Integral of 6*exp(-2x-3y) dy

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

   $$\int 6 e^{- 2 x - 3 y}\, dy = 6 \int e^{- 2 x - 3 y}\, dy$$

   1. Let $u = - 2 x - 3 y$.

      Then let $du = - 3 dy$ and substitute $- \frac{du}{3}$:

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

      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: $- \frac{e^{u}}{3}$

      Now substitute $u$ back in:

      $$- \frac{e^{- 2 x - 3 y}}{3}$$

   So, the result is: $- 2 e^{- 2 x - 3 y}$

2. Now simplify:

   $$- 2 e^{- 2 x - 3 y}$$

3. Add the constant of integration:

   $$- 2 e^{- 2 x - 3 y}+ \mathrm{constant}$$

The answer is:

$$- 2 e^{- 2 x - 3 y}+ \mathrm{constant}$$