Answer
$$
\left( f(x)=10000e^{-0.105t}, \quad T =10, \quad r= 10.5 \% \right)
$$
If $f(t)=10000e^{-0.105t}$ is the rate of continuous money flow at an interest rate $r=0.105$ for $T=10$ years, then the final amount at an interest rate of 10.5% compounded continuously is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(t) e^{-rt}dt.\\
&=e^{0.105(10)} \int_{0}^{10} 10,000 e^{-0.105 t} d t \\
&\approx \$ 176,919.15 \end{aligned}
$$
So, the accumulated amount of money flow at time $T=10$ is $\approx176,919.15.$
Work Step by Step
$$
\left( f(x)=10000e^{-0.105t}, \quad T =10, \quad r= 10.5 \% \right)
$$
If $f(t)=10000e^{-0.105t}$ is the rate of continuous money flow at an interest rate $r=0.105$ for $T=10$ years, then the final amount at an interest rate of 10.5% compounded continuously is
$$
\begin{aligned} A &=e^{rT}\int_{0}^{T} f(t) e^{-rt}dt.\\
&=e^{0.105(10)} \int_{0}^{10} 10,000 e^{-0.105 t} d t \\
&=\left.e^{1.05}\left(\frac{10,000 e^{-0.105 t}}{-0.105}\right)\right|_{0} ^{10} \\
&=\frac{10,000 e^{1.05}}{-0.105}\left(e^{-1.05}-1\right) \\
&\approx-272,157.25(-0.65006) \\
&\approx \$ 176,919.15 \end{aligned}
$$
So, the accumulated amount of money flow at time $T=10$ is $\approx176,919.15.$
