Answer
$$2.59 \mathrm{mi} / \mathrm{s}$$
Work Step by Step
Given
\[
\int \frac{1}{(x+3963)^{2}} d x
\]
Let $x+3963=u, d x=d u$
Antiderivative
\[
\int \frac{1}{u^{2}} d u=\frac{-1}{u}+C=\frac{-1}{x+3963}+C
\]
Evaluate the integral for $a=0, b=200 $. Using the antiderivative:
\[
\begin{array}{l}
v \approx\left[9-191,000 \int_{0}^{200} \frac{1}{(x+3963)^{2}} d x\right]^{1 / 2} \\
=\left[9-191,000\left(-\frac{-1}{3963}+\frac{-1}{200+3963 }\right)\right]^{1 / 2} \\
=\left[9-191,000\left(1.21 \times 10^{-5}\right)\right]^{1 / 2} \\
\quad=(6.6889)^{1 / 2} \approx 2.59 \mathrm{mi} / \mathrm{s}
\end{array}
\]
