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Text
The explain the some properties of indefinite integrals.
I) The process of differentiation and integration are inverses of each other in the sense of the following results :
`d/(dx) int` f(x)dx = f(x)
and `int f '(x) dx = f(x) + C` , where C is any arbitrary constant.
Proof: Let F be any anti derivative of f, i.e.,
`d/(dx)` F(x) = f(x)
Then `int` f(x)dx = F(x) +C
Therefore `d/(dx) int` f(x) dx = `d/(dx)`(f(x)+C)
= `d/(dx)` F(x)=f(x)
Similarly, we note that
f'(x) = `d/(dx)` f(x)
and hence `int`f'(x) dx = f(x) +C
where C is arbitrary constant called constant of integration.
II) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent.
Proof: Let f and g be two functions such that
`d/(dx) int f(x)dx = d/(dx) int g(x)dx`
or `d/dx [int f(x) dx - int g(x) dx]` = 0
Hence ∫f(x) dx - ∫g(x) dx = C, where C is any real number
or `int f(x)dx = int g(x)dx + C `
So the families of curves `{int f(x) dx + C_1, C_1 ∈ R}`
and `{int g(x) dx + C_2 , C_2 ∈ R}` are identical.
Hence, in this sense, `int f(x) dx ` and `int g(x) dx` are equivalent.
III) ∫[f(x) + g(x)]dx = ∫f(x)dx + ∫ g(x) dx
Proof: By Property (I), we have
`d/(dx)int[f(x) + g(x)dx] = f(x) +g(x)` ...(1)
On the otherhand, we find that
`d/(dx)[ int f(x) dx + int g(x) dx] = d/(dx) int f(x) dx + d/(dx) int g(x) dx`
=f(x) + g(x) ...(2)
Thus, in view of Property (II), it follows by (1) and (2) that
`int (f(x) + g(x))dx = int f(x) dx + int g(x) dx .`
IV) For any real number k, `int k f(x) dx = k int f(x) dx`
Proof: By the Property (I),
`d/(dx) int k f(x) dx = k f(x).`
Also `d/(dx) [k int f(x)dx] = k d/(dx) int f(x) dx` = k f(x)
Therefore, using the Property (II), we have `int k f(x) dx = k int f(x) dx .`
V) Properties (III) and (IV) can be generalised to a finite number of functions `f_1, f_2, ..., f_n` and the real numbers, `k_1, k_2, ..., k_n` giving
`int [k_1f_1(x) + k_2f_2(x) + ...+k_nf_n (x)] dx`
= `k_1 int f_1(x) dx +k_2 int f_2 (x) dx + ... + k_n int f_n (x) dx`
