To solve the integral:

$\int_0^1 x(1 - x)^2 \, dx$

we can follow these steps:

Step 1: Expand the Integrand

Rewrite $(1 - x)^2$ by expanding it:

$(1 - x)^2 = 1 - 2x + x^2$

Thus, the integrand $x(1 - x)^2$ becomes:

$x(1 - x)^2 = x(1 - 2x + x^2) = x - 2x^2 + x^3$

Step 2: Set Up the Integral

Now we rewrite the integral as:

$\int_0^1 x(1 - x)^2 \, dx = \int_0^1 (x - 2x^2 + x^3) \, dx$

Step 3: Integrate Term by Term

Now, integrate each term separately:

1. $\int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2}$
2. $\int_0^1 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3}$
3. $\int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}$

Step 4: Combine the Results

Now we put it all together:

$\int_0^1 (x - 2x^2 + x^3) \, dx = \frac{1}{2} - \frac{2}{3} + \frac{1}{4}$

Step 5: Simplify

To combine these fractions, we need a common denominator, which is 12:

$\frac{1}{2} = \frac{6}{12}, \quad \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12}$

Thus,

$\frac{6}{12} - \frac{8}{12} + \frac{3}{12} = \frac{6 - 8 + 3}{12} = \frac{1}{12}$

Final Answer

$\int_0^1 x(1 - x)^2 \, dx = \frac{1}{12}$

Would you like further details, or do you have any questions?

Here are some related questions to expand your understanding:

1. How would the result change if the limits of integration were different?
2. What if the integrand were $x(1 - x)^3$ instead of $x(1 - x)^2$?
3. How does changing $x$ to $x^2$ in the integrand affect the solution?
4. What is the geometric interpretation of this integral?
5. How can integration by parts be used here, and would it simplify the calculation?

Tip: Expanding polynomials in integrals can simplify complex expressions, making integration straightforward by reducing it to basic polynomial integration.