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The integral shown is:

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

### Solution:

To solve this integral, we can use substitution.

Let:

\[
u = 2 - x
\]

Then:

\[
\frac{du}{dx} = -1 \quad \text{or} \quad dx = -du
\]

Also, the limits of integration change as follows:
- When \( x = 0 \), \( u = 2 \).
- When \( x = 1 \), \( u = 1 \).

The integral becomes:

\[
\int_{2}^{1} \frac{(2 - u)^2}{\sqrt{u}} \cdot (-du)
\]

The negative sign allows us to reverse the limits of integration:

\[
\int_{1}^{2} \frac{(2 - u)^2}{\sqrt{u}} \, du
\]

Now expand \( (2 - u)^2 \):

\[
(2 - u)^2 = 4 - 4u + u^2
\]

So the integral is:

\[
\int_{1}^{2} \frac{4 - 4u + u^2}{\sqrt{u}} \, du
\]

This can be broken down into three separate integrals:

\[
\int_{1}^{2} \frac{4}{\sqrt{u}} \, du - \int_{1}^{2} \frac{4u}{\sqrt{u}} \, du + \int_{1}^{2} \frac{u^2}{\sqrt{u}} \, du
\]

Simplifying each term:

1. \( \int_{1}^{2} \frac{4}{\sqrt{u}} \, du = 4 \int_{1}^{2} u^{-\frac{1}{2}} \, du \)
2. \( \int_{1}^{2} \frac{4u}{\sqrt{u}} \, du = 4 \int_{1}^{2} u^{\frac{1}{2}} \, du \)
3. \( \int_{1}^{2} \frac{u^2}{\sqrt{u}} \, du = \int_{1}^{2} u^{\frac{3}{2}} \, du \)

Each of these integrals can be solved individually:

1. \( 4 \int u^{-\frac{1}{2}} \, du = 4 \left[ 2u^{\frac{1}{2}} \right]_{1}^{2} = 4 \left( 2\sqrt{2} - 2 \right) = 8\sqrt{2} - 8 \)
2. \( 4 \int u^{\frac{1}{2}} \, du = 4 \left[ \frac{2}{3}u^{\frac{3}{2}} \right]_{1}^{2} = \frac{8}{3} \left( 2\sqrt{2} - 1 \right) = \frac{16\sqrt{2}}{3} - \frac{8}{3} \)
3. \( \int u^{\frac{3}{2}} \, du = \left[ \frac{2}{5}u^{\frac{5}{2}} \right]_{1}^{2} = \frac{2}{5} \left( 4\sqrt{2} - 1 \right) = \frac{8\sqrt{2}}{5} - \frac{2}{5} \)

Adding these results together:

\[
8\sqrt{2} - 8 - \left( \frac{16\sqrt{2}}{3} - \frac{8}{3} \right) + \left( \frac{8\sqrt{2}}{5} - \frac{2}{5} \right)
\]

To combine these terms, we need a common denominator of 15:

\[
\left( \frac{120\sqrt{2}}{15} - \frac{120}{15} \right) - \left( \frac{80\sqrt{2}}{15} - \frac{40}{15} \right) + \left( \frac{24\sqrt{2}}{15} - \frac{6}{15} \right)
\]

This simplifies to:

\[
\frac{120\sqrt{2} - 80\sqrt{2} + 24\sqrt{2}}{15} - \frac{120 - 40 + 6}{15} = \frac{64\sqrt{2}}{15} - \frac{86}{15}
\]

Thus, the final answer is:

\[
\boxed{\frac{64\sqrt{2} - 86}{15}}
\]

Would you like more details on any part of this solution? Here are five related questions:

1. How do you choose the substitution when solving integrals?
2. What is the geometric interpretation of the integral?
3. How would the result change if the limits of integration were different?
4. What are some other techniques for solving integrals involving square roots?
5. How does integration by parts compare with substitution for similar problems?

**Tip:** Always check the boundaries and limits when changing variables in an integral.

Learn how to solve the integral ∫(x^2 / √(2 - x)) dx from 0 to 1 using substitution and integration techniques. This problem involves advanced calculus concepts and is suitable for college-level students and enthusiasts of integral calculus.

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