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Properties of Indefinite Integral

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Properties of Indefinite Integral - Lesson Summary

Property - I

d/dx ∫ f(x) dx = f(x) and ∫ f '(x) dx = f(x) + C,

where C is any arbitrary constant .

Proof:

Let G be any anti-derivative of f.

d/dx[G(x)] = f(x) ....... (1)

∫ f(x) dx = G(x) + C

d/dx[∫ f(x) dx] = d/dx[G(x) + C]

= d/dx[G(x)]  ['.' d/dx (C) = 0]

= f(x)     [from (1)]

∴ d/dx[∫ f(x)dx] = f(x)

We know that, ∫ d/dx[f(x)]dx = ∫ f ' (x) dx

⇒ ∫ f '(x)dx = f(x) + C

Here, C is called the constant of integration.

Property - II:

Two indefinite integrals having the same derivative lead to the same family of curves and they are equivalent.

Proof:

Let f and g be two functions such that,

d/dx ∫ f(x) dx = d/dx ∫ g(x)dx

⇒d/dx ∫ f(x)dx - d/dx ∫ g(x)dx = 0

⇒d/dx[∫f(x)dx - ∫g(x)dx] = 0

d/dx(Constant) = 0

⇒ ∫ f(x)dx - ∫ g(x) dx = C , arbitrary constant.

∫ f(x)dx = ∫ g(x) dx + C

If C = C2 - C1, then

∫ f(x)dx = ∫ g(x) + C2 - C1

⇒ ∫ f(x)dx + C1 = ∫ g(x) + C2

{∫ f(x)dx + C1,C2 ∈ R} and {∫ g(x)dx + C1,C2 ∈ R} are identical }

∫ f(x)dx and ∫ g(x)dx are equivalent.

NOTE:

The family of curves is

{∫ f(x)dx + C1,C2 ∈ R} ⇔∫ f(x)dx {∫ g(x)dx + C1,C2 ∈ R} ⇔ ∫ g(x)dx

Property - III

If f and g are two functions, then ∫ [f(x) + g(x)]dx = ∫ f(x) dx + ∫ g(x) dx

Proof:

d/dx ∫ f(x) dx = f(x)

d/dx ∫ [f(x) + g(x)] dx = f(x) + g(x) .......(1)

Consider d/dx [∫ f(x) dx + ∫ g(x) dx]

= d/dx ∫ f(x) dx + d/dx ∫ g(x) dx

= f(x) + g(x).....(2)

 d/dx ∫ [f(x) + g(x)] dx =  d/dx [∫ f(x) dx + ∫ g(x) dx]------- (3)

∫ [f(x) + g(x)] dx = ∫ f(x) dx+ ∫ g(x) dx

Hence, the integral of the sum of functions is equal to sum of the integral of the functions.



Property - IV

∫ k f(x) dx = k ∫ f(x) dx, where k is any real number.

Proof:

d/dx ∫ f(x) dx = f(x)

d/dx ∫ kf(x) dx = kf(x)-------(1)

We know that , d/dx(k x) = k d/dx (x)

d/dx[k ∫ f(x)dx] = k d/dx ∫ f(x) dx

= k f(x)  ['.' d/dx ∫ g(x) dx = g(x)]

d/dx[k ∫ f(x)dx] = k f(x).........(2)

d/dx ∫ k f(x)dx = d/dx[k ∫ f(x)dx]

d/dx ∫ k f(x)dx - d/dx[k ∫ f(x)dx] = 0

⇒d/dx[∫ k f(x)dx - k ∫ f(x)dx] = 0

Since the derivative of a constant is equal to zero , we have

∫ kf(x)dx - k ∫ f(x)dx = C

∫ kf(x)dx = k ∫ f(x)dx + C

Hence, ∫ k f(x)dx = k ∫ f(x) dx.

Property - V:

∫[k1f1(x) + k2f2(x) + k3f3(x) + .... +.... knfn(x)]dx = k1 ∫ f1(x) dx + k f2(x) dx + k3 ∫ f3(x) dx + .... +.... kn ∫ fn(x)]dx

Proof:

We combine properties III and IV for a finite number of functions such as f1,f2,f3,....,fn and for real numbers

k1,k2,k3,....,kn.

So, using properties III and IV, we have

∫[k1f1(x) + k2f2(x) + k3f3(x) + .... +.... knfn(x)]dx = k1 ∫ f1(x) dx + k f2(x) dx + k3 ∫ f3(x) dx + .... +.... kn ∫ fn(x)]dx .

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