Integration: Antiderivatives
Antiderivatives and the chain rule
We have seen the chain rule when we differentiated. This can also play a role in integrating functions. We will first look at how we integrate functions of the form #f(ax+b)# using the chain rule. Later, we'll discuss the substitution method, which is a more general form of integration with the consequences of the chain rule.
A function #f(\blue ax+b)# has the function #\frac{1}{\blue a}F(\blue ax+b)+\green C# as an antiderivative, in which #F# is an antiderivative of #f#. Therefore:
\[\displaystyle \int f(\blue ax+b) \; \dd x = \frac{1}{\blue a} F(\blue ax+b)+\green C\]
Example
#\begin{array}{rcl}\displaystyle \int \sin(\blue 3 x+5) \; \dd x&=&\frac{1}{\blue 3} \cdot -\cos(\blue3 x+5) \\ &=& -\frac{1}{3}\cos(3x+5)\end{array}#
#F(x)=# #{{\left(4\cdot x-2\right)^4}\over{16}}#
We use the chain rule for integration to determine the integral, and we note that the function is of the form #g(4\cdot x -2)#, with #g(x)=x^{3}#.
\[\begin{array}{rcl}
\displaystyle \int \left(4\cdot x-2\right)^3 \, \dd x &=&\displaystyle \int g(4 \cdot x -2) \, \dd x\\
&&\phantom{xxx}\blue{\text{rewritten to the form }g(ax+b) \text{ with } g(x)=x^{3}}\\
&=&\displaystyle \frac{1}{4}\cdot G(4 \cdot x -2)\\
&&\phantom{xxx}\blue{\text{chain rule for integration}}\\
&=&\displaystyle \frac{1}{4}\cdot \frac{1}{3+1}(4 \cdot x -2)^{3+1}+C\\
&&\displaystyle \phantom{xxx}\blue{\text{use rule } \int x^{n} \; \dd x =\frac{1}{n+1} x^{n+1} + C}\\
&=&\displaystyle {{\left(4\cdot x-2\right)^4}\over{16}}+C\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]
Because we only asked for one antiderivative, we can now choose #C=0#. This gives:
\[F(x)={{\left(4\cdot x-2\right)^4}\over{16}}\]
We use the chain rule for integration to determine the integral, and we note that the function is of the form #g(4\cdot x -2)#, with #g(x)=x^{3}#.
\[\begin{array}{rcl}
\displaystyle \int \left(4\cdot x-2\right)^3 \, \dd x &=&\displaystyle \int g(4 \cdot x -2) \, \dd x\\
&&\phantom{xxx}\blue{\text{rewritten to the form }g(ax+b) \text{ with } g(x)=x^{3}}\\
&=&\displaystyle \frac{1}{4}\cdot G(4 \cdot x -2)\\
&&\phantom{xxx}\blue{\text{chain rule for integration}}\\
&=&\displaystyle \frac{1}{4}\cdot \frac{1}{3+1}(4 \cdot x -2)^{3+1}+C\\
&&\displaystyle \phantom{xxx}\blue{\text{use rule } \int x^{n} \; \dd x =\frac{1}{n+1} x^{n+1} + C}\\
&=&\displaystyle {{\left(4\cdot x-2\right)^4}\over{16}}+C\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]
Because we only asked for one antiderivative, we can now choose #C=0#. This gives:
\[F(x)={{\left(4\cdot x-2\right)^4}\over{16}}\]
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