1. **State the problem:** Find the area of the region enclosed by the graphs of $y=6x$ and $y=5x^4$. 2. **Find the points of intersection:** Set $6x=5x^4$ to find where the curves meet. $$6x=5x^4 \implies 6x-5x^4=0 \implies x(6-5x^3)=0$$ So, $x=0$ or $6-5x^3=0 \implies 5x^3=6 \implies x=\sqrt[3]{\frac{6}{5}}$. 3. **Determine which function is on top between the intersection points:** For $x$ in $[0, \sqrt[3]{6/5}]$, check $y=6x$ and $y=5x^4$ at a test point, say $x=0.5$: $6(0.5)=3$ and $5(0.5)^4=5(0.0625)=0.3125$, so $6x$ is above $5x^4$. 4. **Set up the integral for the area:** $$\text{Area} = \int_0^{\sqrt[3]{6/5}} (6x - 5x^4) \, dx$$ 5. **Compute the integral:** $$\int_0^{\sqrt[3]{6/5}} 6x \, dx = \left[3x^2\right]_0^{\sqrt[3]{6/5}} = 3\left(\sqrt[3]{\frac{6}{5}}\right)^2$$ $$\int_0^{\sqrt[3]{6/5}} 5x^4 \, dx = \left[\frac{5}{5}x^5\right]_0^{\sqrt[3]{6/5}} = \left(x^5\right)_0^{\sqrt[3]{6/5}} = \left(\sqrt[3]{\frac{6}{5}}\right)^5$$ 6. **Simplify powers:** Recall $\left(\sqrt[3]{a}\right)^n = a^{n/3}$, so $$3\left(\frac{6}{5}\right)^{2/3} - \left(\frac{6}{5}\right)^{5/3}$$ 7. **Calculate numerical value:** Calculate $\left(\frac{6}{5}\right)^{1/3} \approx 1.0408$, then $\left(\frac{6}{5}\right)^{2/3} = (1.0408)^2 \approx 1.0833$, and $\left(\frac{6}{5}\right)^{5/3} = (1.0408)^5 \approx 1.2255$. So area $\approx 3 \times 1.0833 - 1.2255 = 3.2499 - 1.2255 = 2.0244$. 8. **Final answer:** $$\boxed{2.024}$$ (correct to 3 decimal places)