# 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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

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

### Example 6

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

### 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.