1. Problem: Evaluate the definite integral $$\int_{0}^{1} 5 e^{2x-1} dx$$. 2. Step 1: Factor constants and simplify the integrand by writing $5 e^{2x-1}=5 e^{-1} e^{2x}$. 3. Step 2: Pull constant factors outside the integral to get $5 e^{-1}\int_{0}^{1} e^{2x} dx$. 4. Step 3: Compute the antiderivative of $e^{2x}$ which is $\frac{1}{2}e^{2x}$. 5. Step 4: Evaluate the antiderivative at the bounds to obtain $5 e^{-1}\cdot \frac{1}{2}\bigl(e^{2}-1\bigr)$. 6. Step 5: Simplify the expression using $e^{2}e^{-1}=e$ and $e^{0}e^{-1}=e^{-1}$ to get $\frac{5}{2}\bigl(e-e^{-1}\bigr)$. Final answer: $\frac{5}{2}\left(e - e^{-1}\right)$.