Rules of calculation for antiderivatives

Integration: Antiderivatives

Rules of calculation for antiderivatives

As with differentiation, we can give a sum rule and constant rule for finding antiderivatives.

For functions #\blue f# and #\purple g#:

\[\begin{array}{c}
\displaystyle \int \blue{f(x)}+\purple{g(x)} \; {\dd}x = \displaystyle \int \blue{f(x)}\;{\dd}x + \int \purple{g(x)}\;{\dd}x
\end{array}\]

Examples

#\begin{array}{rcl}
\displaystyle \int \blue{2x} + \purple{4} \; {\dd}x &=& \displaystyle \int \blue{2x} \;{\dd}x + \int \purple{4} \;{\dd}x \\
&=& x^2 + 4x + \green {C}
\end{array}#

One constant of integration

Note that we only use one constant of integration #\green C# in the final answer.

When we work out the indefinite integrals in the intermediate step, we get two constants of integration. However, we can combine these constants of integration into one new constant of integration, which is why we only use one in our final answer.

#\begin{array}{rcl}
\displaystyle \int \blue{2x} + \purple{4} \; {\dd}x &=& \displaystyle \int \blue{2x} \;{\dd}x + \int \purple{4} \;{\dd}x \\ \\
&=& x^2 + \green{C_1}+ 4x + \green {C_2} \\ \\ &=& x^2 + 4x + \green {C}
\end{array}#

For a function #\blue f# and a constant #\orange c#:

\[\begin{array}{c}
\displaystyle \int \orange c \cdot \blue{f(x)} \; {\dd}x = \orange c\cdot \displaystyle \int \blue{f(x)}\;{\dd}x
\end{array}\]

Example

#\begin{array}{rcl}
\displaystyle \int \orange{9} \cdot \blue{x^2} \; \dd x &=& \orange{9} \cdot\displaystyle \int \blue{x^2} \; \dd x \\
&=& 9 \cdot \frac{1}{3}x^3 +\green C \\ &=& 3x^3+\green C
\end{array}#

One constant of integration

Note that we use the constant of integration #\green C# in the final answer, instead of #\orange c \cdot \green C#. This is because we can combine these into one new constant of integration.

#\begin{array}{rcl}
\displaystyle \int \orange{9} \cdot \blue{x^2} \; \dd x &=& \orange{9} \cdot\displaystyle \int \blue{x^2} \; \dd x \\
&=& 9 \cdot \frac{1}{3}x^3 +\orange9 \cdot \green{C_1} \\ &=& 3x^3+\green C
\end{array}#

Calculate an antiderivative #F# of the function #f# given by #f(x)=# #2\cdot x^8+4\cdot x^5#.

#F(x)=# #{{2\cdot x^9}\over{9}}+{{2\cdot x^6}\over{3}}#

#\begin{array}{rcl}
\displaystyle \int 2\cdot x^8+4\cdot x^5 \dd x &=&\displaystyle \int 2 \cdot x^{8}\, \dd x + \int 4 \cdot x^5\, \dd x\\
&&\phantom{xxx}\blue{\text{rule }\displaystyle \int f(x) + g(x) \, \dd = \int f(x) \, \dd x + \int g(x) \, \dd x}\\
&=&\displaystyle 2 \cdot \int x^{8}\, \dd x + 4 \cdot \int x^5 \, \dd x\\
&&\phantom{xxx}\displaystyle \blue{\text{twice applied the rule }\int c\cdot f(x)\,\dd x = c \cdot \int f(x) \, \dd x}\\
&=&\displaystyle \frac{2}{8+1}\cdot x^{8+1}+\frac{4}{5+1}\cdot x^{5+1}\\
&&\displaystyle \phantom{xxx}\blue{\text{twice applied the rule }\int x^{n} \; \dd x = \displaystyle\frac{1}{n+1} x^{n+1} + C}\\
&=&\displaystyle {{2\cdot x^9}\over{9}}+{{2\cdot x^6}\over{3}}+C\\
&&\displaystyle\phantom{xxx}\blue{\text{simplified}}
\end{array}#

Because we only asked for one antiderivative, we can now choose #C=0#. This gives:
\[F(x)={{2\cdot x^9}\over{9}}+{{2\cdot x^6}\over{3}}\]

New example