Section 4.2 Integration by Substitution
First, let's go over a few corrolations. Since derivatives and integrals are inverses of each other. The following operations are inverses as well:
RULES
power
\(\longrightarrow\)
power
trig
\(\longrightarrow\)
trig
quotient
\(\longrightarrow\)
N/A
chain
\(\longrightarrow\)
u-substitution
RECALL.
\begin{align*}
\dfrac{d}{dx}\left( (1+ 3x^2)^7 \right) \amp = 7(1 + 3x^2)^6 \cdot 6x\\
\amp = 42x (1 + 3x^2)^6
\end{align*}
IDEA:
\(\) |
\begin{equation*}
\nearrow \dfrac{d}{dx} (\cdots) \searrow
\end{equation*}
|
\(\) |
\(F(x) = (1 + 3x^2)^7\) |
\begin{equation*}
\rightleftharpoons
\end{equation*}
|
\(f(x) = 42x(1 + 3x^2)^7\) |
\(\) |
\begin{equation*}
\nwarrow \int \cdots dx \swarrow
\end{equation*}
|
\(\) |
Example 4.2.1.
Evaluate \(\displaystyle\int 42x (1 + 3x^2)^6 dx\) by u-substitution:
-
choose a \(u = g(x)\) that matches the "inside function"
\begin{equation*} u = g(x) = 1 + 3x^2 \end{equation*} -
define differential \(du\text{;}\) recall \(du = g'(x) dx\)
\begin{equation*} du = 6x dx \end{equation*} -
rewrite original integral in terms of \(u\) and \(du\)
\begin{align*} \int 42x ( 1 + 3x^2)^6 dx \amp = \int \underbrace{(1 + 3x^2)^6}_{u^6} \cdot \underbrace{42x dx}_{7du}\\ \amp = \int 7u^6 du\\ \amp = 7 \left( \dfrac{u^7}{7} \right) + c \\ \amp = (1 + 3x^2)^7 + c \end{align*}