Answer
The integral converges and its value is $\frac{1}{4}e^{36}$.
Work Step by Step
We have
$$
\int_{-\infty}^{9} e^{4x}dx=\frac{1}{4}e^{4x}|_{-\infty}^9\\
=\frac{1}{4}(e^{-\infty}-e^{36})=\frac{1}{4}(0-e^{36})=\frac{1}{4}e^{36}.
$$
Hence, the integral converges and its value is $\frac{1}{4}e^{36}$.
