Integral of yexp(-2y) 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 $\operatorname{dv}{\left(y \right)} = e^{- 2 y}$.

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

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

   1. Let $u = - 2 y$.

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

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

      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: $- \frac{e^{u}}{2}$

      Now substitute $u$ back in:

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

   Now evaluate the sub-integral.

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

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

   1. Let $u = - 2 y$.

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

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

      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: $- \frac{e^{u}}{2}$

      Now substitute $u$ back in:

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

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

3. Now simplify:

   $$- \frac{\left(2 y + 1\right) e^{- 2 y}}{4}$$

4. Add the constant of integration:

   $$- \frac{\left(2 y + 1\right) e^{- 2 y}}{4}+ \mathrm{constant}$$

The answer is:

$$- \frac{\left(2 y + 1\right) e^{- 2 y}}{4}+ \mathrm{constant}$$