Constant of Integration - Necessity of The Constant

Necessity of The Constant

At first glance it may seem that the constant is unnecessary, since it can be set to zero. Furthermore, when evaluating definite integrals using the fundamental theorem of calculus, the constant will always cancel with itself.

However, trying to set the constant equal to zero doesn't always make sense. For example, can be integrated in two different ways:

\begin{align} \int 2\sin(x)\cos(x)\,dx &=& \sin^2(x) + C &=& -\cos^2(x) + 1 + C \\ \int 2\sin(x)\cos(x)\,dx &=& -\cos^2(x) + C &=& \sin^2(x) - 1 + C \end{align}

So setting C to zero can still leave a constant. This means that, for a given function, there is no "simplest antiderivative".

Another problem with setting C equal to zero is that sometimes we want to find an antiderivative that has a given value at a given point (as in an initial value problem). For example, to obtain the antiderivative of that has the value 100 at x = π, then only one value of C will work (in this case C = 100).

This restriction can be rephrased in the language of differential equations. Finding an indefinite integral of a function is the same as solving the differential equation . Any differential equation will have many solutions, and each constant represents the unique solution of a well-posed initial value problem. Imposing the condition that our antiderivative takes the value 100 at x = π is an initial condition. Each initial condition corresponds to one and only one value of C, so without C it would be impossible to solve the problem.

There is another justification, coming from abstract algebra. The space of all (suitable) real-valued functions on the real numbers is a vector space, and the differential operator is a linear operator. The operator maps a function to zero if and only if that function is constant. Consequently, the kernel of is the space of all constant functions. The process of indefinite integration amounts to finding a preimage of a given function. There is no canonical preimage for a given function, but the set of all such preimages forms a coset. Choosing a constant is the same as choosing an element of the coset. In this context, solving an initial value problem is interpreted as lying in the hyperplane given by the initial conditions.

Read more about this topic:  Constant Of Integration

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