Integrals
Integration
* Integration is the inverse process of differentiation.
If \frac{d}{dx} (F(x)) = ɡ(x), then ∫ g (x) dx = F(x) + C
* Where C is an arbitrary constant known as the constant of integration.
* Geometrical interpretation of indefinite integrals:
* An indefinite integral is a collection of a family of curves, each of which is obtained by translating one of the curves parallel to itself.
Properties of Indefinite Integrals
Property – I: \frac{d}{dx} ∫ ƒ (x)dx = ƒ (x) and ∫ ƒ’ (x)dx = ƒ(x)dx = ƒ (x) + C,
Where C is an arbitrary constant.
Property – II: If ƒ and ɡ are two functions, then: \frac{d}{dx} ∫ ƒ(x)dx = \frac{d}{dx} ∫ g(x) dx
Property – III: If ƒ and ɡ are two functions, then ∫ [ ƒ(x) + g (x)] dx = ∫ ƒ (x) dx + ∫ g (x)dx
Property – IV: ∫ kƒ (x) dx = k ∫ ƒ (x)dx where k is any real number.
Property – V: ∫[ k_{1} ƒ_{1} ( x) + k_{2} ƒ_{2} (x) + k_{3} ƒ_{3} (x) + . . . + k_{n} ƒ_{n} (x) ] dx =
k_{1} ∫ ƒ_{1} dx + k_{2} ∫ ƒ_{2} dx + k_{3} ∫ ƒ_{3} dx + . . . . + k_{n} ∫ ƒ_{n} dx
Comparison between Differentiation and Integration
* Differentiation and integration operate on functions.
* Differentiation and integration satisfy the property of linearity.
* All functions are not differentiable and all functions are not integrable.
* A function has unique derivative, but does not have a unique integral.
* When a polynomial function is differentiated, the result is a polynomial function whose degree is one less than the degree of the original function.
* When a polynomial function is integrated, the result is a polynomial function whose degree is one more than the degree of the original function.
* The derivative of a function can be found at a point, whereas the integral of a function can be found over an interval.
* Derivatives are useful in finding a physical quantity like the velocity of a moving particle. Integrals are useful in finding the displacement when the velocity of a particle at a time t is given.
* The processes of differentiation and integration involve limits.
* The processes of differentiation and integration are inverses of each other.
Integration by Substitution
Integrals of the form, ∫ ƒ [ g(x) ‘ (x)dx
∫ ƒ [ g (x)’ (x)dx Can be solved using the method of substitution.
| integral | Solution |
| \int \tan x \, dx | log | sec x | + C |
| \int \cot x \, dx | log | sin x | + C |
| \int \sec x \, dx | log | sec x + tan x | + C |
| \int \cosec x \, dx | log | cosec x - cot x | + C |
Integration Using Trigonometric Identities
Steps to solve integrals involving trigonometric functions:
* Use standard trigonometric identities to convert the integrated into functions
whose integrals can be evaluated easily.
* Simplify the integrand.
* Perform the integration. Use method of substation, if required.
Integrals of Some Particular Functions – I
| Integral | Solution |
| \int \frac{dx}{ x^{2} - a^{2} } | \frac{1}{2a} log | \frac{x - a}{x + a} | + C |
| \int \frac{dx}{ a^{2} - x^{2} } | \frac{1}{2a} log | \frac{ a + x}{a - x} | + C |
| \int \frac{dx}{ x^{2} + a^{2} } | \frac{1}{a} tan^{- 1} \frac{x}{a} + C |
Integrals of Some Particular Functions – II
\int \frac{1}{ \sqrt{ x^{2} - a^{2} } } dx = log | x + \sqrt{ x^{2} - a^{2} } | + C
\int \frac{1}{ \sqrt{ a^{2} - x^{2} } } dx = sin^{- 1} ( \frac{x}{a} )+ C
\int \frac{1}{ \sqrt{ x^{2} + a^{2} } } dx = log | x + \sqrt{ x^{2} + a^{2} } | + C
Integrals of Some Particular Functions – III
\int \frac{px + q}{ ax^{2} + bx + c} dx\int \frac{px + q}{ \sqrt{ ax^{2} + bx + c } } dx
Integration by Partial Function
* Proper function \frac{P(x)}{Q(x)}
Degree P(x) < Degree Q(x)
* Improper function \frac{P(x)}{Q(x)}
Degree P(x) > Degree Q(x)
| Form of the Rational Function | Form of the Partial Function |
| \frac{px + q}{( x - a) ( x - b)} , a ≠ b | \frac{A}{ x - a} + \frac{B}{x - b} |
| \frac{px + q}{ (x - a)^{2} } | \frac{A}{ x - a} + \frac{B}{ (x - a)^{2} } |
| \frac{ px^{2} + qr + r }{( x - a) ( x - b) (x - c)} | \frac{A}{x - a} + \frac{B}{x - b} + \frac{C}{x - c} |
| \frac{ px^{2} + qx + r }{ (x - a)^{2} ( x - b) } | \frac{A}{x - a} + \frac{B}{ (x - a)^{2} } + \frac{C}{x - b} |
| \frac{ px^{2} + qx + r }{ ( x - a) ( x^{2} + bx + c )} | \frac{A}{x - a} + + \frac{Bx + C}{ x^{2} + bx + c} |
| Where x^{2} + bx + c cannot be factorized further | |
Integration by Parts
Method of integration by parts:
∫ ƒ(x) g(x) dx = ƒ (x) ∫ g(x) dx − ∫ [ ƒ’ (x) ∫ g(x) dx ] dx
* The ILATE precedence rule is used for determining the first function.
Integrals Involving Exponential Functions
∫ e^{x} [ ƒ’ (x) ] dx = e^{x} ƒ (x) + C
Integrals of Some More Types
∫ \sqrt{ x^{2} - a^{2} } dx= \frac{x}{2} \sqrt{ x^{2} - a^{2} } - \frac{ a^{2} }{2} log | x + \sqrt{ x^{2} + a^{2} } | + C
∫ \sqrt{ x^{2} + a^{2} } dx= \frac{x}{2} \sqrt{ x^{2} + a^{2} } + \frac{ a^{2} }{2} log | x + \sqrt{ x^{2} + a^{2} } | + C
∫ \sqrt{ a^{2} - x^{2} } dx= \frac{x}{2} \sqrt{ a^{2} - x^{2} } + \frac{ a^{2} }{2} sin^{-1} \frac{x}{a} + C
Definite Integration
The area of the region bounded by the curve y = ƒ(x), the X-axis and the coordinate x = a and x = b, where a ≤ x ≤ b is, \int_{a}^{b} ƒ(x) dx
A = \int_{a}^{b} ƒ(x) dx = (b – a) lim_{n \rightarrow ∞} × \frac{1}{n} [ ƒ(a) + ƒ(a + h) + … + ƒ(a + (n – 1)h)]
Where h = \frac{b - a}{n} ; as n → ∞
Fundamental Theorem of Calculus
* First fundamental theorem on integral calculus:
* Let the area function defined by A (x) = \int_{a}^{x} ƒ(x) dx, for all x ≥ a,
* Where the function assumed to continuous on
[a, b]. Then A’(x), for all x ∈ [a, b]
* Second fundamental theorem on integral calculus:
* Let ƒ be a continuous function of x defined on closed interval [a, b] and ɡ be another function such that ɡ’(x) = ƒ(x) for all x in the domain of ƒ.
Then: \int_{a}^{b} ƒ(x) dx = ɡ(b) – ɡ(a)
Evaluation of Definite Integration by Method of the Substitution
* Step to evaluate \int_{a}^{b} ƒ(x) dx by the method of substitution:
Step 1: Consider the integral without taking the given limits.
Step 2: Substitute y = ƒ(x) or x = ɡ(y) to reduce the given integral to a known form.
Step 3: Integrate the new integrate with respect to the new variable without placing the constant of integration.
Step 4: Write the answer in terms of the original variable by re-substituting the new variable.
Or
Keep the integral in the new variable itself and change the l limits of the integral accordingly.
Steps to evaluate \int_{a}^{b} ƒ(x) dx by the method of substitution:
Step 5: Find the values of the answers obtained in the previous step at the given upper and lower limits.
Step 6: Subtract the value at the lower limit from the value of the upper limit to obtain the required definite integral.
Properties of Definite Integrals – I
Property -1:
\int_{a}^{b} ƒ(x) dx = \int_{a}^{b} ƒ(t) dt
Property 2:
\int_{a}^{b} ƒ(x) dx = – \int_{b}^{a} ƒ(x) dx also, \int_{b}^{a} ƒ(x) dx = 0
Property 3:
If a < b, then \int_{a}^{b} ƒ(x) dx = \int_{a}^{c} ƒ(x) dx + \int_{c}^{b} ƒ(x) dx
Properties of Definite Integrals – II
Property 4:
\int_{a}^{b} ƒ(x) dx = \int_{a}^{b} ƒ(a + b − x ) dx
Property 5:
\int_{0}^{b} ƒ(x) dx – \int_{0}^{a} ƒ(a − x ) dx
Properties of Definite Integrals – III
1. Property
\int_{0}^{2a} ƒ(x) dx = \int_{0}^{a} ƒ(x) dx + \int_{0}^{a} ƒ(2a − bx) dx
2. Property
\int_{0}^{2a} ƒ(x) dx = { 2\int_{0}^{a} ƒ(x) dx, if (2a – x) = ƒ(x) 0, if ƒ(2a – x) = – ƒ(x)}
3. Property
\int_{- a}^{a} ƒ(x) dx = 2\int_{0}^{a} ƒ(x) dx, if ƒ is an even function, i.e. ƒ (–x) = ƒ(x)
\int_{- a}^{a} ƒ(x) dx = 0 if ƒ is an odd function, i.e. (–x) = – ƒ(x)