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Solution:given, I(t)=200 e^{0.1 t}(a) To find capital formation from the end of the 3^{\text {rd }} year to end of the 10^{\text {th }} year so\begin{aligned}& \int_{3}^{10} 200 e^{0.1 t} d t \\= & 200 \int_{3}^{10} e^{0.1 t} d t \\= & \left.200 \frac{e^{0.1 t}}{0.1}\right|_{3} ^{10} \\= & 200\left[\frac{e^{0.1(10)}}{0.1}-\frac{e^{0.1(3)}}{0.1}\right] \\= & 200\left[\frac{e^{1}}{0.1}-\frac{0.3}{0.1}\right]=200\left[\frac{e^{1}-e^{0.3}}{0.1}\right] \\= & 2000\left[e^{1}-e^{0.3}\right] \\= & 2000[1.3684230208830] \\= & 2736.846\end{aligned}(b)To find Number of years required before the capital stock exceeds \$ 50000\begin{aligned}& \int_{0}^{t} 200 e^{0.1 t} d t=50000 \\= & 200\left[\frac{e^{0.1 t}}{0.1}\right]_{0}^{t}=50000 \\= & 200\left[\frac{e^{0.1 t}}{0.1}-\frac{e^{0.1(0)}}{0.1}\right]=50000\end{aligned}\begin{array}{l}=200\left[\frac{e^{0.1 t}-e^{0}}{0.1}\right]=50000 \\=2000\left[e^{0.1 t}-1\right]=50000 \\=e^{0.1 t}-1=25 \\e^{0.1 t}=25+1 \\e^{0.1 t}=26 \text { on bottsides }\end{array}Apply natural log on bottsides\begin{aligned}\ln \left(e^{0.1 t}\right) & =\ln (26) \\0.1 t & =\ln (26) \\0.1 t & =3.2580965380214 \\t & =32.580965380214 \\t & =32.58\end{aligned}The number of years required before the capital stock exceeds \$ 50000 is 32.58 years ...