Class 12 Maths Integrals Integration as an Inverse Process of Differentiation

Integration as an Inverse Process of Differentiation

Integration is the inverse process of differentiation. Let f(x) be a function. Then the collection of all primitives is called the indefinite integral of f(x) and denoted by ʃ f(x) dx.

Thus, d[ф(x) + C]/dx = f(x) => ʃ f(x) dx = ф(x) + C

Where ф(x) is primitive of f(x) and C is an arbitrary constant known as constant of integration.

    Class_12_Maths_Integrals_Figure_2                            

The following is a list to find integrals of other functions.

  Class_12_Maths_Integrals_Formulas_Of_Integration                                  

                             

Some properties of indefinite integral

(i) The process of differentiation and integration are inverses of each other in the sense of the following results :

d[ʃ f(x) dx]/dx = f(x)

and  ʃ f’(x) dx = f(x) + C, where C is any arbitrary constant.

(ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent.

i.e. d[ʃ f(x) dx]/dx = d[ʃ g(x) dx]/dx

=> d[ʃ f(x) dx - ʃ g(x) dx]/dx = 0

=> ʃ f(x) dx - ʃ g(x) dx = C, where C is any real number

=> ʃ f(x) dx = ʃ g(x) dx + C

So, the families of curve [ʃ f(x) dx + C1, C1 є R] and [ʃ g(x) dx + C2, C2 є R] are identical.

Hence, ʃ f(x) dx and ʃ g(x) dx are equivalent.

(iii) ʃ [f(x) + g(x)] dx = ʃ f(x) dx + ʃ g(x) dx

(iv) For any real number k, ʃ [k * f(x)] dx = k * ʃ f(x) dx

 (v) ʃ [k1 * f1(x) + k2 * f2(x) + . . . . .+ kn * fn(x)] dx = k1 * ʃ f1(x) dx + k2 * ʃ f2(x) dx + . . . . . .+ kn * ʃ fn(x) dx

Problem: Find the integral of the following functions:

(a) ʃ (ax2 + bx + c) dx  (b) ʃ (2x2 + ex) dx

Solution:

(a) ʃ (ax2 + bx + c) dx = ʃ ax2 dx + ʃ bx dx + ʃ c dx

                                     = aʃ x2 dx + b ʃ x dx + c ʃ dx

                                     = ax3/3 + bx2/2 + cx + C

So, ʃ (ax2 + bx + c) dx = ax3/3 + bx2/2 + cx + C

(b) ʃ (2x2 + ex) dx = ʃ 2x2 dx + ʃ ex dx

                              = 2ʃ x2 dx + ʃ ex dx

                              = 2x3/3 + ex + C

So, ʃ (2x2 + ex) dx = 2x3/3 + ex + C

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