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288

Z

 

 

 

 

CHAPTER 6. TECHNIQUES OF INTEGRATION

58.

x2 + 4 + 13 dx

59.

Z

x2

+ 2x + 5

dx

60.

Z

4x2 − 1

 

 

 

2x + 7

 

 

 

 

x + 3

 

 

 

 

 

dx

 

 

Z

 

x + 4

 

 

Z

 

x

2

 

 

 

 

Z

e2xdx

61.

 

dx

62.

 

+

dx

 

63.

 

 

 

 

 

(5 − e2x + e4x)1/2

9x2 + 16

16

9x2

 

 

Z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

64.

 

e3xdx

 

 

 

 

 

 

 

 

 

 

 

 

 

(e6x + 4e3x + 3)1/2

 

 

 

 

 

 

 

 

 

 

 

6.6Integration by Partial Fractions

A polynomial with real coe cients can be factored into a product of powers of linear and quadratic factors. This fact can be used to integrate rational functions of the form P (x)/Q(x) where P (x) and Q(x) are polynomials that have no factors in common. If the degree of P (x) is greater than or equal to the degree of Q(x), then by long division we can express the rational function by

P (x) = q(x) + r(x)

Q(x) Q(x)

where q(x) is the quotient and r(x) is the remainder whose degree is less than the degree of Q(x). Then Q(x) is factored as a product of powers of linear and quadratic factors. Finally r(x)/Q(x) is split into a sum of fractions of the form

 

 

A1

 

 

+

 

A2

+ · · · +

 

An

 

 

 

ax + b

 

(ax + b)2

(ax + b)n

 

 

and

 

 

 

 

B2x + c2

 

 

 

Bmx + cm

 

B1x + c1

 

 

 

 

 

+ · · · +

 

 

 

+

 

 

 

.

 

ax2 + bx + c

 

(ax2 + bxc)2

(ax2 + bx + c)m

Many calculators and computer algebra systems, such as Maple or Mathematica, are able to factor polynomials and split rational functions into partial fractions. Once the partial fraction split up is made, the problem of integrating a rational function is reduced to integration by substitution using linear or trigonometric substitutions. It is best to study some examples and do some simple problems by hand.

Exercises 6.6 Evaluate each of the following integrals:

6.7.

FRACTIONAL POWER SUBSTITUTIONS

289

1.

Z

(x − 1)(x − 2)(x + 4)

2.

Z

(x − 4)(10 + x)

 

 

 

dx

 

 

 

 

dx

 

 

 

 

 

 

3.

Z

(x − a)(x − b)

4.

Z

(x − a)(b − x)

 

 

 

dx

 

 

 

 

dx

 

 

 

 

 

 

5.

Z

(x2 + 1)(x2 + 4)

6.

Z

(x − 1)(x2 + 1)

 

 

 

dx

 

 

 

 

dx

 

 

 

 

 

 

7.

Z

x2 − 5x + 6

8.

Z

(x + 3)(x + 4)

 

 

 

2x dx

 

 

 

 

x dx

 

 

 

 

 

 

9.

Z

(x + 2)(x2 + 4) dx

10.

Z

 

(x + 3)(x2 + 1)

 

 

 

x + 1

 

 

 

 

(x + 2)dx

 

11.

Z

 

(x2 + 4)(x2 + 9)

12.

Z

 

(x2 − 4)(x2

− 9)

 

 

 

2 dx

 

 

 

 

dx

 

 

 

 

 

 

13.

Z

 

x2 dx

 

14.

Z

 

x dx

 

 

 

 

 

 

 

(x2 + 4)(x2 + 9)

 

(x2 − 4)(x2 − 9)

15.

Z

 

x4 − 16

16.

Z

 

x4 − 81

 

 

 

 

 

 

 

 

 

dx

 

 

 

x dx

 

 

 

 

 

 

6.7Fractional Power Substitutions

If the integrand contains one or more fractional powers of the form xs/r, then the substitution, x = un, where n is the least common multiple of the denominators of the fractional exponents, may be helpful in computing the integral. It is best to look at some examples and work some problems by hand.

Exercises 6.7 Evaluate each of the following integrals using the given substitution.

 

Z

4x3/2

 

Z

dx

 

1.

 

dx; x = u6

2.

 

; x = u3

1 + x1/3

1 + x1/3

290

Z

 

CHAPTER 6.

TECHNIQUES OF INTEGRATION

3.

1 + e2x ; u2

= 1 + e2x

4.

Z

xx3

− 8; u2

= x3 − 8

 

 

dx

 

 

 

 

dx

 

 

 

Evaluate each of the following by using an appropriate substitution:

 

Z

 

x dx

 

Z

 

 

2dx

 

5.

 

 

 

 

 

 

 

 

6.

xx + 4

 

 

 

 

 

x + 2

 

 

 

 

7.

Z

4 +1x dx

8.

Z

1 + x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

x dx

 

 

 

 

Z

 

 

 

 

 

 

 

 

 

Z

 

 

x2/3

 

 

 

 

x

 

 

 

 

9.

1 + 3

 

 

 

 

10.

 

 

 

 

 

 

 

 

8

+ x1/2

 

x

 

 

 

 

11.

Z

 

x2/31+ 1 dx

12.

Z

1

+

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

13.

Z

 

1 + x2/3

14.

Z

2

+

 

 

 

 

 

 

 

x dx

 

 

 

 

x dx

 

 

 

1

+

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z

 

 

 

 

 

 

 

Z

 

 

 

 

 

 

 

 

1 + x3/2

 

 

1 − x3/2

15.

 

 

1 − x

dx

16.

 

1 +

 

 

 

x

dx

 

 

 

 

 

 

 

 

 

 

 

 

 

6.8Tangent x/2 Substitution

If the integrand contains an expression of the form (a+b sin x) or (a+b cos x), then the following theorem may be helpful in evaluating the integral.

Theorem 6.8.1 Suppose that u = tan(x/2). Then

 

sin x =

 

2u

, cos x =

1 − u2

and

dx =

 

2

du.

 

 

 

1 + u2

1 + u2

1 + u2

 

 

 

 

 

 

 

 

 

 

 

 

 

Furthermore,

Z

 

 

 

 

 

 

Z

 

 

 

 

 

 

 

Z

a + b sin x

 

a + b

1+u2

 

 

a(1 + u2) + 2bu

 

 

 

 

dx

=

 

(2/(1 + u2))du

=

2du

 

 

 

 

 

 

 

 

(2/(1 +

 

2u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z

a + b cos x

Z

 

u ))

Z

 

 

2

 

 

u )

 

a + b

1+u2

 

 

a(1 + u ) + b(1

 

 

dx

=

 

 

 

2 du

 

=

 

 

du

 

 

 

.

 

 

 

 

 

1 u2

 

 

2

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6.9. NUMERICAL INTEGRATION

291

Proof. The proof of this theorem is left as an exercise.

Exercises 6.8

1.

Prove Theorem 6.8.1

 

 

 

 

 

 

 

 

Evaluate the following integrals:

 

Z

 

 

 

 

 

 

2.

Z

2 + sin x

3.

sin x + cos x

 

 

 

dx

 

 

 

 

dx

 

4.

Z

sin x − cos x

5.

Z

2 sin x + 3 cos x

 

 

 

dx

 

 

 

 

dx

6.

Z

2 − sin x

7.

Z

3 + cos x

 

 

 

dx

 

 

 

 

dx

 

8.

Z

3 − cos x

9.

Z

sin x + cos x

 

 

 

dx

 

 

 

 

sin x dx

 

10.

Z

 

cos x dx

11.

Z

 

(1 + sin x)dx

 

 

 

 

 

 

 

 

 

sin x − cos x

 

(1 − sin x)

 

Z

 

1 + cos x

 

Z

 

2 + cos x

12.

Z

 

1 − cos x

dx

13.

Z

 

2 − cos x

dx

 

 

 

 

 

2 − sin x

 

 

3 + cos x

14.

 

 

2 + cos x

dx

15.

 

 

2 − sin x

dx

Z

 

 

 

 

 

16.

 

1 + sin x + cos x

 

 

 

 

 

 

 

 

 

 

 

dx

 

 

 

 

 

 

 

 

6.9Numerical Integration

Not all integrals can be computed in the closed form in terms of the elementary functions. It becomes necessary to use approximation methods. Some of the simplest numerical methods of integration are stated in the next few theorems.

292

 

 

 

CHAPTER 6. TECHNIQUES OF INTEGRATION

Theorem 6.9.1 (Midpoint Rule) If f, f0

and f00 are continuous on [a, b],

then there exists some c such that a < c < b and

 

 

 

 

 

 

 

 

b

f(x)dx = (b − a)f

a + b

+

f00(c)

 

 

 

 

 

 

Za

 

 

 

 

 

 

 

 

 

(b − a)3.

 

 

 

2

24

 

 

 

Proof. The proof of this theorem is omitted.

 

 

 

 

 

 

 

Theorem 6.9.2 (Trapezoidal Rule) If f, f0 and f00

are continuous on [a, b],

then there exists some c such that a < c < b and

 

 

 

 

 

 

 

 

b

 

 

1

 

 

 

 

f00(c)

 

 

 

 

 

 

 

 

 

 

 

 

Za

f(x)dx = (b − a)

 

(f(a) + f(b)) −

 

 

(b − a)3.

 

2

12

 

 

Proof. The proof of this theorem is omitted.

 

 

 

 

 

 

 

Theorem 6.9.3 (Simpson’s Rule) If f, f0, f00, f(3)

and f(4) are continuous

on [a, b], then there exists some c such that a < c < b and

 

 

b f(x)dx =

 

b − a

f(a) + 4f

a + b

+ f(b)

f(4)(c)

(b

a)5.

 

 

 

 

Za

 

6

 

 

2

 

2880

 

 

These basic numerical formulas can be applied on each subinterval [xi, xi+1]

of a partition

P = {a = x0

< x1

< · · · < xn

= b}

of the interval

[a, b]

to get composite numerical methods. We assume that h = (b − a)/n,

xi =

a + ih, i = 0, 1, 2, · · · , n.

 

 

 

 

 

 

 

Proof. The proof of this theorem is omitted.

 

 

 

 

 

Theorem 6.9.4 (Composite Trapezoidal Rule) If f, f0

and f00 are continu-

ous on [a, b], then there exists some c such that a < c < b and

 

Z

a

2

"

 

i

#

 

12

 

 

b

 

 

 

n−1

 

b

− a h2f00(c).

 

 

 

 

 

i=1

 

 

 

f(x)dx = h f(a) + 2

f(x ) + f(b)

 

 

 

 

 

 

 

X

 

 

 

 

 

Proof. The proof of this theorem is omitted.

Theorem 6.9.5 (Composite Simpson’s Rule) If f, f0, f00, f(3) and f(4) are continuous on [a, b], then there exists some c such that a < c < b and

b

h

 

n/2−1

 

 

n/2

 

 

b − a

h4f(4)

 

f(x)dx =

f(a) + 2

f(x

 

) + 4

f(x

 

) + f(b)

(c).

3

 

2i

2i−1

180

Za

 

 

i=1

 

X

 

 

 

 

 

 

X

 

 

 

 

 

 

 

i=1

where n is an even natural number.

6.9. NUMERICAL INTEGRATION

293

Proof. The proof of this theorem is omitted.

Remark 22 In practice, the composite Trapezoidal and Simpson’s rules can be applied when the value of the function is known at the subdivision points xi, i = 0, 1, 2, · · · , n.

Exercises 6.9 Approximate the value of each of the following integrals for a given value of n and using

 

 

 

 

n

 

 

 

Xi n

(a)

Left-hand end point approximation:

f(xi−1)(xi − xi−1)

 

 

 

 

=1

 

 

n

 

 

Xi

(b)

Right-hand end point approximation:

f(xi)(xi − xi−1)

 

X

f

 

=1

(xi − xi−1)

 

xi−1 + xi

(c)

Mid point approximation: i=1

2

(d)Composite Trapezoidal Rule

(e)Composite Simpson’s Rule

 

Z1

3

1

 

 

 

 

 

Z2

4

 

1

 

 

1.

 

 

 

dx, n = 10

2.

 

 

 

 

dx, n = 10

 

 

x

 

 

 

 

 

 

 

x

3.

Z0

1

1

+ x dx, n = 10

4.

Z1

2

 

 

1 + x2 dx, n = 10

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z0

1

 

 

+

 

 

 

Z0

2

 

 

 

 

 

 

 

 

 

x

 

 

 

 

 

 

 

5.

 

 

1

dx, n = 10

6.

 

 

x3 dx, n = 10

 

 

1 + x

 

 

7.

Z0

2

(x2 − 2x) dx, n = 10

8.

Z0

1

(1 + x2)1/2dx, n = 10

9.

Z0

1

(1 + x3)1/2dx, n = 10

10.

Z0

1

 

(1 + x4)1/2dx, n = 10