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To determine the area of the figure bounded by the lines given:

1. \( y = \sqrt[3]{x^2} \)
2. \( y = 0 \)
3. \( x = 1 \)

we need to set up the definite integral of \( y = \sqrt[3]{x^2} \) from \( x = 0 \) to \( x = 1 \).

The integral to calculate the area is:

\[ \int_{0}^{1} \sqrt[3]{x^2} \, dx \]

To solve this, let's use the substitution method:

Let \( u = x^2 \), then \( du = 2x \, dx \) or \( dx = \frac{du}{2x} \).

Thus, the integral becomes:

\[ \int_{0}^{1} (u^{1/3}) \frac{du}{2\sqrt{u}} = \frac{1}{2} \int_{0}^{1} u^{1/3} u^{-1/2} \, du = \frac{1}{2} \int_{0}^{1} u^{-1/6} \, du \]

This can be integrated using the power rule:

\[ \frac{1}{2} \int_{0}^{1} u^{-1/6} \, du = \frac{1}{2} \left[ \frac{u^{5/6}}{5/6} \right]_{0}^{1} = \frac{1}{2} \left( \frac{6}{5} \left[ u^{5/6} \right]_{0}^{1} \right) = \frac{1}{2} \cdot \frac{6}{5} \left( 1^{5/6} - 0^{5/6} \right) = \frac{1}{2} \cdot \frac{6}{5} = \frac{3}{5} = 0.6 \]

Therefore, the area of the figure is 0.6.

Would you like any further details or explanations?

Here are some questions you might want to ask:
1. How do you find the area under a curve?
2. What is the power rule for integration?
3. How does substitution work in integration?
4. What are definite integrals?
5. Can you show another example of finding an area using integrals?

**Tip:** When integrating, always double-check your substitutions and limits to ensure accuracy.

Learn how to calculate the area bounded by the curves y = cuberoot(x^2) and y = 0 using definite integrals. This problem involves integral calculus, specifically evaluating the integral of cuberoot(x^2) from x = 0 to x = 1. Suitable for advanced high school students studying calculus.

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