To evaluate the integral:

$\int 2x^2 \sqrt{4 - x^3} \, dx$

We'll start by applying a substitution to simplify the expression inside the square root.

Step 1: Substitution

Let:

$u = 4 - x^3$

Then, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = -3x^2 \quad \text{or} \quad du = -3x^2 \, dx$

Now, we need to match the differential in the integral. Since we have $2x^2 \, dx$, divide both sides of $du = -3x^2 \, dx$ by $-3$:

$x^2 \, dx = -\frac{1}{3} \, du$

So:

$2x^2 \, dx = -\frac{2}{3} \, du$

Step 2: Substituting into the Integral

Substitute $u = 4 - x^3$ and $2x^2 \, dx = -\frac{2}{3} \, du$ into the integral:

$\int 2x^2 \sqrt{4 - x^3} \, dx = -\frac{2}{3} \int \sqrt{u} \, du$

Step 3: Solve the Integral

The integral $\int \sqrt{u} \, du$ is straightforward:

$\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

Thus:

$-\frac{2}{3} \int \sqrt{u} \, du = -\frac{2}{3} \times \frac{2}{3} u^{3/2} = -\frac{4}{9} u^{3/2}$

Step 4: Substitute Back $u = 4 - x^3$

Now substitute back $u = 4 - x^3$:

$-\frac{4}{9} (4 - x^3)^{3/2}$

Final Answer:

The evaluated integral is:

$-\frac{4}{9} (4 - x^3)^{3/2} + C$

where $C$ is the constant of integration.

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Do you have any questions or want further details on any step?

Here are 5 related questions to expand your understanding:

1. How would you solve the same integral with different limits?
2. What is the importance of substitution in solving integrals like this?
3. Can you apply integration by parts to this problem?
4. What happens if the power inside the square root changes?
5. How do we know that $u$-substitution is appropriate for this problem?

Tip: When applying substitution, always check that the differential matches up, so you can correctly transform the integral.