Use a change of variables or the table to evaluate the following definite integral. \( \int_{0}^{1} 6 e^{3 x} \mathrm{dx} \) \( \int_{0}^{1} 6 e^{3 x} \mathrm{dx}=\square \) (Type an exact answer.) (To view the table of general integration formulas.

To evaluate the definite integral \[ \int_{0}^{1} 6 e^{3x} \, dx, \] we can use a change of variables. Let's set \[ u = 3x. \] Then, the differential \(du\) is given by \[ du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}. \] Next, we need to change the limits of integration. When \(x = 0\), \[ u = 3 \cdot 0 = 0, \] and when \(x = 1\), \[ u = 3 \cdot 1 = 3. \] Now we can rewrite the integral in terms of \(u\): \[ \int_{0}^{1} 6 e^{3x} \, dx = \int_{0}^{3} 6 e^{u} \cdot \frac{du}{3}. \] This simplifies to: \[ \int_{0}^{3} 2 e^{u} \, du. \] Now we can evaluate the integral: \[ \int 2 e^{u} \, du = 2 e^{u} + C. \] Now we will evaluate this from \(0\) to \(3\): \[ \left[ 2 e^{u} \right]_{0}^{3} = 2 e^{3} - 2 e^{0} = 2 e^{3} - 2. \] Thus, the value of the definite integral is: \[ \int_{0}^{1} 6 e^{3x} \, dx = 2 e^{3} - 2. \] So, the final answer is: \[ \int_{0}^{1} 6 e^{3x} \, dx = 2 e^{3} - 2. \]