Find the area of the region enclosed by the three graphs y = x^2, y = 8 x, and y = 1

To find the area of the region enclosed by the three graphs \(y = x^2\), \(y = 8x\), and \(y = 1\), we first need to determine the points of intersection of these curves, as these points will define the boundaries of the enclosed region.

### Step 1: Find the Points of Intersection

1. **Intersection of \(y = x^2\) and \(y = 8x\):**

   Set \(x^2 = 8x\):

   \[
   x^2 - 8x = 0
   \]

   Factor the equation:

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

   So, the solutions are:

   \[
   x = 0 \quad \text{and} \quad x = 8
   \]

2. **Intersection of \(y = x^2\) and \(y = 1\):**

   Set \(x^2 = 1\):

   \[
   x = \pm 1
   \]

3. **Intersection of \(y = 8x\) and \(y = 1\):**

   Set \(8x = 1\):

   \[
   x = \frac{1}{8}
   \]

Now, we have the points of intersection:
- \(x = 0\) and \(x = 8\) from \(y = x^2\) and \(y = 8x\),
- \(x = 1\) and \(x = -1\) from \(y = x^2\) and \(y = 1\),
- \(x = \frac{1}{8}\) from \(y = 8x\) and \(y = 1\).

### Step 2: Determine the Intervals and Functions for the Enclosed Region

The area of interest is between \(x = 0\) and \(x = 1\), as this is where all three graphs intersect and define the enclosed region:

- From \(x = 0\) to \(x = \frac{1}{8}\), the region is enclosed between \(y = 1\) (above) and \(y = 8x\) (below).
- From \(x = \frac{1}{8}\) to \(x = 1\), the region is enclosed between \(y = 1\) (above) and \(y = x^2\) (below).

### Step 3: Set Up the Integral for the Area

1. **Area from \(x = 0\) to \(x = \frac{1}{8}\):**

\[
A_1 = \int_{0}^{\frac{1}{8}} (1 - 8x) \, dx
\]

2. **Area from \(x = \frac{1}{8}\) to \(x = 1\):**

\[
A_2 = \int_{\frac{1}{8}}^{1} (1 - x^2) \, dx
\]

### Step 4: Compute the Integrals

Compute \(A_1\):

\[
A_1 = \int_{0}^{\frac{1}{8}} (1 - 8x) \, dx = \left[ x - 4x^2 \right]_{0}^{\frac{1}{8}}
\]

Evaluate:

\[
A_1 = \left( \frac{1}{8} - 4\left(\frac{1}{8}\right)^2 \right) - (0 - 0)
\]

\[
A_1 = \frac{1}{8} - 4 \cdot \frac{1}{64} = \frac{1}{8} - \frac{1}{16} = \frac{2 - 1}{16} = \frac{1}{16}
\]

Compute \(A_2\):

\[
A_2 = \int_{\frac{1}{8}}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{\frac{1}{8}}^{1}
\]

Evaluate:

\[
A_2 = \left(1 - \frac{1^3}{3}\right) - \left(\frac{1}{8} - \frac{\left(\frac{1}{8}\right)^3}{3}\right)
\]

\[
A_2 = \left(1 - \frac{1}{3}\right) - \left(\frac{1}{8} - \frac{1}{1536}\right)
\]

Simplify each part:

\[
A_2 = \frac{2}{3} - \left(\frac{192}{1536} - \frac{1}{1536}\right)
\]

\[
A_2 = \frac{2}{3} - \frac{191}{1536}
\]

Convert \(\frac{2}{3}\) to a fraction with denominator 1536:

\[
\frac{2}{3} = \frac{1024}{1536}
\]

So,

\[
A_2 = \frac{1024 - 191}{1536} = \frac{833}{1536}
\]

### Step 5: Add the Areas

Total area \(A\) is:

\[
A = A_1 + A_2 = \frac{1}{16} + \frac{833}{1536}
\]

Convert \(\frac{1}{16}\) to a fraction with denominator 1536:

\[
\frac{1}{16} = \frac{96}{1536}
\]

So,

\[
A = \frac{96}{1536} + \frac{833}{1536} = \frac{929}{1536}
\]

### Conclusion

The area of the region enclosed by the graphs \(y = x^2\), \(y = 8x\), and \(y = 1\) is \(\frac{929}{1536}\) square units.

Would you like any more details or have any questions?

Here are some related questions you might find interesting:
1. How do you find the area between two curves?
2. What are the conditions for two curves to enclose a region?
3. How can you use symmetry to simplify area calculations between curves?
4. What methods can be used to find the volume of a solid of revolution?
5. How does changing the order of integration affect the computed area?

**Tip:** When dealing with areas between curves, sketching the region first can help identify the correct bounds and functions to use.

Learn how to find the area of the region enclosed by the graphs y = x^2, y = 8x, and y = 1 using calculus. This problem involves finding intersections of curves, setting up integrals, and calculating areas between curves. Suitable for advanced high school students or anyone studying integral calculus.

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