cos2 x sin x dx

3.

∫

3π/4

π/2

sin5 x cos3 x dx 4.

∫

cos5 x sin4 x dx

5.

∫ π/2

0

sin2 3x dx 6.

∫

cos2 x sin2 x dx

7.

∫

(1− sin 2x)2 dx 8.

∫ π/4

0

sin4 x cos2 x dx

9.

∫

sin3 x√cos x dx 10.

∫

cos2 x tan3 x dx

11.

∫

1− sin x

cosxdx 12.

∫

tan2 x dx

13.

∫

sec4 x dx 14.

∫

tan5 x dx

15.

∫

tanx sec3 x dx 16.

∫ π/4

0

tan4 x sec2 x dx

17.

∫

tan3 x sec x dx 18.

∫ π/3

0

tan3 x sec x dx

19.

∫

sec2 x

cot xdx 20.

∫ π/2

π/6

cot2 x dx

21.

∫

cot3 x csc3 x dx 22.

∫

cot2 ω csc4 ω dω

23.

∫

csc x dx 24.

∫

sin x

tanxdx

25.

∫

1− tan2 x

sec2 xdx 26.

∫

sin 5x sin 2x dx

27.

∫

cos 7θ cos 5θ dθ 28.

∫

sin 4x cos 5x dx


1. 1

5sin5 x+ C 2. −1

3cos3 x+ C 3. − 11

3844. 1

5sin5 x− 2

7sin7 x+ 1

9sin9 x+ C

5. π4

6. 1

8x− 1

32sin 4x+ C 7. 3

2x+ cos 2x− 1

8sin 4x+ C 8. 1

192(3π − 4)

9. [27cos3 x− 2

3cos x]

√cosx+ C 10. 1

2cos2 x− ln | cosx|+ C 11. ln(1 + sin x) + C

12. tanx− x+ C 13. 1

3tan3 x+ tan x+ C 14. 1

4sec4 x− tan2 x+ ln | sec x| + C

15. 1

3sec3 x+ C 16. 1

517. 1

3sec3 x− sec x+ C 18. 38

1519. 1

2tan2 x+ C

20.√3− π

321. −1

5csc5 x+ 1

3csc3 x+ C 22. −1

3cot3 ω − 1

5cot5 ω + C


23. ln | csc x− cotx|+C 24. sin x+C 25. 1

2sin 2x+C 26. 1

6sin 3x− 1

14sin 7x+C

27. 1

4sin 2θ + 1

24sin 12θ + C 28. 1

2cos x− 1

18cos 9x+ C

7.3 Trigonometric Substitutions

If an integral contains a term of the form√a2 − x2,

√a2 + x2 or

√x2 − a2, for some

a > 0, you can often evaluate the integral by making a substitution involving a trigfunction.

First, suppose that an integrand contains a term of the form√a2 − x2, for some

a > 0. If we let x = a sin θ, where −π2≤ θ ≤ π

2, then we can eliminate the square root,

as follows.

Notice that we now have

√a2 − x2 =

√

a2 − (a sin θ)2 =√

a2(1− sin2 θ)

= a√cos2 θ = a| cos θ| = a cos θ

since for −π2≤ θ ≤ π

2, cos θ > 0.

Evaluate

∫

x3

√4− x2

dx.

Example 7.15.

Solution


Next, suppose that an integrand contains a term of the form√a2 + x2, for some

a > 0. If we let x = a tan θ, where −π2< θ < π

2, then we can eliminate the square root,

as follows.

Notice that in this case, we have

√a2 + x2 =

√

a2 + (a tan θ)2 =√

a2(1 + tan2 θ)

= a√sec2 θ = a| sec θ| = a sec θ

since for −π2< θ < π

2, sec θ > 0.

Evaluate the integral

∫

1√9 + x2

dx.

Example 7.16.

Solution


Finally, suppose that an integrand contains a term of the form√x2 − a2, for some

a > 0. If we let x = a sec θ, where θ ∈[

0, π2

)

∪[

π, 3π2

)

, then we can eliminate the squareroot, as follows.

Notice that in this case, we have

√x2 − a2 =

√

(a sec θ)2 − a2 =√

a2(sec2 θ − 1)

= a√tan2 θ = a| tan θ| = a tan θ

since for θ ∈[

0, π2

)

∪[

π, 3π2

)

, tan θ ≥ 0.

Evaluate the integral

∫

√9x2 − 1

xdx.

Example 7.17.

Solution


We summarize the three trigonometric substitutions presented here in the followingtable.

Expression Substitution Identity

√a2 − x2 x = a sin θ, −π

2≤ θ ≤ π

21− sin2 θ = cos2 θ

√a2 + x2 x = a tan θ, −π

2< θ < π

21 + tan2 θ = sec2 θ

√x2 − a2 x = a sec θ, θ ∈

[

0, π2

)

∪[

π, 3π2

)

sec2 θ − 1 = tan2 θ

Exercise 7.3


1.

∫

2

√2

1

t3√t2 − 1

dt 2.

∫

2√x2 − 4

dx

3.

∫

1

x2√25− x2

dx 4.

∫

dx

x√x2 + 3

dx

5.

∫

x2

√x2 + 9

dx 6.

∫ √1− 4x2 dx

7.

∫ √x2 + 16 dx 8.

∫

√9x2 − 4

xdx

9.

∫

x2

(a2 − x2)3/2dx 10.

∫

x√x2 − 7

dx

11.

∫

3

0

dx√9 + x2

12.

∫

1

0

x√x2 + 8 dx

13.

∫

2/3

0

x3√4− 9x2 dx 14.

∫ √2x− x2 dx

15.

∫

1√9x2 + 6x− 8

dx 16.

∫

dx

(x2 + 2x+ 2)2

17.

∫

et√9− e2t dt 18.

∫

dx√x2 + a2

dx


1. π24+

√3

8− 1

42. −4 − x2

2x+C 3. −

√25− x2

25x+C 4. ( 1√

3) ln

∣

∣

∣

∣

∣

(√x2 + 3−

√3)

x

∣

∣

∣

∣

∣

+C

5. 1

3(x2 + 4)3/2 − 4

√x2 + 4 + C 6. 1

4sin−1(2x) + 1

2x√1− 4x2 + C

7. 1

2x√x2 + 4 + 2 ln

∣

∣

∣

∣

∣

x

2+

√x2 + 4

2

∣

∣

∣

∣

∣

+ C 8.√9x2 − 4− 2 sec−1(3x

2) + C

9.x√

a2 − x2− sin−1

(

xa

)

+C 10.√x2 − 7+C 11. ln(1+

√2) 12. 1

3(x2+4)3/2+C


13. 64

121514. 1

2[sin−1(x− 1) + (x− 1)

√2x− x2] + C

15. 1

3ln∣

∣3x+ 1 +√9x2 + 6x− 8

∣

∣+ C 16. 1

2

[

tan−1(x− 1) +(x+ 1)

(x2 + 2x+ 2)

]

+ C

17. 1

2

[

et√9− e2t + 9 sin−1( e

t

3)]

+ C 18. ln(x+√x2 + a2) + C

7.4 Integrating Rational Functions by Partial Frac-

tions

In this section we show how to integrate any rational function by expressing it as a sumof simpler fractions, called partial fractions.

Observe that

3

x+ 2− 2

x− 5=

3(x− 5)− 2(x+ 2)

(x+ 2)(x− 5)=

x− 19

x2 − 3x− 10.

To integrate the function on the right side of this equation, we have∫

x− 19

x2 − 3x− 10dx =

∫(

3

x+ 2− 2

x− 5

)

dx

= 3 ln |x+ 2| − 2 ln |x− 5|+ C

To see how the method of partial fractions work in general, let’s consider a rationalfunction

f(x) =P (x)

Q(x)(7.1)

where P and Q are polynomial. It’s possible to express f as a sum of simpler fractionsprovided the degree of P

(

deg(P ))

is less than the degree of Q(

deg(Q))

. Such a rationalfunction is called proper.

If f is improper, that is, deg(P ) ≥ deg(Q), then we must take the preliminary stepof dividing Q into P (by long division) until the remainder R(x) is obtained such thatdeg(R) < deg(Q).

The division statement is

f(x) =P (x)

Q(x)= S(x) +

R(x)

Q(x)

where S and R are also polynomial.

Find

∫

x3 − 4x2 + 5x− 2

x− 3dx.

Example 7.18.

Solution


The next step is to factor the denominator Q(x) as far as possible. And the thirdstep is to express the proper rational function R(x)/Q(x) as a sum of partial factionsof the form

A

(ax+ b)ior

Ax+B

(ax2 + bx+ c)j

CASE I: The denominator Q(x) is a product of distinct linear factors.

This means that we can write

Q(x) = (a1x+ b1)(a2x+ b2) · · · (anx+ bn)

where no factor is repeated.

In this case the partial fraction theorem states that there exist constants A1, A2, . . .,Ak such that

R(x)

Q(x)=

R(x)

(a1x+ b1)(a2x+ b2) · · · (akx+ bk)

=A1

a1x+ b1+

A2

a2x+ b2+ · · ·+ Ak

akx+ bk(7.2)

These constant can be determined as in the following example.

Evaluate

∫

2x+ 2

x2 − 6x+ 8dx.

Example 7.19.

Solution


Evaluate

∫

2x2 + 5x− 1

x3 + x2 − 2xdx.

Example 7.20.

Solution

CASE II: Q(x) is a product of distinct linear factors, some of which are

repeated.

Suppose the first linear factor (a1x+ b1) is repeated r times; that is, (a1x+ b1)r occurs

in the factorization of Q(x). Then instead of the single term A1/(a1x+ b1) in Equation(7.2), we would use

A11

a1x+ b1+

A12

(a1x+ b1)2+ · · ·+ A1r

(a1x+ b1)r. (7.3)

For example,

x2 − 5

x2(x+ 1)3=

A

x+

B

x2+

C

(x+ 1)+

D

(x+ 1)2+

E

(x+ 1)3


Find

∫

x− 1

x3 + 4x2 + 4xdx.

Example 7.21.

Solution

CASE III: Q(x) contains irreducible quadratic factors, none of which is re-

peated.

If Q(x) has the factor ax2 + bx+ c, where b2 − 4ac < 0, then, in addition to the partialfractions in Equations (7.2) and (7.3), the expression for R(x)/Q(x) will have a term ofthe form

Ax+B

ax2 + bx+ c(7.4)

where A and B are constants to be determined.


For instance, the fraction given by

f(x) =x

(x+ 2)(x2 + 1)(x2 + 2)

has a partial fraction decomposition of the form

x

(x+ 2)(x2 + 1)(x2 + 2)=

A

x+ 2+

Bx+ C

x2 + 1+

Dx+ E

x2 + 2

The term given in (7.4) can be integrate by completing the square and using the formula

∫

1

x2 + a2dx =

1

atan−1

(x

a

)

+ C

Evaluate

∫

3x2 − 7x+ 10

(x− 2)(x2 + 4)dx.

Example 7.22.

Solution


CASE IV: Q(x) contains a repeated irreducible quadratic factors.

If Q(x) has the factor (ax2 + bx + c)r, where b2 − 4ac < 0, then instead of the singlepartial fraction (7.4), the sum

A1x+B1

ax2 + bx+ c+

A2x+B2

(ax2 + bx+ c)2+ · · ·+ Arx+Br

(ax2 + bx+ c)r(7.5)

occurs in the partial fraction decomposition of R(x)/Q(x). Each of the term of (7.5) canbe integrated by first completing the square.

Evaluate

∫

1− x− 2x2 − x3

x(x2 + 1)2dx.

Example 7.23.

Solution


Exercise 7.4


1.

∫

x− 5

x2 − 1dx 2.

∫

6x

x2 − x+ 2dx

3.

∫

x+ 1

x2 − x− 6dx 4.

∫ −x+ 5

x3 − x2 − 2xdx

5.

∫

x3 + x+ 2

x2 + 2x− 8dx 6.

∫ −3x− 1

x3 − xdx

7.

∫

2x+ 3

(x+ 2)2dx 8.

∫

x− 1

x3 + 4x2 + 4xdx

9.

∫

x+ 4

x3 + 3x2 + 2xdx 10.

∫

x+ 2

x3 + xdx

11.

∫

4x− 2

x4 − 1dx 12.

∫

3x2 − 6

x2 − x− 2dx

13.

∫

2x+ 3

x2 + 2x+ 1dx 14.

∫

x2 + 2x+ 1

x3 + xdx

15.

∫

4x2 + 3

x3 + x2 + xdx 16.

∫

3x3 + 1

x3 − x2 + x− 1dx


1. 3 ln |x+ 1| − 2 ln |x− 1|+ C 2. 2 ln |x+ 1|+ 4 ln |x− 2|+ C

3. 1

5ln |x+ 2|+ 4

5ln |x− 3|+ C 4. 2 ln |x+ 1|+ 1

2ln |x− 2| − 5

2ln |x|+C

5. 11 ln |x+ 4|+ 2 ln |x− 2|+ 1

2x2 − 2x+ C

6. ln |x+ 1| − 2 ln |x− 1|+ ln |x|+ C 7. 2 ln |x+ 2| − (x+ 2)−1 + C

8. 1

4ln |x+ 2| − 3

2(x+ 2)−1 − 1

4ln |x|+ C 9. ln |x+ 2| − 3 ln |x+ 1|+ 2 ln |x|+ C

10. − ln(x2 + 1) + tan−1 x+ 2 ln |x|+C

11. 3

2ln |x+ 1|+ 1

2ln |x− 1| − ln(x2 + 1) + tan−1 x+ C

12. 3x+ ln |x+ 1|+ 2 ln |x− 2|+ C 13. 2 ln |x+ 1| − (x+ 1)−1 + C

14. 2 tan−1 x+ ln |x|+ C 15. 3 ln |x|+ 1

2ln |x2 + x+ 1| − 7√

3tan−1

(

2x+1√3

)

+ C

16. 3x+ 2 ln |x− 1|+ 1

2ln(x2 + 1)− 2 tan−1 x+ C


7.5 Numerical Integration

We know that if f is continuous on the closed interval [a, b], then the definite integral∫ b

a

f(x) dx must exist. Existence is one thing; evaluation is a different matter. There are

many definite integral that can not be evaluated by the methods we have learned—thatis, by use of the Fundamental Theorem of Calculus. For example, the indefinite integralof such integrands as

e−x2

, sin(x2),√1− x4,

sin x

x

can not be expressed algebraically of elementary functions. Then one must settle forsome kind of numerical approximation of the integral. In this section, we develop somenew methods of approximating definite integrals.

Since a definite integral is the limit of a sequence of Riemann sums, any Riemannsum serves as an approximation of the integral,

∫ b

a

f(x) dx ≈n

∑

i=1

f(x∗i )∆x

where ∆x = (b − a)/n and x∗i is any point chosen from the subinterval [xi−1, xi for

i = 1, 2, . . . , n. We denote this by writing

∫ b

a

f(x) dx ≈(

b− a

n

)

[f(x∗1) + f(x∗

2) + · · ·+ f(x∗n)] (7.6)

If we denote the values of f at the endpoints of the subintervals by

y0 = f(a), y1 = f(x1), y2 = f(x2), . . . , yn−1 = f(xn−1), yn = f(xn)

and the values of f at the midpoints of the subintervals by

ym1, ym2

, . . . , ymn

then it follows from (7.6) that the left endpoint, right endpoint, and midpoint approxi-mations discussed in Section 5.3 can be expressed as shown in Figure below.

b

b

b

bb

b

x

y

a b

y0 y1 y2 yn−1

Left endpoint approximation:

∫ b

a

f(x) dx ≈(

b− a

n

)

[y0 + y1 + · · ·+ yn−1]


b

b

bb

b

b

x

y

a b

y1 y2 yn−1 yn

Right endpoint approximation:

∫ b

a

f(x) dx ≈(

b− a

n

)

[y1 + y2 + · · ·+ yn]

bb

bb

bb

x

y

a b

ym1ym2

ymn

Midpoint approximation:

∫ b

a

f(x) dx ≈(

b− a

n

)

[ym1+ ym2

+ · · ·+ ymn]

Although we originally obtained these results for nonnegative functions in the contextof approximating areas, they are applicable to any function that is continuous on [a, b].

Trapezoidal Approximation

It will be convenient in this section to denote the left endpoint, right endpoint, andmidpoint approximations with n subintervals by Ln, Rn, and Mn, respectively. Of thethree approximations, the midpoint approximation is most widely used in applications.If we take the average of Ln and Rn, then we obtain another important approximationdenoted by

Tn =1

2(Ln +Rn)

called the trapezoidal approximation:

Trapezoidal Approximation∫ b

a

f(x) dx ≈(

b− a

2n

)

[y0 + 2y1 + · · ·+ 2yn−1 + yn]

The name “trapezoidal approximatio” results from the fact that in the case where fis nonnegative on the interval of integration, the approximation Tn is the sum of thetrapezoidal areas shown in Figure below.


b

b

b

bb

b

b

x

y

a b

y0 y1 y2 yn−1

Trapezoidal approximation

Compute the trapezoidal approximation with n = 4 for

∫

1

0

3x2 dx.

Example 7.24.

Solution Since n = 4, ∆x =1− 0

4=

1

4, and

x0 = 0 f(x0) = f(0) = 0

x1 =1

4f(x1) = f

(

1

4

)

=3

8

x2 =1

2f(x2) = f

(

1

2

)

=12

8

x3 =3

4f(x3) = f

(

3

4

)

=27

8

x4 = 1 f(x4) = f (1) = 3

Thus,

T4 ≈ 1− 0

(2)(4)

[

f(0) + 2f

(

1

4

)

+ 2f

(

1

2

)

+ 2f

(

3

4

)

+ f(1)

]

=1

8

(

0 +3

8+

12

8+

27

8+ 3

)

=66

64≈ 1.03125. z

Simpson’s Rule

Recall that in the trapezoidal approximation, we take a regular partition of the interval[a, b] and for each subinterval [xi−1, xi] connect the points (xi−1, f(xi−1)) and (xi, f(xi))with a straight line segment. We then approximate the area under the curve by thearea under the line segment (i.e. the area of a trapezoid). We suggest the followingalternative to the trapezoidal approximation. Again, construct a regular partition of theinterval [a, b]:

a = x0 < x1 < x2 < · · · < xn = b,

where

xi − xi−1 =b− a

n= ∆x.


for each i = 1, 2, . . . , n and where n is an even number. Instead of connecting each pairof points with a straight line segment, we connect each set of three consecutive points,(xi−2, f(xi−2)), (xi−1, f(xi−1)) and (xi, f(xi)) for i = 2, 4, . . . , n with a parabola.

b b

b

x

y

xi−2 xi−1 xi

That is, we look for the quadratic function p(x) whose graph passes through these threepoints, so that

p(xi−2) = f(xi−2), p(xi−1) = f(xi−1), and p(xi) = f(xi).

We can use this to approximate the value of the integral of f on the interval [xi−1, xi].We have

∫ xi

xi−2

f(x) dx ≈∫ xi

xi−2

p(x) dx

=xi − xi−2

6[f(xi−2) + 4f(xi−1) + f(xi)]

=b− a

3n[f(xi−2) + 4f(xi−1) + f(xi)]

Adding together the integral over each subinterval [xi−1, xi] for i = 2, 4, . . . , n, we get

∫ b

a

f(x) dx ≈ b− a

3n[f(x0) + 4f(x1) + f(x2)] +

b− a

3n[f(x2) + 4f(x3) + f(x4)] + · · ·

+b− a

3n[f(xn−2) + 4f(xn−1) + f(xn)]

=b− a

3n[f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + · · ·+ 4f(xn−1) + f(xn)].

This is called the (n+ 1)-point Simpson’s rule, Sn.

Simpson’s rule∫ b

a

f(x) dx ≈ Sn =b− a

3n[f(x0) + 4f(x1) + 2f(x2) + 4f(x3)

+2f(x4) + · · ·+ 4f(xn−1) + f(xn)]

Approximate the value of

∫

1

0

3x2 dx using Simpson’s rule with n = 4.

Example 7.25.


Solution We have

S4 =1− 0

(3)(4)

[

f(0) + 4f

(

1

4

)

+ 2f

(

1

2

)

+ 4f

(

3

4

)

+ f(1)

]

= 1,

which is in fact, the exact value. Notice that this is far more accurate than the trape-zoidal approximation. z

The following example illustrates an integral that we do not know how to computeexactly.

Compute Simpson’s rule approximation with n = 4 for

∫

2

0

√x2 + 1 dx.

Example 7.26.

Solution We have

S4 =2− 0

(3)(4)

[

f(0) + 4f

(

1

2

)

+ 2f (1) + 4f

(

3

2

)

+ f(2)

]

=

(

1

6

)

[

1 + 4

√

5

4+ 2

√2 + 4

√

13

4+√5

]

≈ 2.95795560. z


7.6 Improper Integrals

In defining a definite integral

∫ b

a

f(x) dx we dealt with a function f defined on a finite

interval [a, b] and we assumed that f does not have an infinite discontinuity.

In this section we extend the concept of the definite integral to the case where theinterval is infinite and also to the case where f has an infinite discontinuity in [a, b].

In either case the integral is called an improper integral.

Type I: Infinite Intervals

(a) If

∫ t

a

f(x) dx exists for every number t ≥ a, then

∫ ∞

a

f(x) dx = limt→∞

∫ t

a

f(x) dx

provided this limit exists (as a finite number).

(b) If

∫ b

t

f(x) dx exists for every number t ≤ b, then

∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx

provided this limit exists (as a finite number).

The improper integrals

∫ t

a

f(x) dx and

∫ b

t

f(x) dx are called convergent if the

corresponding limit exists and divergent if the limit does not exist.

(c) If both

∫ ∞

a

f(x) dx and

∫ a

−∞f(x) dx are convergent, then we define

∫ ∞

−∞f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞

a

f(x) dx

In part (c) any real number a can be used.

Definition 7.1.


Determine whether the integral

∫ ∞

1

1

xdx is convergent or divergent.

Example 7.27.

Solution

Evaluate

∫

0

−∞xex dx.

Example 7.28.

Solution


Evaluate

∫ ∞

−∞

1

1 + x2dx.

Example 7.29.

Solution


Type II: Discontinuous Integrands

(a) If f continuous on [a, b) and is discontinuous at b, then

∫ b

a

f(x) dx = limt→b−

∫ t

a

f(x) dx

if this limit exists (as a finite number).

(b) If f continuous on (a, b] and is discontinuous at a, then

∫ b

a

f(x) dx = limt→a+

∫ b

t

f(x) dx

if this limit exists (as a finite number).

The improper integrals

∫ b

a

f(x) dx is called convergent if the corresponding

limit exists and divergent if the limit does not exist.

(c) If f has a discontinuity at c, where a < c < b, and both

∫ c

a

f(x) dx and∫ b

c

f(x) dx are convergent, then we define

∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx

Definition 7.2.

Find

∫

5

2

1√x− 2

dx.

Example 7.30.

Solution


Determine whether the integral

∫ π/2

0

sec x dx converges or diverges.

Example 7.31.

Solution

Evaluate

∫

3

0

1

x− 1dx if possible.

Example 7.32.

Solution


Exercise 7.6

Determine whether each integral is convergent or divergent. Evaluate those that areconvergent.

1.

∫ ∞

1

1

(3x+ 1)2dx 2.

∫ −1

−∞

1√2− w

dw

3.

∫ ∞

0

e−x dx 4.

∫ ∞

−∞x3 dx

5.

∫ ∞

−∞xe−x2

dx 6.

∫ ∞

0

1

(x+ 2)(x+ 3)dx

7.

∫ ∞

0

cosx dx 8.

∫

1

−∞xe2x dx

9.

∫ ∞

1

ln x

xdx 10.

∫ ∞

−∞

x

1 + x2dx

11.

∫ ∞

1

ln x

x2dx 12.

∫

3

0

1√xdx

13.

∫

0

−1

1

x2dx 14.

∫ π/4

0

csc2 t dt

15.

∫

3

−2

1

x4dx 16.

∫ π

0

sec x dx

17.

∫

2

−2

1

x2 − 1dx 18.

∫

2

0

z2 ln z dz


1. 1

22. D 3. 1 4. D 5. 0 6. − ln 2

37. D 8. e2

49. D 10. D 11. 1

12. 2√3 13. D 14. D 15. D 16. D 17. D 18. 8

3ln 2− 8

9

Chapter 7 Techniques of...

Documents

Transcript of Chapter 7 Techniques of...