To find the area of the region bounded by the curves \( y = x^4 \) and \( y = 8x \), we need to:

1. **Find the points of intersection** of the two curves.
2. **Set up the integral** that represents the area between the curves.
3. **Evaluate the integral**.

### Step 1: Find the Points of Intersection

To find the points where the curves intersect, we set the equations equal to each other:

\[
x^4 = 8x
\]

Rearrange this equation:

\[
x^4 - 8x = 0
\]

Factor out an \( x \):

\[
x(x^3 - 8) = 0
\]

This gives us:

\[
x = 0 \quad \text{or} \quad x^3 = 8
\]

Solve for \( x \) from \( x^3 = 8 \):

\[
x = 2
\]

Thus, the points of intersection are at \( x = 0 \) and \( x = 2 \).

### Step 2: Set Up the Integral

To find the area between the curves from \( x = 0 \) to \( x = 2 \), we take the integral of the difference between the two functions \( y = 8x \) and \( y = x^4 \). Since \( y = 8x \) is above \( y = x^4 \) in this interval, we set up the integral as:

\[
\text{Area} = \int_{0}^{2} (8x - x^4) \, dx
\]

### Step 3: Evaluate the Integral

Now, let's evaluate the integral:

\[
\int_{0}^{2} (8x - x^4) \, dx
\]

Integrate term by term:

\[
\int (8x - x^4) \, dx = \int 8x \, dx - \int x^4 \, dx
\]

\[
= 4x^2 - \frac{x^5}{5} \Bigg|_{0}^{2}
\]

Now, plug in the limits of integration:

\[
= \left(4(2)^2 - \frac{(2)^5}{5}\right) - \left(4(0)^2 - \frac{(0)^5}{5}\right)
\]

Simplify each term:

\[
= \left(4 \cdot 4 - \frac{32}{5}\right) - (0 - 0)
\]

\[
= 16 - \frac{32}{5}
\]

Convert 16 to a fraction with a common denominator:

\[
= \frac{80}{5} - \frac{32}{5}
\]

\[
= \frac{48}{5}
\]

Thus, the area of the region bounded by the curves \( y = x^4 \) and \( y = 8x \) is:

\[
\boxed{\frac{48}{5}}
\]

Would you like more details on any part of the solution, or do you have any questions?

### Additional Related Questions:

1. How do the graphs of \( y = x^4 \) and \( y = 8x \) look on the same coordinate plane?
2. What is the volume of the solid obtained by rotating the area bounded by \( y = x^4 \) and \( y = 8x \) around the x-axis?
3. What is the volume of the solid obtained by rotating the area bounded by \( y = x^4 \) and \( y = 8x \) around the y-axis?
4. If we consider a region bounded by \( y = x^n \) and \( y = 8x \), how does the area change with different \( n \)?
5. How would you find the centroid of the region bounded by these curves?

**Tip:** Always verify which function is on top and which is on the bottom over the interval when setting up the integral for the area between two curves.

Learn how to calculate the area between the curves y = x^4 and y = 8x using integral calculus. Step-by-step solution included. Suitable for advanced high school students and calculus enthusiasts.

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