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# Arbitrary constant of integration

Let us assume a mathematicsfunction F (x) is an integration of a function f (x), then the set of all integrals of f (x) is given by F (x) + cwhere, c be any real constant. Thatmeans a function has single derivative but multiple integrals.Reason behind the arbitrary constant is we know that the derivative of a constant function is zero. According to the second fundamental theorem derivative of the integral of a function is a function itself.Let us consider a function f(x).Integral f (x) = F (x) + c according to the definition of integral.d/dx integral f(x) dx = d/dx F(x) + d/dx c = F’(x) + 0since F’(x) = f(x) we get, d/dx integral f(x) dx = f(x)once we get one integral of a function we can find set of all integrals by adding or subtracting different constants to the integral.

For example; integral of sin x dx = -cos x + c. Here ‘c’ can have any real value.Since, integral of sinx dx = -cos x + c , derivative of –cos x + c will return a function sin x for any value of ‘c’.Let us try the same for c = 0, -1 and 2. At c = 0, d/ dx –cosx = - (-sin x) = sin x and at c = -1, d/dx [-cos x – 1] = d/dx –cos x – d/dx 1 = - (-sin x) – 0 = sin x. Now at c = 2, d/dx [-cos x + 2] = d/dx –cos x + d/dx 2 = - (-sin x) + 0 = sin x. We get same derivative function as derivative of a constant term is always zero.Hence for integrals of any function we get a set of functions which differ by a constant term always.

Integral f(x) dx = F(x) + c1 = F(x) + c2. Here the difference between two integrals is abs (c1 – c2). Hence we get multiple integrals for f(x) as: F(x), F(x) + c1, F(x) – c1….. and so on.Here ‘c’ is called as the arbitrary constant of integration or integration constant.It plays a vital role while finding indefinite integral, since it represents a set of all possible integrals for a given function f (x).Arbitrary constant in definite integral:Definition of definite integral: integral from x = a to x = b f(x) dx = F(b) – F(a). We don’t use arbitrary constant in definite integrals as it is zero or already known in this case.Since we have two bounds of integral we have enough information to solve integral for a constant. Thus there is no need to put ‘c’ while finding definite integrals.F(b) – F(a) suggests that two integrals of f(x) at x = a and x = b differ by a constant value.

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