# Calculus

## Integration of Functions # Antiderivatives and Initial Value Problems

### Definition of Antiderivative

If $$F\left( x \right)$$ and $$f\left( x \right)$$ are functions defined on an interval $$I$$ and

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

for all $$x \in I,$$ then $$F\left( x \right)$$ is called an antiderivative of $$f\left( x \right).$$

Finding an antiderivative is the reverse operation to differentiation.

For example, the antiderivative of $${x^2}$$ is $${\large{\frac{{{x^3}}}{3}}\normalsize}$$ 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 $$\large{\frac{{{x^3}}}{3}}\normalsize + 5,$$ $$\large{\frac{{{x^3}}}{3}}\normalsize – 2$$ and any function $$\large{\frac{{{x^3}}}{3}}\normalsize + C$$ 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

${\frac{{dy}}{{dx}} = f\left( x \right)\;\;}\kern0pt{\text{or}\;\;y^\prime\left( x \right) = f\left( x \right).}$

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 $$\large{\frac{{dy}}{{dx}}}\normalsize = f\left( x \right).$$

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 $$f\left( x \right) = \large{\frac{1}{{{x^4}}}}\normalsize.$$

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

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

### Example 6

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

### Example 7

Solve the initial value problem $$\large{\frac{{dy}}{{dx}}}\normalsize = \large{\frac{2}{{x + 1}}}\normalsize,$$ $$y\left( 0 \right) = 2.$$

### Example 8

Solve the initial value problem $$\large{\frac{{dr}}{{d\theta }}}\normalsize = \cos \large{\frac{\theta }{2}}\normalsize,$$ $$r\left( {\large{\frac{\pi }{3}}\normalsize} \right) = 2.$$

### Example 9

Solve the initial value problem $$\large{\frac{{dz}}{{dt}}}\normalsize = \cos t – 2\sin t,$$ $$z\left( 0 \right) = 5.$$

### Example 10

A function $$y\left( x \right)$$ is given by the differential equation $$\large{\frac{{dy}}{{dx}}}\normalsize = \large{\frac{1}{x}}\normalsize + 2x$$ with the initial condition $$y\left( 1 \right) = 0.$$ Find the function value when $$x = e.$$

### Example 11

A ball dropped from the top of a tall building has velocity $$v\left( t \right) = -10t – 5 \,{\large{\frac{\text{m}}{\text{s}}}\normalsize}.$$ Find the height of the building given that the ball strikes the ground after $$t = 4$$ seconds.

### Example 1.

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

Solution.

Notice that the derivative of the function $${\large{\frac{1}{{{x^3}}}}}\normalsize$$ 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 $$f\left( x \right) = \large{\frac{1}{{\sqrt{x}}}}\normalsize.$$

Solution.

Notice that

${\left( {\sqrt{{{x^2}}}} \right)^\prime = \left( {{x^{\frac{2}{3}}}} \right)^\prime }={ \frac{2}{3}{x^{\frac{2}{3} – 1}} }={ \frac{2}{3}{x^{ – \frac{1}{3}}} }={ \frac{2}{{3{x^{\frac{1}{3}}}}} }={ \frac{2}{{3\sqrt{x}}}.}$

Clearly that an antiderivative is written as

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

because

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

### 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 $$\large{\frac{{dy}}{{dx}}}\normalsize = {x^2} – 1,$$ $$y\left( 3 \right) = 7.$$

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 $${\large{\frac{{dy}}{{dx}}}\normalsize = 2x – \large{\frac{1}{{{x^2}}}}\normalsize,}$$ $${x \ne 0,}$$ $${y\left( 1 \right) = 5.}$$

Solution.

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

${\left( {{x^2}} \right)^\prime = 2x\;\;}\kern0pt{\text{and}\;\;\left( {\frac{1}{x}} \right)^\prime = – \frac{1}{{{x^2}}},}$

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

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

### Example 7.

Solve the initial value problem $$\large{\frac{{dy}}{{dx}}}\normalsize = \large{\frac{2}{{x + 1}}}\normalsize,$$ $$y\left( 0 \right) = 2.$$

Solution.

Find the general antiderivative of $$\large{\frac{2}{{x + 1}}}\normalsize:$$

$y = 2\ln \left| {x + 1} \right| + C.$

Use the initial condition $$y\left( 0 \right) = 2$$ to evaluate the constant $$C:$$

${2\ln \left| {0 + 1} \right| + C = 2,}\;\; \Rightarrow {2 \cdot \ln 1 + C = 2,}\;\; \Rightarrow {2 \cdot 0 + C = 2,}\;\; \Rightarrow {C = 2.}$

So the particular solution is written as

$y = 2\ln \left| {x + 1} \right| + 2.$

### Example 8.

Solve the initial value problem $$\large{\frac{{dr}}{{d\theta }}}\normalsize = \cos \large{\frac{\theta }{2}}\normalsize,$$ $$r\left( {\large{\frac{\pi }{3}}\normalsize} \right) = 2.$$

Solution.

We first write the general antiderivative of the function $$\cos \large{\frac{\theta }{2}}\normalsize:$$

$r\left( \theta \right) = 2\sin \frac{\theta }{2} + C.$

Substitute the initial condition $$r\left( {\large{\frac{\pi }{3}}\normalsize} \right) = 2$$ to find the constant $$C:$$

${2\sin \frac{{\frac{\pi }{3}}}{2} + C = 2,}\;\; \Rightarrow {2\sin \frac{\pi }{6} + C = 2,}\;\; \Rightarrow {2 \cdot \frac{1}{2} + C = 2,}\;\; \Rightarrow {1 + C = 2,} \Rightarrow {C = 1.}$

Hence, the particular solution of the differential equation is given by

$r\left( \theta \right) = 2\sin \frac{\theta }{2} + 1.$

### Example 9.

Solve the initial value problem $$\large{\frac{{dz}}{{dt}}}\normalsize = \cos t – 2\sin t,$$ $$z\left( 0 \right) = 5.$$

Solution.

The general antiderivative of the function $$\cos t – 2\sin t$$ is

$z\left( t \right) = \sin t + 2\cos t + C.$

Substitute the initial condition $$z\left( 0 \right) = 5$$ to evaluate the constant $$C:$$

${\sin 0 + 2\cos 0 + C = 5,}\;\; \Rightarrow {0 + 2 \cdot 1 + C = 5,}\;\; \Rightarrow {C = 3.}$

So the solution of the initial value problem is given by

$z\left( t \right) = \sin t + 2\cos t + 3.$

### Example 10.

A function $$y\left( x \right)$$ is given by the differential equation $$\large{\frac{{dy}}{{dx}}}\normalsize = \large{\frac{1}{x}}\normalsize + 2x$$ with the initial condition $$y\left( 1 \right) = 0.$$ Find the function value when $$x = e.$$

Solution.

Write the general antiderivative of the function $$\large{\frac{1}{x}}\normalsize + 2x:$$

$y = \ln x + {x^2} + C.$

Using the initial condition $$y\left( 1 \right) = 0,$$ determine the constant $$C:$$

${y\left( 1 \right) = 0,}\;\; \Rightarrow {\ln 1 + {1^2} + C = 0,}\;\; \Rightarrow {0 + 1 + C = 0,}\;\; \Rightarrow {C = – 1.}$

Then the function $$y\left( x \right)$$ is given by

$y = \ln x + {x^2} – 1.$

Compute the function value at $$x = e:$$

${\require{cancel}y\left( e \right) = \ln e + {e^2} – 1 }={ \cancel{1} + {e^2} – \cancel{1} }={ {e^2}.}$

### Example 11.

A ball dropped from the top of a tall building has velocity $$v\left( t \right) = -10t – 5 \,{\large{\frac{\text{m}}{\text{s}}}\normalsize}.$$ Find the height of the building given that the ball strikes the ground after $$t = 4$$ seconds.

Solution.

Let $$H$$ be the height of the building. To get the vertical position of the ball $$h\left( t \right),$$ we determine the general antiderivative of the velocity function:

$h\left( t \right) = C – 5{t^2} – 5t.$

The constant $$C$$ is found from the initial condition $$h\left( 0 \right) = H.$$ Hence

$h\left( t \right) = H – 5{t^2} – 5t.$

After $$t = 4$$ seconds, the height $$h$$ is equal to $$0,$$ so we have

${H – 5 \cdot {4^2} – 5 \cdot 4 = 0,}\;\; \Rightarrow {H = 80 + 20 = 100\,\text{m}}.$