Integral of -y*exp(-y/2)/2 dy

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

   $$\int \frac{- y e^{\frac{\left(-1\right) y}{2}}}{2}\, dy = \frac{\int \left(- y e^{\frac{\left(-1\right) y}{2}}\right)\, dy}{2}$$

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

      $$\int \left(- y e^{\frac{\left(-1\right) y}{2}}\right)\, dy = - \int y e^{\frac{\left(-1\right) y}{2}}\, dy$$

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(y \right)} = y$ and let $\operatorname{dv}{\left(y \right)} = e^{- \frac{y}{2}}$.

         Then $\operatorname{du}{\left(y \right)} = 1$.

         To find $v{\left(y \right)}$:

         1. Let $u = - \frac{y}{2}$.

            Then let $du = - \frac{dy}{2}$ and substitute $- 2 du$:

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

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

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

            Now substitute $u$ back in:

            $$- 2 e^{- \frac{y}{2}}$$

         Now evaluate the sub-integral.

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

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

         1. Let $u = - \frac{y}{2}$.

            Then let $du = - \frac{dy}{2}$ and substitute $- 2 du$:

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

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

               $$\text{False}$$

               1. The integral of the exponential function is itself.

                  $$\int e^{u}\, du = e^{u}$$

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

            Now substitute $u$ back in:

            $$- 2 e^{- \frac{y}{2}}$$

         So, the result is: $4 e^{- \frac{y}{2}}$

      So, the result is: $2 y e^{- \frac{y}{2}} + 4 e^{- \frac{y}{2}}$

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

2. Now simplify:

   $$\left(y + 2\right) e^{- \frac{y}{2}}$$

3. Add the constant of integration:

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

The answer is:

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