Sometimes the integral of an algebraic fraction can be found by first expressing the algebraic fraction as the sum of its partial fractions
28 août 2004 · Sometimes the integral of an algebraic fraction can be found by first expressing the algebraic fraction as the sum of its partial fractions
Definition: A set of all antiderivatives of the function f on Integration of rational functions (partial fraction decomposition)
Finding antiderivatives of fractions How does the method of partial fractions enable any rational function to be antidifferentiated?
Therefore, when we are finding the antiderivative of a function we must 1 a fraction with a variable raised to an exponent in the denominator such as
If F is an antiderivative of f, then ? f(x)dx = F(x) + c is called the (general) (3) Treating the derivative as if it were a fraction, solve for dx:
whose antiderivatives we can easily find The technique of partial fractions is a method of decomposing rational functions, and is very useful for preparing
Indefinite integrals (antiderivatives) of rational functions can always be found Partial Fraction Expansion: Expand the proper rational function using
If we can integrate this new function of u, then the antiderivative of the Now consider the following simple algebra of fractions:
14403_2integral_calculus.pdf
Chapter 7: Antiderivatives.
1/23
Indefinite integral.
Definition:Letfbe a function defined on the intervalI.
FunctionFdefined onIsuch that
F
0(x) =f(x)8x2I
is called antider ivativeor pr imitivefunction of the function fon the intervalI. Remark:LetFbe an antiderivative offon intervalIand
G(x) =F(x) +cfor allx2I;
wherecis a constant. ThenGis the antiderivative offonI. Definition:A set of all antiderivatives of the functionfon intervalIis calledindefinite integ ralof fonIand we denote itZ f(x)dx=fF(x) +c;c2R;Fis an antiderivative offonIg: Convention:We will not distinguish between primitive functions and indefinite integrals. 2/23
Existence of antiderivatives.
Theorem: Existence of antiderivatives
Letfbe continuous onI:Thenfhas an antiderivative onI. Remark:According to the previous theorem each continuous function on interval has an antiderivative. But in some particular case we are not able to find an explicit formula of this function.
So we define it using integrals. E.g.:
Z e x2dx;Zsinxx dx;::: 3/23
Basic antiderivatives.
R xndx=xn+1n+1n2R;n6= 1 R1x dx=lnjxj
Rsinxdx= cosx
Rcosxdx=sinx
Raxdx=axlnaa>0;a6=1
R11+x2dx=arctgx
R 1p
1 x2dx=arcsinx
R 1cos
2xdx=tgx
R1sin
2xdx= cotgx
R 1p x
2+adx=lnjx+px
2+aja6=0
R f0(x)f(x)dx=lnjf(x)j 4/23
Properties of antiderivatives
Theorem:It holds
(i)Z k f(x)dx=kZ f(x)dx;wherekis a constant. (ii) Z (f(x)g(x))dx=Z f(x)dxZ g(x)dx 5/23
Methods for computing antiderivatives.
Integration by parts (per-partes).
Integration by change of variable (substitution
method).Integration of rational functions (partial fraction decomposition). 6/23
Per-partes
Theorem:
Suppose that functionsuandvhave continuous derivatives on
I. Then
Z u(x)v0(x)dx=u(x)v(x) Z u
0(x)v(x)dx
onI: 7/23
Substitution method
Theorem:Let functionf(t)be continuous on(a;b)and denote Fantiderivative offon(a;b):Suppose that function'(x); ': (;) !(a;b)has continuous derivative on(;):Then (i) Z f('(x))'0(x)dx=F('(x)); forx2(;). (ii)
Fur thermore,let f or8x2(;)be'0(x)6=0. Then
Z f(t)dt=F(' 1(t)); fort2(a;b). Remark:In practice we use equalityRf('(x))'0(x)dx=Rf(t)dt;wheret='(x)is used substitution. 8/23
Integration of rational functions
long division of polynomials factorization of polynomials decomposition of the pure rational function into the partial fractionsintegration of partial fractions 9/23
Polynomial Factorization.
Recall that polynomial is:
P n(x) =anxn+an 1xn 1++a1x+a0; wheren2N,a0;a1;:::;an2Rare coefficients andan6=0. P
0(x) =a0is a polynomial of zero-degree.
A root of the polynomialPn(x)forn1 is a number2Csuch thatPn() =0.
Theorem:Ifis root of the polynomialPn(x)then
P n(x) = (x )Q(x); whereQ(x)is a polynomial of degree(n 1). Definition:is arepeated root of order kof polynomialP(x)if
P(x) = (x )kQ(x);
Q(x)is a polynomial,Q()6=0.
10/23
Polynomial Factorization (2).
Remark:is repeated root of orderkof polynomialP(x),
P() =0;P0() =0;:::;Pk 1() =0 aPk()6=0:
Theorem:Each polynomial of degreenhas exactlynroots while each root is considered with its multiplicity (some roots can be complex numbers). 11/23
Rational functions.
Definition:A functionf(x) =P(x)Q(x), whereP(x)andQ(x)are polynomials, is called r ationalfunction . If degreeP(x)is smaller than degreeQ(x)the functionfis called pure or proper r ationalfunction and is called improper r ationalfunction otherwise . Theorem:Each rational function is a sum of polynomial and pure rational function. Remark:We already know how to integrate a polynomial. Now we will see how to integrate a pure rational function - partial fraction decomposition (expansion). 12/23
Partial Fraction Decomposition.
1Decompose the denominator into factors.2Factor(ax+b)corresponds in fraction decomposition to a
fractionA(ax+b); whereAis a suitable real constant.3Factors(ax+b)k,k=2;3;:::correspond in fraction decomposition to fractions A
1(ax+b);A2(ax+b)2;:::;Ak(ax+b)k;
whereA1;A2;:::;Akare suitable real constants. 13/23
Partial Fraction Decomposition.
4Factor(ax2+bx+c)(with strictly complex roots)
corresponds in fraction decomposition to a fraction
Ax+B(ax2+bx+c);
whereA;Bare suitable real constants.5Factors(ax2+bx+c)k,k=2;3;:::(with strictly complex roots) correspond in fraction decomposition to fractions A
1x+B1(ax2+bx+c);A2x+B2(ax2+bx+c)2;:::;Akx+Bk(ax2+bx+c)k;
whereA1;B1;;:::;Ak;Bkare suitable real constants. 14/23
Integration of rational functions
Using suitable substitution the integration of partial fractions corresponds to following primitive functions1
RA(ax+b)dxsubstitution!logarithm function2
RAk(ax+b)kdxsubstitution!linear rational function3 RAx+B(ax2+b x+c)dxsubstitution!arctangent function 15/23
Chapter 8: Definite Integral.
16/23
Definite Integral (1)
Definition:Letf(x)be defined on intervalIandF(x)be its antiderivative onI. Leta;b2I. Thendefinite integ ralof ffroma tobis real numberF(b) F(a):We write Z b a f(x)dx=F(b) F(a): a-lo werlimit of integ ration b-upper limit of integ ration
Notation:[F(x)]ba=F(b) F(a)
17/23
Definite Integral (2)
Remark:
(i) Definite integ raldoes notdepend on a choice of an antiderivative. (ii) If fis continuous onIandx02Ithen an antiderivative
G(x)of functionf(x)onIcan be expressed in form
G(x) =Z
x x
0f(t)dt;x2I:
G(x)is such of primitive functions thatG(x0) =0.
18/23
Properties
Theorem:It holds
(i)Z b a kf(x)dx=kZ b a f(x)dx (ii) Z b a f(x)g(x)dx=Z b a f(x)dxZ b a g(x)dx (iii) Z b a f(x)dx=Z c a f(x)dx+Z b c f(x)dx,c2(a;b) (iv) Z b a f(x)dx= Z a b f(x)dx 19/23
Geometrical interpretation
Theorem:Letfbecontin uous,nonnegativ efunction on inter val [a;b]. Then a planar figure enclosed by thex axis, the graph of the functionfand linesx=a,x=bhas an area P=Z b a f(x)dx: Remark:Letfbecontin uous,negativ efunction on inter val [a;b]. Then P= Z b a f(x)dx: Theorem:Letfandgbecontin uousfunctions on inter val[a;b] and let8x2[a;b] :g(x)f(x). Then an area of planar figure enclosed by the graphs of functionsfandgand linesx=a, x=bis given by P=Z b a (f(x) g(x))dx: 20/23
Methods
Theorem: Per partes
Let functionsu(x)andv(x)have continuous derivatives on interval[a;b]:Then Z b a u(x)v0(x)dx= [u(x)v(x)]ba Z b a u0(x)v(x):
Theorem: Substitution
Letf(t)be continuous on[a;b]and let functiont='(x)have continuous, nonz ero der ivativeon inter val[;]and map interval[;]on interval[a;b]such that'() =aand '() =b:Then Z f('(x))'0(x)dx=Z b a f(t)dt Z b a f(t)dt=Z f('(x))'0(x)dx 21/23
Improper integrals
Definition:Let functionf(x)be continuous on open interval (a;b)(a= 1orb=1is allowed). LetF(x)be an antiderivative off(x)on interval(a;b):Then by improper integralZb a f(x)dx we understand the difference [F(x)]ba=limx!b F(x) limx!a+F(x); if the right side of equality is well defined. In case that the right side is a finite number, we say that integ ral converges and it div erges otherwise . 22/23
Numerical integration
Theorem: Trapezoidal method
Let functionf(x)be continuous on[a;b]. Divide this interval [a;b]intonsubintervals of the same length with endpoints a=x1