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Chapter 7: Antiderivatives.
1/23
Indefinite integral.
Definition: Let f be a function defined on the interval I.Function F defined on I such that
F ′(x) = f (x) ∀ x ∈ I
is called antiderivative or primitive function of the function f onthe interval I.
Remark: Let F be an antiderivative of f on interval I and
G(x) = F (x) + c for all x ∈ I,
where c is a constant. Then G is the antiderivative of f on I.
Definition: A set of all antiderivatives of the function f oninterval I is called indefinite integral of f on I and we denote it∫
f (x)dx = {F (x) + c; c ∈ R, F is an antiderivative of f on I}.
Convention: We will not distinguish between primitivefunctions and indefinite integrals.
2/23
Existence of antiderivatives.
Theorem: Existence of antiderivativesLet f be continuous on I. Then f has an antiderivative on I.
Remark: According to the previous theorem each continuousfunction on interval has an antiderivative. But in some particularcase we are not able to find an explicit formula of this function.So we define it using integrals. E.g.:∫
e−x2dx ,
∫sin x
xdx , . . .
3/23
Basic antiderivatives.∫xn dx = xn+1
n+1 n ∈ R, n 6= −1∫ 1x dx = ln |x |∫sin x dx = − cos x∫cos x dx = sin x∫ax dx = ax
ln a a > 0, a 6= 1∫ 11+x2 dx = arctg x∫ 1√
1−x2dx = arcsin x∫ 1
cos2 x dx = tg x∫ 1sin2 x
dx = −cotg x∫ 1√x2+a
dx = ln |x +√
x2 + a| a 6= 0∫ f ′(x)f (x) dx = ln |f (x)|
4/23
Properties of antiderivatives
Theorem: It holds
(i)∫
k f (x)dx = k∫
f (x)dx , where k is a constant.
(ii)∫(f (x)± g(x))dx =
∫f (x)dx ±
∫g(x)dx
5/23
Methods for computing antiderivatives.
Integration by parts (per-partes).
Integration by change of variable (substitutionmethod).
Integration of rational functions (partial fractiondecomposition).
6/23
Per-partes
Theorem:Suppose that functions u and v have continuous derivatives onI. Then ∫
u(x) v ′(x)dx = u(x) v(x)−∫
u′(x) v(x)dx
on I.
7/23
Substitution method
Theorem: Let function f (t) be continuous on (a,b) and denoteF antiderivative of f on (a,b). Suppose that function ϕ(x),ϕ : (α, β) −→ (a,b) has continuous derivative on (α, β). Then
(i) ∫f (ϕ(x))ϕ′(x)dx = F (ϕ(x)),
for x ∈ (α, β).(ii) Furthermore, let for ∀x ∈ (α, β) be ϕ′(x) 6= 0. Then∫
f (t)dt = F (ϕ−1(t)),
for t ∈ (a,b).Remark: In practice we use equality∫
f (ϕ(x))ϕ′(x)dx =∫
f (t)dt , where t = ϕ(x) is usedsubstitution.
8/23
Integration of rational functions
long division of polynomials
factorization of polynomials
decomposition of the pure rational function into the partialfractions
integration of partial fractions
9/23
Polynomial Factorization.
Recall that polynomial is:
Pn(x) = anxn + an−1xn−1 + · · ·+ a1x + a0 ,
where n ∈ N, a0,a1, . . . ,an ∈ R are coefficients and an 6= 0.P0(x) = a0 is a polynomial of zero-degree.A root of the polynomial Pn(x) for n ≥ 1 is a number α ∈ C suchthat Pn(α) = 0.
Theorem: If α is root of the polynomial Pn(x) then
Pn(x) = (x − α)Q(x) ,
where Q(x) is a polynomial of degree (n − 1).
Definition: α is a repeated root of order k of polynomial P(x) if
P(x) = (x − α)kQ(x),
Q(x) is a polynomial, Q(α) 6= 0.10/23
Polynomial Factorization (2).
Remark: α is repeated root of order k of polynomial P(x)⇔
P(α) = 0, P ′(α) = 0, . . . ,Pk−1(α) = 0 a Pk (α) 6= 0.
Theorem: Each polynomial of degree n has exactly n rootswhile each root is considered with its multiplicity (some rootscan be complex numbers).
11/23
Rational functions.
Definition: A function f (x) =P(x)Q(x)
, where P(x) and Q(x) are
polynomials, is called rational function.If degree P(x) is smaller than degree Q(x) the function f iscalled pure or proper rational functionand is called improper rational function otherwise.
Theorem: Each rational function is a sum of polynomial andpure rational function.
Remark: We already know how to integrate a polynomial. Nowwe will see how to integrate a pure rational function - partialfraction decomposition (expansion).
12/23
Partial Fraction Decomposition.
1 Decompose the denominator into factors.2 Factor (a x + b) corresponds in fraction decomposition to a
fractionA
(a x + b),
where A is a suitable real constant.3 Factors (a x + b)k , k = 2,3, . . . correspond in fraction
decomposition to fractions
A1
(a x + b),
A2
(a x + b)2 , . . . ,Ak
(a x + b)k ,
where A1, A2, . . . ,Ak are suitable real constants.
13/23
Partial Fraction Decomposition.
4 Factor (a x2 + b x + c) (with strictly complex roots)corresponds in fraction decomposition to a fraction
A x + B(a x2 + b x + c)
,
where A, B are suitable real constants.5 Factors (a x2 + b x + c)k , k = 2,3, . . . (with strictly complex
roots) correspond in fraction decomposition to fractions
A1x + B1
(a x2 + b x + c),
A2x + B2
(a x2 + b x + c)2 , . . . ,Akx + Bk
(a x2 + b x + c)k ,
where A1, B1, , . . . ,Ak , Bk are suitable real constants.
14/23
Integration of rational functions
Using suitable substitution the integration of partial fractionscorresponds to following primitive functions
1∫ A
(a x+b)dx substitution→ logarithm function
2∫ Ak
(a x+b)k dx substitution→ linear rational function
3∫ A x+B
(a x2+b x+c)dx substitution→ arctangent function
15/23
Chapter 8: Definite Integral.
16/23
Definite Integral (1)
Definition: Let f (x) be defined on interval I and F (x) be itsantiderivative on I. Let a,b ∈ I. Then definite integral of f from ato b is real number F (b)− F (a). We write∫ b
af (x)dx = F (b)− F (a) .
a - lower limit of integrationb - upper limit of integration
Notation: [F (x)]ba = F (b)− F (a)
17/23
Definite Integral (2)
Remark:(i) Definite integral does not depend on a choice of an
antiderivative.
(ii) If f is continuous on I and x0 ∈ I then an antiderivativeG(x) of function f (x) on I can be expressed in form
G(x) =∫ x
x0
f (t)dt , x ∈ I.
G(x) is such of primitive functions that G(x0) = 0.
18/23
Properties
Theorem: It holds
(i)∫ b
akf (x)dx = k
∫ b
af (x)dx
(ii)∫ b
af (x)± g(x)dx =
∫ b
af (x)dx ±
∫ b
ag(x)dx
(iii)∫ b
af (x)dx =
∫ c
af (x)dx +
∫ b
cf (x)dx , c ∈ (a,b)
(iv)∫ b
af (x)dx = −
∫ a
bf (x)dx
19/23
Geometrical interpretation
Theorem: Let f be continuous, nonnegative function on interval[a,b ]. Then a planar figure enclosed by the x−axis, the graphof the function f and lines x = a, x = b has an area
P =
∫ b
af (x)dx .
Remark: Let f be continuous, negative function on interval[a,b ]. Then
P = −∫ b
af (x)dx .
Theorem: Let f and g be continuous functions on interval [a,b ]and let ∀x ∈ [a,b ] : g(x) ≤ f (x). Then an area of planar figureenclosed by the graphs of functions f and g and lines x = a,x = b is given by
P =
∫ b
a(f (x)− g(x))dx .
20/23
Methods
Theorem: Per partesLet functions u(x) and v(x) have continuous derivatives oninterval [a,b ]. Then∫ b
au(x)v ′(x)dx = [u(x)v(x)]ba −
∫ b
au′(x)v(x) .
Theorem: SubstitutionLet f (t) be continuous on [a,b ] and let function t = ϕ(x) havecontinuous, nonzero derivative on interval [α, β ] and mapinterval [α, β ] on interval [a,b ] such that ϕ(α) = a andϕ(β) = b. Then∫ β
αf (ϕ(x))ϕ′(x)dx =
∫ b
af (t)dt∫ b
af (t)dt =
∫ β
αf (ϕ(x))ϕ′(x)dx
21/23
Improper integrals
Definition: Let function f (x) be continuous on open interval(a,b) (a = −∞ or b =∞ is allowed). Let F (x) be anantiderivative of f (x) on interval (a,b). Then by improperintegral ∫ b
af (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 integralconverges and it diverges otherwise.
22/23
Numerical integration
Theorem: Trapezoidal methodLet function f (x) be continuous on [a,b ]. Divide this interval[a,b ] into n subintervals of the same length with endpointsa = x1 < x2 < · · · < xn = b. Denote h = b−a
n a step ofsubdivision, f (xi) = yi for i = 0,1, . . .n. Then∫ b
af (x)dx .
=h2(y0 + 2y1 + 2y2 + · · ·+ 2yn−1 + yn) .
23/23