Integral of -(y*(4-y))+((y-4)*y) dy

1. Integrate term-by-term:

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

      $$\int \left(- y \left(4 - y\right)\right)\, dy = - \int y \left(4 - y\right)\, dy$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = - y$.

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

            $$\int \left(u^{2} + 4 u\right)\, du$$

            1. Integrate term-by-term:

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

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

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

                  $$\int 4 u\, du = 4 \int u\, du$$

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

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

                  So, the result is: $2 u^{2}$

               The result is: $\frac{u^{3}}{3} + 2 u^{2}$

            Now substitute $u$ back in:

            $$- \frac{y^{3}}{3} + 2 y^{2}$$

         Method #2

         1. Rewrite the integrand:

            $$y \left(4 - y\right) = - y^{2} + 4 y$$

         2. Integrate term-by-term:

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

               $$\int \left(- y^{2}\right)\, dy = - \int y^{2}\, dy$$

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

                  $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

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

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

               $$\int 4 y\, dy = 4 \int y\, dy$$

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

                  $$\int y\, dy = \frac{y^{2}}{2}$$

               So, the result is: $2 y^{2}$

            The result is: $- \frac{y^{3}}{3} + 2 y^{2}$

      So, the result is: $\frac{y^{3}}{3} - 2 y^{2}$

   1. Rewrite the integrand:

      $$y \left(y - 4\right) = y^{2} - 4 y$$

   2. Integrate term-by-term:

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

         $$\int y^{2}\, dy = \frac{y^{3}}{3}$$

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

         $$\int \left(- 4 y\right)\, dy = - 4 \int y\, dy$$

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

            $$\int y\, dy = \frac{y^{2}}{2}$$

         So, the result is: $- 2 y^{2}$

      The result is: $\frac{y^{3}}{3} - 2 y^{2}$

   The result is: $\frac{2 y^{3}}{3} - 4 y^{2}$

2. Now simplify:

   $$\frac{2 y^{2} \left(y - 6\right)}{3}$$

3. Add the constant of integration:

   $$\frac{2 y^{2} \left(y - 6\right)}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{2 y^{2} \left(y - 6\right)}{3}+ \mathrm{constant}$$