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 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

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