Integral of x^2(1-x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - x$.

      Then let $du = - dx$ and substitute $du$:

      $$\int \left(- u^{3} - u^{2}\right)\, du$$

      1. Integrate term-by-term:

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u^{3}\right)\, du = - \int u^{3}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

               $$\int u^{3}\, du = \frac{u^{4}}{4}$$

            So, the result is: $- \frac{u^{4}}{4}$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

            1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

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

            So, the result is: $- \frac{u^{3}}{3}$

         The result is: $- \frac{u^{4}}{4} - \frac{u^{3}}{3}$

      Now substitute $u$ back in:

      $$- \frac{x^{4}}{4} + \frac{x^{3}}{3}$$

   Method #2

   1. Rewrite the integrand:

      $$x^{2} \left(1 - x\right) = - x^{3} + x^{2}$$

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- x^{3}\right)\, dx = - \int x^{3}\, dx$$

         1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

         So, the result is: $- \frac{x^{4}}{4}$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

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

      The result is: $- \frac{x^{4}}{4} + \frac{x^{3}}{3}$

2. Now simplify:

   $$\frac{x^{3} \left(4 - 3 x\right)}{12}$$

3. Add the constant of integration:

   $$\frac{x^{3} \left(4 - 3 x\right)}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{3} \left(4 - 3 x\right)}{12}+ \mathrm{constant}$$