1.7. Integration using substitution#
When we know an integral
(1.160)#\[\begin{equation}
\int e^x \d{x} = e^x + c
\end{equation}\]
we can evaluate related integrals like:
(1.161)#\[\begin{equation}
\int e^{-x} \d{x} = -e^{-x} + c
\end{equation}\]
by making a substitution. Specifically, we define a new variable u, calculate its differential, and then replace both. For example, in this case, let \(u = -x\). Then, \(\d{u} = -\d{x}\). Equivalently, \(x = -u\) and \(\d{x} = -\d{u}\) so
(1.162)#\[\begin{equation}
\int e^{-x} \d{x} = -\int e^u \d{u} = -e^u + c = -e^{-x} + c
\end{equation}\]
Example: u-substitution
Evaluate the integral
(1.163)#\[\begin{equation}
\int x e^{x^2} \d{x}
\end{equation}\]
Let \(u=x^2\) so \(\d{u} = 2x \d{x}\). Then,
(1.164)#\[\begin{align}
\int e^{x^2} x\d{x} = \int e^u \cdot \frac{1}{2}\d{u} = \frac{1}{2} e^u + c
\end{align}\]
Plugging u back in gives the final answer:
(1.165)#\[\begin{equation}
\int xe^{x^2} \d{x} = \frac{1}{2}e^{x^2} + c
\end{equation}\]