Integral Calculus Chapter Notes | Quantitative Aptitude for CA Foundation PDF Download

Integration

We know 

Integration is the inverse operation of differentiation and is denoted by the symbol ∫ .
Hence, from equation (1), it follows that

i.e. Integral of xn with respect to variable x is equal to

Thus if we differentiate we can get back xⁿ
Again if we differentiate and c being a constant, we get back the same xⁿ . 

 Hence  + c and this c is called the constant of integration.

Integral calculus was primarily invented to determine the area bounded by the curves dividing the entire area into infinite number of infinitesimal small areas and taking the sum of all these small areas.

Basic Formulas

Note: In the answer for all integral sums we add +c (constant of integration) since the differentiation of constant is always zero.

Elementary Rules:

Examples : Find 

Solution:

Examples : Evaluate the following integral:

(i) 

(ii) 

(iii) 

(iv)

(v)

Method of Substitution (Change of Variable)

Example:

We put (2x + 3) = t 
⇒ so 2 dx = dt or dx = dt / 2

Therefore 

This method is known as Method of Substitution

Example:

We put (x² +1) = t

so 2x dx = dt or x dx = dt / 2

=

Important Standard Formulae

Examples: (a)

(b)

(c)

Integration by Parts

where u and v are two different functions of x

Evaluate:

(i) ∫ xe^x dx
Integrating by parts we see

(ii)

Integrating by parts,

iii)

=

Method of Partial Fraction

Type I :

Example:

Solution: let

=

[Here degree of the numerator must be lower than that of the denominator; the denominator contains non–repeated linear factor]

so 3x + 2 = A (x – 3) + B (x – 2)

We put x = 2 and get

3.2 + 2 = A (2–3) + B (2–2) ⇒ A = –8 we put x = 3 and get

3.3 +2 = A (3–3) + B (3–2) ⇒ B= 11

= –log(x–2)+11log (x -3) + c

Type II:
Example:

Solution: let

or 3x + 2 = A (x – 2) (x – 3) + B (x – 3) +C (x – 2)²
Comparing coefficients of x2 , x and the constant terms of both sides, we find
A+C = 0 …………(i)
–5A + B – 4C = 3 ……(ii)
6A – 3B + 4C = 2 …….(iii)
By (ii) + (iii) A – 2B = 5 ..…….(iv)
(i) – (iv) 2B + C = –5 …….(v)
From (iv) A = 5 + 2B
From (v) C = –5 – 2B
From (ii) –5 ( 5 + 2B) + B – 4 (– 5 – 2B) = 3 or – 25 – 10B + B + 20 + 8B = 3
or – B – 5 = 3
or B = – 8, A = 5 – 16 = – 11, from (iv) C = – A = 11

Therefore

Type III:
Example:

Solution: Let

so 3x² –2x +5 = A (x² + 5 ) + (Bx +C) (x–1)

Equating the coefficients of x² , x and the constant terms from both sides we get

Example:

Solution: 

Example : Find the equation of the curve where slope at (x, y) is 9x and which passes through the origin.

Solution: dy / dx = 9x 

Since it passes through the origin, c = 0; thus required eqn . is 9x² = 2y.

Definite Integration

‘b’ is called the upper limit and ‘a’ the lower limit of integration. We shall first deal with indefinite integral and then take up definite integral.

Example:

Solution: 

Note: In definite integration the constant (c) should not be added

Example:

Solution:

Now,

Important Properties of definite Integral 

Example:

Solution: Let I =

or I = 2/2 = 1

Example: Evaluate

Solution:

let x⁵ = t so that 5x⁴ dx = dt

Therefore,