Calculus

Integration of Functions

Integration of Functions Logo

Antiderivatives and Initial Value Problems

Definition of Antiderivative

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

\[{F^\prime\left( x \right) = f\left( x \right)}\]

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

\[\left( {\frac{{{x^3}}}{3}} \right)^\prime = \frac{1}{3}\left( {{x^3}} \right)^\prime = \frac{1}{3} \cdot 3{x^2} = {x^2}.\]

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

\[\left( {\frac{{{x^3}}}{3} + C} \right)^\prime = \left( {\frac{{{x^3}}}{3}} \right)^\prime + C^\prime = {x^2} + 0 = {x^2}.\]

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

\[{F\left( x \right) + C,}\]

where \(C\) is an arbitrary constant.

Initial Value Problems

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

\[\frac{{dy}}{{dx}} = f\left( x \right)\;\;\text{or}\;\;y^\prime\left( x \right) = f\left( x \right).\]

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

Example 1.

Find an antiderivative of the function \[f\left( x \right) = \frac{1}{{{x^4}}}.\]

Solution.

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

\[\left( {\frac{1}{{{x^3}}}} \right)^\prime = \left( {{x^{ - 3}}} \right)^\prime = - 3{x^{ - 3 - 1}} = - 3{x^{ - 4}} = - \frac{3}{{{x^4}}}.\]

Hence, an antiderivative has the form

\[F\left( x \right) = - \frac{1}{{3{x^3}}}.\]

We can check this by differentiation:

\[F^\prime\left( x \right) = \left( { - \frac{1}{{3{x^3}}}} \right)^\prime = - \frac{1}{3}\left( {{x^{ - 3}}} \right)^\prime = - \frac{1}{3} \cdot \left( { - 3} \right){x^{ - 4}} = {x^{ - 4}} = \frac{1}{{{x^4}}} = f\left( x \right).\]

Example 2.

Find an antiderivative of the function \[f\left( x \right) = {e^{2x}}.\]

Solution.

It is easy to see that

\[\left( {{e^{2x}}} \right)^\prime = {e^{2x}} \cdot \left( {2x} \right)^\prime = 2{e^{2x}}.\]

So an antiderivative is given by

\[F\left( x \right) = \frac{{{e^{2x}}}}{2}.\]

We can check this by differentiation:

\[F^\prime\left( x \right) = \left( {\frac{{{e^{2x}}}}{2}} \right)^\prime = \frac{1}{2}\left( {{e^{2x}}} \right)^\prime = \frac{1}{2} \cdot 2{e^{2x}} = {e^{2x}} = f\left( x \right).\]

Example 3.

Find an antiderivative of the function \[f\left( x \right) = \frac{1}{{\sqrt[3]{x}}}.\]

Solution.

Notice that

\[\left( {\sqrt[3]{{{x^2}}}} \right)^\prime = \left( {{x^{\frac{2}{3}}}} \right)^\prime = \frac{2}{3}{x^{\frac{2}{3} - 1}} = \frac{2}{3}{x^{ - \frac{1}{3}}} = \frac{2}{{3{x^{\frac{1}{3}}}}} = \frac{2}{{3\sqrt[3]{x}}}.\]

Clearly that an antiderivative is written as

\[F\left( x \right) = \frac{{3\sqrt[3]{{{x^2}}}}}{2},\]

because

\[F^\prime\left( x \right) = \left( {\frac{{3\sqrt[3]{{{x^2}}}}}{2}} \right)^\prime = \frac{3}{2}\left( {\sqrt[3]{{{x^2}}}} \right)^\prime = \frac{3}{2} \cdot \frac{2}{{3\sqrt[3]{x}}} = \frac{1}{{\sqrt[3]{x}}} = f\left( x \right).\]

Example 4.

Find an antiderivative of the function \[f\left( x \right) = {3^{ - x}}.\]

Solution.

It is known that

\[\left( {{3^x}} \right)^\prime = {3^x}\ln 3.\]

Hence

\[\left( {{3^{ - x}}} \right)^\prime = - {3^{ - x}}\ln 3.\]

Therefore an antiderivative is given by

\[F\left( x \right) = - \frac{1}{{\ln 3}} \cdot {3^{ - x}} = - \frac{{{3^{ - x}}}}{{\ln 3}}.\]

We can check the result by differentiation:

\[F^\prime\left( x \right) = \left( { - \frac{{{3^{ - x}}}}{{\ln 3}}} \right)^\prime = - \frac{1}{{\ln 3}}\left( {{3^{ - x}}} \right)^\prime = - \frac{1}{{\ln 3}} \cdot \left( { - {3^{ - x}}\ln 3} \right) = {3^{ - x}}.\]

Example 5.

Solve the initial value problem \[\frac{{dy}}{{dx}} = {x^2} - 1, y\left( 3 \right) = 7.\]

Solution.

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

\[y = \frac{{{x^3}}}{3} - x + C.\]

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

\[\frac{{{3^3}}}{3} - 3 + C = 7,\;\; \Rightarrow 6 + C = 7,\;\; \Rightarrow C = 1.\]

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

\[y = \frac{{{x^3}}}{3} - x + 1.\]

Example 6.

Solve the initial value problem \[\frac{{dy}}{{dx}} = 2x - \frac{1}{{{x^2}}}, x \ne 0, y\left( 1 \right) = 5.\]

Solution.

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

\[\left( {{x^2}} \right)^\prime = 2x\;\; \text{and}\;\;\left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\]

then

\[y = {x^2} + \frac{1}{x} + C.\]

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

\[y\left( {x = 1} \right) = {1^2} + \frac{1}{1} + C = 5,\;\;\Rightarrow 2 + C = 5,\;\;\Rightarrow C = 3.\]

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

\[y = {x^2} + \frac{1}{x} + 3.\]

See more problems on Page 2.

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