Answer
The capital value is equal $ \$ 200,000.$
Work Step by Step
Suppose income from an investment starts (at time 0) at \$6000 a year and increases linearly and continuously at a rate of \$200 a year,so
$$
R(t)=(6000+200t)
$$
The capital value at an interest rate of $5\%$ compounded continuously (also $r=0.05$) is given by the following integral:
$$
\int_{0}^{\infty} (6000+200 t) e^{-0.05 t} d t
$$
First, we evaluate the indefinite integral as follows:
$$
\int (6000+200 t) e^{-0.05 t} d t
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=(6000+200 t), \quad\quad dv= e^{-0.05 t} d t } \\ {d u= 200 dt , \quad\quad v= \frac{1}{-0.05}e^{-0.05 t} }\end{array}\right] , \text { then }\\
$$
$ \Rightarrow$
$$
\begin{array}{l}{\int(6000+200 t) e^{-0.05 t} d t} \\ {\quad=-20 e^{-0.05 t}(6000+200 t)+\int 4000 e^{-0.05 t} d t} \\ {\quad=-20 e^{-0.05 t}(6000+200 t)-80,000 e^{-0.05 t}+C}\end{array}
$$
Now we will evaluate the improper integral ,by definition, as follows:
$$
\begin{split}
\int_{0}^{\infty} &(6000+200 t) e^{-0.05 t} d t =\lim _{b \rightarrow \infty} \int_{0}^{b}(6000+200 t) e^{-0.05 t} d t\\
&\left.=\lim _{b \rightarrow \infty}\left[-20 e^{-0.05 t}(6000+200 t)- 80,000 e^{-0.05 t}\right)\right]\left.\right|_{0} ^{b}\\
&=\left.\lim _{b \rightarrow \infty}\left[e^{-0.05 t}((-20)(6000+200 t)-80,000)\right]\right|_{0} ^{b} \\
&= \lim _{b \rightarrow \infty}\left[e^{-0.05 b}(-120,000-4000 b-80,000)\right.+200,000]\\
&=0+200,000 \\
&=\$ 200,000.
\end{split}
$$
So, The capital value is equal $ \$ 200,000.$
