Find the area between the curves Y=x³ Y= x^1/3

To find the area between the curves \(y = x^{1/3}\) and \(y = x^3\):

1. Find the points of intersection:

x^{3} = x^{1/3}
x^{9} = x
x(x^8 - 1) = 0
x(x - 1)(x + 1)(x^{6} + x^4 + x^2 + 1) = 0
x= 0, +1, -1

Intersections are at \( x = -1, 0, 1 \).

2. Compute the integral:

A = \int_{-1}^{1} (x^{1/3} - x^3) \, dx

Using symmetry:

A = 2 \int_{0}^{1} (x^{1/3} - x^3) \, dx

3. Split into two integrals:

A = 2 \left( \int_{0}^{1} x^{1/3} \, dx - \int_{0}^{1} x^3 \, dx \right)

4. Solve the integrals:

\int_{0}^{1} x^{1/3} \, dx = \left[ \frac{3}{4}x^{4/3} \right]_{0}^{1} = \frac{3}{4}
\int_{0}^{1} x^3 \, dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{4}

5. Compute the difference:

2 \left( \frac{3}{4} - \frac{1}{4} \right) = 2 \times \frac{1}{2}= 1

6. Area \( A = 1 \)

Final answer:
A = 1