1. **Problem Statement:** Find the area of the shaded region bounded by the curves $y=1$, $y=\frac{1}{4}x$, and $y=\frac{1}{36}x^2$. 2. **Understanding the curves:** - $y=1$ is a horizontal line. - $y=\frac{1}{4}x$ is a straight line through the origin with slope $\frac{1}{4}$. - $y=\frac{1}{36}x^2$ is a parabola opening upwards. 3. **Find intersection points:** - Intersection of $y=1$ and $y=\frac{1}{4}x$: $$1=\frac{1}{4}x \implies x=4$$ - Intersection of $y=1$ and $y=\frac{1}{36}x^2$: $$1=\frac{1}{36}x^2 \implies x^2=36 \implies x=6$$ - Intersection of $y=\frac{1}{4}x$ and $y=\frac{1}{36}x^2$: $$\frac{1}{4}x=\frac{1}{36}x^2 \implies \cancel{x} \left(\frac{1}{4}-\frac{1}{36}x\right)=0 \implies x=0 \text{ or } \frac{1}{4}=\frac{1}{36}x$$ $$\frac{1}{4}=\frac{1}{36}x \implies x=9$$ 4. **Determine the region boundaries:** - The shaded region is between $x=0$ and $x=4$ where $y=\frac{1}{4}x$ is above $y=\frac{1}{36}x^2$ and below $y=1$. 5. **Set up the integral for the area:** - The area between two curves $y=f(x)$ and $y=g(x)$ from $a$ to $b$ is: $$\text{Area} = \int_a^b (f(x)-g(x)) \, dx$$ - Here, between $x=0$ and $x=4$, the top curve is $y=\frac{1}{4}x$ and the bottom curve is $y=\frac{1}{36}x^2$. 6. **Write the integral:** $$\int_0^4 \left(\frac{1}{4}x - \frac{1}{36}x^2\right) dx$$ 7. **Calculate the integral:** $$\int_0^4 \frac{1}{4}x \, dx - \int_0^4 \frac{1}{36}x^2 \, dx = \left[\frac{1}{8}x^2\right]_0^4 - \left[\frac{1}{108}x^3\right]_0^4$$ $$= \frac{1}{8} \times 16 - \frac{1}{108} \times 64 = 2 - \frac{64}{108} = 2 - \frac{16}{27} = \frac{54}{27} - \frac{16}{27} = \frac{38}{27}$$ 8. **Final answer:** The area of the shaded region is $\boxed{\frac{38}{27}}$.