cálculo matemático ingeniería

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712 ESSENTIAL FORMULAE

Integral Calculus

Standard integrals

y!

y dx

axn axn+1

n + 1+ c

(except where n = !1)

cos ax1a

sin ax + c

sin ax !1a

cos ax + c

sec2 ax1a

tan ax + c

cosec2 ax !1a

cot ax + c

cosec ax cot ax !1a

cosec ax + c

sec ax tan ax1a

sec ax + c

eax 1a

eax + c

1x

ln x + c

tan ax1a

ln ( sec ax) + c

cos2 x12

"x + sin 2x

2

#+ c

sin2 x12

"x ! sin 2x

2

#+ c

tan2 x tan x ! x + c

cot2 x !cot x ! x + c

1$

(a2 ! x2)sin!1 x

a+ c

$(a2 ! x2)

a2

2sin!1 x

a+ x

2

$(a2 ! x2) + c

1(a2 + x2)

1a

tan!1 xa

+ c

y!

y dx

1$

(x2 + a2)sinh!1 x

a+ c or

ln

%x +

$(x2 + a2)a

&

+ c

$(x2 + a2)

a2

2sinh!1 x

a+ x

2

$(x2 + a2) + c

1$

(x2 ! a2)cosh!1 x

a+ c or

ln

%x +

$(x2 ! a2)a

&

+ c

$(x2 ! a2)

x2

$(x2 ! a2) ! a2

2cosh!1 x

a+ c

t = tan!

2substitution

To determine! 1

a cos ! + b sin ! + cd! let

sin ! = 2t(1 + t2)

cos ! = 1 ! t2

1 + t2 and

d! = 2 dt(1 + t2)

Integration by parts

If u and v are both functions of x then:

'u

dv

dxdx = uv !

'v

dudx

dx

Reduction formulae'

xnex dx = In = xnex ! nIn!1'

xn cos x dx = In = xn sin x + nxn!1 cos x

!n(n ! 1)In!2

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ESSENTIAL FORMULAE 713

! !

0xn cos x dx = In = !n!n!1 ! n(n ! 1)In!2

!xn sin x dx = In = !xn cos x + nxn!1 sin x

!n(n ! 1)In!2!

sinn x dx = In = !1n

sinn!1 x cos x + n ! 1n

In!2

!cosn x dx = In = 1

ncosn!1 sin x + n ! 1

nIn!2

! !/2

0sinn x dx =

! !/2

0cosn x dx = In = n ! 1

nIn!2

!tann x dx = In = tann!1 x

n ! 1! In!2

!(ln x)n dx = In = x( ln x)n ! nIn!1

With reference to Fig. FA4.

0 x 5 a x 5 b x

y

y 5 f (x)

A

Figure FA4

Area under a curve:

area A =! b

ay dx

Mean value:

mean value = 1b ! a

! b

ay dx

R.m.s. value:

r.m.s. value =

"##$%

1b ! a

! b

ay2 dx

&

Volume of solid of revolution:

volume =! b

a!y2 dx about the x-axis

Centroids

With reference to Fig. FA5:

x =

! b

axy dx

! b

ay dx

and y =12

! b

ay2 dx

! b

ay dx

Area A

y 5 f (x)

C

yx

0 x 5 a x 5 b x

y

Figure FA5

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Second moment of area and radius of gyration

Shape Position of axis Second moment Radius ofof area, I gyration, k

Rectangle (1) Coinciding with bbl3

31!3length l

(2) Coinciding with llb3

3b!3

breadth b

(3) Through centroid,bl3

121!12parallel to b

(4) Through centroid,lb3

12b!12parallel to l

Triangle (1) Coinciding with bbh3

12h!6Perpendicular

(2) Through centroid,bh3

36h!18

height h

parallel to basebase b

(3) Through vertex,bh3

4h!2parallel to base

Circle (1) Through centre,!r4

2r!2radius r perpendicular to plane

(i.e. polar axis)

(2) Coinciding with diameter!r4

4r2

(3) About a tangent5!r4

4

!5

2r

Semicircle Coinciding with!r4

8r2radius r diameter

Theorem of Pappus

With reference to Fig. FA5, when the curve is rotatedone revolution about the x-axis between the limitsx = a and x = b, the volume V generated is given by:V = 2!Ay.

Parallel axis theorem:

If C is the centroid of area A in Fig. FA6 then

Ak2BB = Ak2

GG + Ad2 or k2BB = k2

GG + d2

G B

CArea A

d

G B

Figure FA6

Ess-For-H8152.tex 19/7/2006 18: 2 Page 715


Perpendicular axis theorem:

If OX and OY lie in the plane of area A in Fig. FA7,

then Ak2OZ = Ak2

OX + Ak2OY or k2

OZ = k2OX + k2

OY

Z

Area A

O

X

Y

Figure FA7

Numerical integration

Trapezoidal rule

!ydx !

"width ofinterval

# $12

"first + lastordinates

#

+"

sum of remainingordinates

#%

Mid-ordinate rule!

ydx !"

width ofinterval

# "sum of

mid-ordinates

#

Simpson’s rule

!ydx ! 1

3

"width ofinterval

# $"first + lastordinate

#

+ 4"

sum of evenordinates

#

+ 2"

sum of remainingodd ordinates

#%

Differential Equations

First order differential equations

Separation of variables

Ifdydx

= f (x) then y =!

f (x) dx

Ifdydx

= f (y) then!

dx =!

dyf (y)

Ifdydx

= f (x) · f (y) then!

dyf (y)

=!

f (x) dx

Homogeneous equations

If Pdydx

= Q, where P and Q are functions of bothx and y of the same degree throughout (i.e. ahomogeneous first order differential equation) then:

(i) Rearrange Pdydx

= Q into the formdydx

= QP

(ii) Make the substitution y = vx (where v is afunction of x), from which, by the product rule,dydx

= v(1) + xdv

dx

(iii) Substitute for both y anddydx

in the equationdydx

= QP

(iv) Simplify, by cancelling, and then separate the

variables and solve using thedydx

= f (x) · f (y)method

(v) Substitute v = yx

to solve in terms of the originalvariables.

Linear first order

Ifdydx

+ Py = Q, where P and Q are functions ofx only (i.e. a linear first order differential equation),then

(i) determine the integrating factor, e&

P dx

(ii) substitute the integrating factor (I.F.) intothe equation

y (I.F.) =!

(I.F.) Q dx

(iii) determine the integral&

(I.F.)Q dx

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Numerical solutions of first order differentialequations

Euler’s method: y1 = y0 + h(y!)0

Euler-Cauchy method: yP1 = y0 + h(y!)0

and yC1 = y0 + 12

h[(y!)0 + f (x1, yp1 )]

Runge-Kutta method:

To solve the differential equationdydx

= f (x, y) giventhe initial condition y = y0 at x = x0 for a range ofvalues of x = x0(h)xn:

1. Identify x0, y0 and h, and values of x1, x2, x3, . . .

2. Evaluate k1 = f (xn, yn) starting with n = 0

3. Evaluate k2 = f!

xn + h2

, yn + h2

k1

"

4. Evaluate k3 = f!

xn + h2

, yn + h2

k2

"

5. Evaluate k4 = f(xn + h, yn + hk3)

6. Use the values determined from steps 2 to 5 toevaluate:

yn+1 = yn + h6{k1 + 2k2 + 2k3 + k4}

7. Repeat steps 2 to 6 for n = 1, 2, 3, . . .

Second order differential equations

If ad2ydx2 + b

dydx

+ cy = 0 (where a, b and c areconstants) then:

(i) rewrite the differential equation as(aD2 + bD + c)y = 0

(ii) substitute m for D and solve the auxiliaryequation am2 + bm + c = 0

(iii) if the roots of the auxiliary equation are:

(a) real and different, say m = ! and m = "then the general solution is

y = Ae!x + Be"x

(b) real and equal, say m = ! twice, then thegeneral solution is

y = (Ax + B)e!x

(c) complex, say m = ! ± j", then the generalsolution is

y = e!x(A cos "x + B sin "x)

(iv) given boundary conditions, constants A and Bcan be determined and the particular solutionobtained.

If ad2ydx2 + b

dydx

+ cy = f (x) then:

(i) rewrite the differential equation as(aD2 + bD + c)y = 0.

(ii) substitute m for D and solve the auxiliaryequation am2 + bm + c = 0.

(iii) obtain the complimentary function (C.F.), u, asper (iii) above.

(iv) to find the particular integral, v, first assume aparticular integral which is suggested by f (x),but which contains undetermined coefficients(See Table 51.1, page 482 for guidance).

(v) substitute the suggested particular integral intothe original differential equation and equaterelevant coefficients to find the constantsintroduced.

(vi) the general solution is given by y = u + v.(vii) given boundary conditions, arbitrary constants

in the C.F. can be determined and the particularsolution obtained.

Higher derivatives

y y(n)

eax an eax

sin ax an sin#

ax + n#

2

$

cos ax an cos#

ax + n#

2

$

xa a!(a " n)!x

a"n

sinh axan

2{[1 + ("1)n] sinh ax

+ [1 " ("1)n] cosh ax}

cosh axan

2{[1 " ("1)n] sinh ax

+[1 + ("1)n] cosh ax}

ln ax ("1)n"1 (n " 1)!xn

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Leibniz’s theorem

To find the n’th derivative of a product y = uv:

y(n) = (uv)(n) = u(n)v + nu(n!1)v(1)

+n(n ! 1)2! u(n!2)v(2)

+n(n ! 1)(n ! 2)3! u(n!3)v(3) + · · ·

Power series solutions of second order differentialequations.

(a) Leibniz-Maclaurin method

(i) Differentiate the given equation n times,using the Leibniz theorem,

(ii) rearrange the result to obtain the recurrencerelation at x = 0,

(iii) determine the values of the derivatives atx = 0, i.e. find (y)0 and (y")0,

(iv) substitute in the Maclaurin expansion fory = f (x),

(v) simplify the result where possible and applyboundary condition (if given).

(b) Frobenius method

(i) Assume a trial solution of the form:y = xc{a0 + a1x + a2x2 + a3x3 + · · · +

arxr + · · · } a0 #= 0,

(ii) differentiate the trial series to find y"and y"",

(iii) substitute the results in the given differentialequation,

(iv) equate coefficients of corresponding pow-ers of the variable on each side of theequation: this enables index c and coeffi-cients a1, a2, a3, . . . from the trial solution,to be determined.

Bessel’s equation

The solution of x2 d2ydx2 + x

dydx

+ (x2 ! v2)y = 0

is:

y = Axv

!1 ! x2

22(v + 1)

+ x4

24 $ 2!(v + 1)(v + 2)

! x6

26 $ 3!(v + 1)(v + 2)(v + 3)+ · · ·

"

+ Bx!v

!1 + x2

22(v ! 1)+ x4

24 $ 2!(v ! 1)(v ! 2)

+ x6

26 $ 3!(v ! 1)(v ! 2)(v ! 3)+ · · ·

"

or, in terms of Bessel functions and gammafunctions:

y = AJv(x) + BJ!v(x)

= A#x

2

$v!

1!(v + 1)

! x2

22(1!)!(v + 2)

+ x4

24(2!)!(v + 4)! · · ·

"

+ B#x

2

$!v!

1!(1 ! v)

! x2

22(1!)!(2 ! v)

+ x4

24(2!)!(3 ! v)! · · ·

"

In general terms:

Jv(x) =#x

2

$v%%

k=0

(!1)kx2k

22k(k!)!(v + k + 1)

and J!v(x) =#x

2

$!v%%

k=0

(!1)kx2k

22k(k!)!(k ! v + 1)

and in particular:

Jn(x) =#x

2

$n!

1n! ! 1

(n + 1)!#x

2

$2

+ 1(2!)(n + 2)!

#x2

$4! · · ·

"

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J0(x) = 1 ! x2

22(1!)2 + x4

24(2!)2

! x6

26(3!)2 + · · ·

and J1(x) = x2

! x3

23(1!)(2!) + x5

25(2!)(3!)

! x7

27(3!)(4!) + · · ·

Legendre’s equation

The solution of (1!x2)d2ydx2 !2x

dydx

+k(k+1)y = 0

is:

y = a0

!1 ! k(k + 1)

2! x2

+ k(k + 1)(k ! 2)(k + 3)4! x4 ! · · ·

"

+ a1

!x ! (k ! 1)(k + 2)

3! x3

+ (k ! 1)(k ! 3)(k + 2)(k + 4)5! x5 ! · · ·

"

Rodrigue’s formula

Pn(x) = 12nn!

dn(x2 ! 1)n

dxn

Statistics and Probability

Mean, median, mode and standard deviation

If x = variate and f = frequency then:

mean x =#

fx#f

The median is the middle term of a ranked set ofdata.The mode is the most commonly occurring value ina set of data.

Standard deviation

! =

$%%&'# (

f (x ! x)2)

#f

*

for a population

Binomial probability distribution

If n = number in sample, p = probability of theoccurrence of an event and q = 1 ! p, then theprobability of 0, 1, 2, 3, . . . occurrences is given by:

qn, nqn!1p,n(n ! 1)

2! qn!2p2,

n(n ! 1)(n ! 2)3! qn!3p3, . . .

(i.e. successive terms of the (q + p)n expansion).

Normal approximation to a binomial distribution:

Mean = np Standard deviation ! = "(npq)

Poisson distribution

If " is the expectation of the occurrence of an eventthen the probability of 0, 1, 2, 3, . . . occurrences isgiven by:

e!", "e!", "2 e!"

2! , "3 e!"

3! , . . .

Product-moment formula for the linear correlationcoefficient

Coefficient of correlation r =#

xy+,-#

x2. -#

y2./

where x = X ! X and y = Y ! Y and (X1, Y1),(X2, Y2), . . . denote a random sample from a bivari-ate normal distribution and X and Y are the meansof the X and Y values respectively.

Normal probability distribution

Partial areas under the standardized normal curve —see Table 58.1 on page 561.

Student’s t distribution

Percentile values (tp) for Student’s t distribution with# degrees of freedom — see Table 61.2 on page 587.

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Chi-square distribution

Percentile values (!2p) for the Chi-square distribu-

tion with " degrees of freedom—see Table 63.1 onpage 609.

!2 = ! "(o ! e)2

e

#where o and e are the observed

and expected frequencies.

Symbols:

Population

number of members Np, mean µ, standard devia-tion #.

Sample

number of members N , mean x, standard deviation s.

Sampling distributions

mean of sampling distribution of means µxstandard error of means #xstandard error of the standard deviations #s.

Standard error of the means

Standard error of the means of a sample distribu-tion, i.e. the standard deviation of the means ofsamples, is:

#x = #"N

$%Np ! NNp ! 1

&

for a finite population and/or for sampling withoutreplacement, and

#x = #"N

for an infinite population and/or for sampling withreplacement.

The relationship between sample mean andpopulation mean

µx = µ for all possible samples of size N are drawnfrom a population of size Np.

Estimating the mean of a population (! known)

The confidence coefficient for a large sample size,(N # 30) is zc where:

Confidence Confidencelevel % coefficient zc

99 2.5898 2.3396 2.0595 1.9690 1.64580 1.2850 0.6745

The confidence limits of a population mean basedon sample data are given by:

x ± zc#"N

$%Np ! NNp ! 1

&

for a finite population of size Np, and by

x ± zc#"N

for an infinite population

Estimating the mean of a population (! unknown)

The confidence limits of a population mean basedon sample data are given by: µx ± zc#x.

Estimating the standard deviation of a population

The confidence limits of the standard deviation of apopulation based on sample data are given by:s ± zc#s.

Estimating the mean of a population based on asmall sample size

The confidence coefficient for a small sample size(N < 30) is tc which can be determined usingTable 61.1, page 582. The confidence limits of apopulation mean based on sample data is given by:

x ± tcs"(N ! 1)

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Laplace Transforms

Function Laplace transformsf (t) L{f (t)} =

! !0 e"st f (t) dt

1 1s

k ks

eat 1s"a

sin at as2+a2

cos at ss2+a2

t 1s2

tn(n = positve integer) n!sn+1

cosh at ss2"a2

sinh at as2"a2

e"attn n!(s+a)n+1

e"at sin !t !(s+a)2+!2

e"at cos !t s+a(s+a)2+!2

e"at cosh !t s+a(s+a)2"!2

e"at sinh !t !(s+a)2"!2

The Laplace transforms of derivatives

First derivative

L"

dydx

#= sL{y} ! y(0)

where y(0) is the value of y at x = 0.

Second derivative

L"

dydx

#= s2L{y} ! sy(0) ! y"(0)

where y#(0) is the value ofdydx

at x = 0.

Fourier Series

If f (x) is a periodic function of period 2" then itsFourier series is given by:

f (x) = a0 +!$

n=1

(an cos nx + bn sin nx)

where, for the range "" to +":

a0 = 12"

% "

""f (x) dx

an = 1"

% "

""f (x) cos nx dx (n = 1, 2, 3, . . . )

bn = 1"

% "

""f (x) sin nx dx (n = 1, 2, 3, . . . )

If f (x) is a periodic function of period L then itsFourier series is given by:

f (x) = a0 +#$

n=1

&an cos

'2!nx

L

(+ bn sin

) 2!nxL

*+

where for the range "L2

to +L2

:

a0 = 1L

% L/2

"L/2f (x) dx

an = 2L

% L/2

"L/2f (x) cos

) 2"nxL

*dx (n = 1, 2, 3, . . . )

bn = 2L

% L/2

"L/2f (x) sin

) 2"nxL

*dx (n = 1, 2, 3, . . . )

Complex or exponential Fourier series

f (x) =!$

n="!cne j 2"nx

L

where cn = 1L

% L2

" L2

f (x)e"j 2"nxL dx

For even symmetry,

cn = 2L

% L2

0f (x) cos

) 2"nxL

*dx

For odd symmetry,

cn = "j2L

% L2

0f (x) sin

) 2"nxL

*dx

cálculo matemático ingeniería

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