Section 5.3
Evaluate $\int e^{9x-2}dx$ using $u$ substitution.
Can somebody verify this for me? Thanks!
Let $u=9x-2$. Then $\frac{du}{dx}=9$ and so $\frac{du}{9}=dx$. Thus we have:
$\int e^{9x-2}dx$
$= \int e^u \frac{du}{9}$
$= \frac{1}{9} \int e^u du$
$=\frac{1}{9} e^u + C$
$=\frac{1}{9}e^{9x-2}+C$