Calculus

Integration of Functions

Integration of Functions Logo

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 \(f\left( x \right) = \large{\frac{1}{{\sqrt[3]{x}}}}\normalsize.\)

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 \(f\left( x \right) = \large{\frac{1}{{\sqrt[3]{x}}}}\normalsize.\)

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

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

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