Answer
$$
910 mi
$$
Work Step by Step
The object will reach its maximum distance when its speed equals zero. Evaluate the integral for $a=0, b=k, $ when the total result is equal to 0, using the antiderivative.
\[
\begin{array}{l}
{\left[9-191,000 \int_{0}^{k} \frac{1}{(x+3963)^{2}} d x\right]^{1 / 2}=0} \\
{\left[9-191,000\left(-\frac{-1}{3963}+\frac{-1}{k+3963}\right)\right]^{1 / 2}=0}
\end{array}
\]
\[
\left[\begin{array}{cc}
\left.9-191,000 \cdot \frac{-3963+(k+3963)}{3963(k+3963)}\right]^{1 / 2}=0 \\
9-191,000 \cdot \frac{k}{3963(k+3963) }=0 \\
\frac{191,000 k}{3963(3963+k) }=9
\end{array}\right.
\]
This simplifies as:
\[
\begin{aligned}
&3963(k+3963)9=191,000 k \\
k & \approx 910 \mathrm{mi}
\end{aligned}
\]
