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Step 1The solution is given by using two methods integral method and Laplace transform method as follows :Step 2Ans::f=e^{t}, g=e^{-5 t}\begin{array}{l}(f \neq g)(t)=\int_{0}^{t} f(\tau) \cdot g(t-t) d t \\=\int_{0}^{t} e^{t} \cdot e^{-5(t-\tau)} d t=\int_{0}^{t} e^{t} \cdot e^{-5 t} \cdot e^{5 t} d \tau \\=e^{-5 t} \int_{0}^{t} e^{\tau} \cdot e^{5 t} d t \Rightarrow e^{-5 t} \int_{0}^{t} e^{\tau+s t} d \tau \\\Rightarrow e^{-s t} \int_{0}^{t} e^{6 t} d t=\frac{e^{-5 t}}{}\left[\frac{e^{6 \tau}}{6}\right]_{0}^{t} \\\Rightarrow e^{-5 t}\left[\frac{e^{6 t}-1}{6}\right]=\frac{e^{t}-e^{-5 t}}{6}-\frac{e^{-5 t}}{6} \\F(s)=L\{f(t)\}=L\left\{e^{t}\right\}=\frac{1}{s-1}=F(s) \\G(s)=L\{g(t)\}=L\left\{e^{-5 t}\right\}=\frac{1}{s+5}=G(s) \\\Rightarrow \quad f G)\end{array}\begin{array}{l}F(s)=L\{f(t)\}=L\left\{e^{t}\right\}=\frac{1}{s-1}=F(s) \\G(s)=L\{g(t)\}=L\left\{e^{-5 t}\right\}=\frac{1}{s+5}=G(s)\end{array}Then (f * G)(t)=L^{-1}\{f(s) \cdot G(s)\}=L^{-1}\left\{\frac{1}{(s-1)} \times \frac{1}{(s+s)}\right\}.aow by partial benetion.\begin{array}{l}\frac{1}{(s-1)(s+5)}=\frac{\left(\frac{1}{6}\right)}{(s-1)}+\frac{\left(-\frac{1}{6}\right)}{s+5} \\\Rightarrow L^{-1}\left\{\frac{1}{(s-1)(s+5)}\right\}=\frac{1}{6} i^{-1}\left\{\frac{1}{s-1}\right\}-\frac{1}{6} L^{-1}\left\{\frac{1}{s+5}\right\} \\=\frac{1}{6} e^{t}-\frac{1}{6} e^{-5 t}=\frac{e^{t}}{6}-\frac{e^{-5 t}}{6} \\=\frac{e^{t}-e^{-5 t}}{6} \\\end{array} ...