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

To find the area enclosed by the curves \( y = x^4 \) and \( y^2 = 8x \), we first need to determine the points of intersection between these curves.

1. **Find the points of intersection:**

   Set \( y = x^4 \) equal to \( y = \sqrt{8x} \) (from \( y^2 = 8x \)):

   \[
   x^4 = \sqrt{8x}
   \]

   Squaring both sides to eliminate the square root:

   \[
   (x^4)^2 = (\sqrt{8x})^2
   \]

   \[
   x^8 = 8x
   \]

   This simplifies to:

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

   Factor out \( x \):

   \[
   x(x^7 - 8) = 0
   \]

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

   For \( x^7 = 8 \):

   \[
   x = \sqrt[7]{8} = 2^{2/7}
   \]

   Therefore, the points of intersection are \( x = 0 \) and \( x = 2^{2/7} \).

2. **Set up the integral to find the area:**

   The area \( A \) enclosed by the curves is given by:

   \[
   A = \int_{x_1}^{x_2} \left( \sqrt{8x} - x^4 \right) dx
   \]

   Where \( x_1 = 0 \) and \( x_2 = 2^{2/7} \).

3. **Evaluate the integral:**

   Let's compute the integral:

   \[
   A = \int_{0}^{2^{2/7}} \left( \sqrt{8x} - x^4 \right) dx
   \]

   To compute this, we split it into two integrals:

   \[
   A = \int_{0}^{2^{2/7}} \sqrt{8x} \, dx - \int_{0}^{2^{2/7}} x^4 \, dx
   \]

   Compute each integral:

   - The first integral:

     \[
     \int \sqrt{8x} \, dx = \int \sqrt{8} \cdot \sqrt{x} \, dx = \frac{2\sqrt{8}}{3} x^{3/2}
     \]

     Evaluating from 0 to \( 2^{2/7} \):

     \[
     \frac{2\sqrt{8}}{3} \left( 2^{2/7} \right)^{3/2}
     \]

   - The second integral:

     \[
     \int x^4 \, dx = \frac{x^5}{5}
     \]

     Evaluating from 0 to \( 2^{2/7} \):

     \[
     \frac{(2^{2/7})^5}{5}
     \]

   Finally, subtract the two results to find the area \( A \):

   \[
   A = \frac{2\sqrt{8}}{3} \left( 2^{2/7} \right)^{3/2} - \frac{(2^{2/7})^5}{5}
   \]

4. **Numerical evaluation:**

   Now, we'll calculate the numerical value of \( A \):

   \[
   A \approx \frac{2\sqrt{8}}{3} \cdot 2^{3/7} - \frac{2^{10/7}}{5}
   \]

   Let's calculate this numerically:

   \[
   A \approx \frac{2\cdot2\sqrt{2}}{3} \cdot 2^{3/7} - \frac{2^{10/7}}{5}
   \]

   \[
   A \approx \frac{4\sqrt{2}}{3} \cdot 2^{3/7} - \frac{2^{10/7}}{5}
   \]

   Calculate each part:

   - \( \frac{4\sqrt{2}}{3} \cdot 2^{3/7} \approx 2.759 \)
   - \( \frac{2^{10/7}}{5} \approx 0.946 \)

   So,

   \[
   A \approx 2.759 - 0.946 = 1.813
   \]

Thus, the area enclosed by the curves is approximately \( \boxed{1.813} \) square units.

Would you like more details or have any questions? 

Here are 5 related questions:
1. How would the area change if the curves were modified?
2. Can we find the area enclosed between \( y = x^4 \) and another curve?
3. What happens to the area if we consider \( y = -x^4 \) and \( y^2 = 8x \)?
4. How do we interpret the intersections graphically?
5. Could this method be extended to polar coordinates?

**Tip:** When solving integrals, always double-check the bounds of integration, as they determine the region you are analyzing.

Learn how to find the area enclosed by the curves y=x^4 and y^2=8x using calculus. Step-by-step solution includes finding intersection points, setting up and evaluating the integral, and calculating the numerical answer. Suitable for advanced high school and undergraduate students.

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