Find the area enclosed by the curve y=x⁴ and y²=8x give numerical answer 

To find the area enclosed by the curves \( y = x^4 \) and \( y^2 = 8x \), we'll follow these steps:

### 1. Find the Points of Intersection
First, let's find the points where the curves intersect by solving the equations simultaneously.

Given:
\[
y = x^4
\]
\[
y^2 = 8x
\]

Substitute \( y = x^4 \) into \( y^2 = 8x \):
\[
(x^4)^2 = 8x
\]
\[
x^8 = 8x
\]

We can factor this equation:
\[
x(x^7 - 8) = 0
\]

So, \( x = 0 \) or \( x^7 = 8 \).

For \( x = 0 \), \( y = 0^4 = 0 \).

For \( x^7 = 8 \):
\[
x = \sqrt[7]{8} = 2^{2/7}.
\]
So the intersection points are \( (0, 0) \) and \( \left(2^{2/7}, \left(2^{2/7}\right)^4\right) \).

### 2. Set Up the Integral for the Area
The area between the curves is given by:
\[
\text{Area} = \int_{a}^{b} \left(f(x) - g(x)\right) dx,
\]
where \( f(x) \) is the upper function and \( g(x) \) is the lower function.

Here:
- \( f(x) = \sqrt{8x} \) (from \( y^2 = 8x \))
- \( g(x) = x^4 \)

The integral is then:
\[
\text{Area} = \int_{0}^{2^{2/7}} \left(\sqrt{8x} - x^4\right) dx
\]

### 3. Evaluate the Integral
Now, let's compute the integral.

1. **Integral of \( \sqrt{8x} \)**:
   \[
   \int \sqrt{8x} \, dx = \int \sqrt{8} \cdot \sqrt{x} \, dx = \frac{2\sqrt{8}}{3}x^{3/2}
   \]
   Evaluated from \( 0 \) to \( 2^{2/7} \):
   \[
   \frac{2\sqrt{8}}{3}\left(2^{2/7}\right)^{3/2}
   \]

2. **Integral of \( x^4 \)**:
   \[
   \int x^4 \, dx = \frac{x^5}{5}
   \]
   Evaluated from \( 0 \) to \( 2^{2/7} \):
   \[
   \frac{(2^{2/7})^5}{5}
   \]

Finally, the area is:
\[
\text{Area} = \frac{2\sqrt{8}}{3}\left(2^{2/7}\right)^{3/2} - \frac{(2^{2/7})^5}{5}
\]

### 4. Compute the Numerical Value
Let's calculate the numerical value of the area.

\[
\text{Area} = \frac{2\sqrt{8}}{3} \times 2^{3/7} - \frac{2^{10/7}}{5}
\]

Simplifying and calculating:

1. \( \frac{2\sqrt{8}}{3} \times 2^{3/7} = \frac{4\sqrt{2}}{3} \times 2^{3/7} \approx 1.885 \)
2. \( \frac{2^{10/7}}{5} \approx 0.916 \)

Thus:
\[
\text{Area} \approx 1.885 - 0.916 \approx 0.969 \text{ square units.}
\]

The area enclosed by the curves \( y = x^4 \) and \( y^2 = 8x \) is approximately \( 0.969 \) square units.

Would you like to explore more about this problem, or have any specific questions?

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### Relative Questions:
1. What are the general steps to find the area between two curves?
2. How does the order of the functions in the integral affect the sign of the area?
3. What is the significance of the points of intersection in setting up integrals?
4. How do you handle curves that intersect at more than two points?
5. What are the benefits of using numerical methods for evaluating such integrals?

### Tip:
When calculating the area between curves, ensure the upper and lower bounds are correctly identified to avoid errors in the integration process.

Learn how to find the area enclosed by the curves y=x^4 and y^2=8x with a detailed step-by-step solution. This problem involves calculus concepts such as integration and the fundamental theorem of calculus. Suitable for advanced college-level students.

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