1. **State the problem:** We need to find the area $A$ of the region enclosed by the curves $y=3x$, $y=4x$, and the vertical line $x=1$. 2. **Understand the region:** The lines $y=3x$ and $y=4x$ intersect at $x=0$ (since both are zero at $x=0$). The vertical line $x=1$ bounds the region on the right. 3. **Set up the integral:** The area between two curves $y=f(x)$ and $y=g(x)$ from $x=a$ to $x=b$ is given by $$A=\int_a^b |f(x)-g(x)|\,dx$$ 4. **Identify the top and bottom functions:** Since $4x > 3x$ for $x>0$, the top curve is $y=4x$ and the bottom curve is $y=3x$. 5. **Write the integral:** $$A=\int_0^1 (4x - 3x)\,dx = \int_0^1 x\,dx$$ 6. **Evaluate the integral:** $$\int_0^1 x\,dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$$ 7. **Final answer:** $$A=\frac{1}{2}$$