Antiderivatives and Initial Value Problems

Definition of Antiderivative

If F (x) and f (x) are functions defined on an interval x and

for all x ∈ I, then F (x) is called an antiderivative of f (x).

Finding an antiderivative is the reverse operation to differentiation.

For example, the antiderivative of x² is \({\frac{{{x^3}}}{3}}\) because

Note that the functions \(\frac{{{x^3}}}{3} + 5,\) \(\frac{{{x^3}}}{3} - 2\) and any function \(\frac{{{x^3}}}{3} + C\) are also antiderivatives of \({x^2}\) because

Hence, if \(F\left( x \right)\) is an antiderivative of \(f\left( x \right)\) on an interval \(I,\) then the most general antiderivative of \(f\left( x \right)\) on \(I\) is

where \(C\) is an arbitrary constant.

Initial Value Problems

Finding an antiderivative of \(f\left( x \right)\) is equivalent to solving differential equation

A differential equation with an initial condition \(y\left( {{x_0}} \right) = {y_0}\) is called an initial value problem.

The most general antiderivative \(F\left( x \right) + C\) of the function \(f\left( x \right)\) gives the general solution of the differential equation \(\frac{{dy}}{{dx}} = f\left( x \right).\)

The particular solution of the initial value problem is a function that satisfies both the differential equation and the initial condition. To find the particular solution, we must apply the initial condition and determine the constant \(C.\)

Solved Problems

Solution.

Notice that the derivative of the function \({\frac{1}{{{x^3}}}}\) is given by

Hence, an antiderivative has the form

We can check this by differentiation:

Solution.

It is easy to see that

So an antiderivative is given by

We can check this by differentiation:

Solution.

Notice that

Clearly that an antiderivative is written as

because

Solution.

It is known that

Hence

Therefore an antiderivative is given by

We can check the result by differentiation:

Solution.

The general antiderivative of \({x^2} - 1\) is

Substitute the initial condition \(y\left( 3 \right) = 7\) to determine the value of \(C:\)

Hence, the solution of the initial value problem is given by

Solution.

First we write the general solution (general antiderivative) of the differential equation. As

then

Determine the constant \(C\) using the initial condition \(y\left( 1 \right) = 5:\)

Hence, the particular solution of the initial value problem has the form