The integral appears to be a triple integral of the form:

$\int_0^2 \int_0^3 \int_0^{2x+y} e^{x+y+z} \, dz \, dy \, dx$

Let us evaluate this step by step:

Step 1: Evaluate the inner integral with respect to $z$:

$\int_0^{2x+y} e^{x+y+z} \, dz$

The exponential function's argument $e^{x+y+z}$ is independent of $z$ in its partial derivative. So:

$\int e^{x+y+z} \, dz = \frac{1}{1} e^{x+y+z} + C$

Apply limits $z = 0$ to $z = 2x+y$:

$\left[ e^{x+y+z} \right]_0^{2x+y} = e^{x+y+(2x+y)} - e^{x+y+0}$

Simplify:

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

Step 2: Evaluate the middle integral with respect to $y$:

$\int_0^3 \left( e^{3x+2y} - e^{x+y} \right) \, dy$

Split the integral:

$\int_0^3 e^{3x+2y} \, dy - \int_0^3 e^{x+y} \, dy$

First term:

$\int_0^3 e^{3x+2y} \, dy = e^{3x} \int_0^3 e^{2y} \, dy = e^{3x} \cdot \frac{1}{2} e^{2y} \Big|_0^3$

$= e^{3x} \cdot \frac{1}{2} \left( e^{6} - 1 \right)$

Second term:

$\int_0^3 e^{x+y} \, dy = e^x \int_0^3 e^y \, dy = e^x \cdot e^y \Big|_0^3$

$= e^x \cdot \left( e^3 - 1 \right)$

Thus:

$\int_0^3 \left( e^{3x+2y} - e^{x+y} \right) \, dy = \frac{e^{3x}}{2} \left( e^6 - 1 \right) - e^x \left( e^3 - 1 \right)$

Step 3: Evaluate the outer integral with respect to $x$:

$\int_0^2 \left[ \frac{e^{3x}}{2} \left( e^6 - 1 \right) - e^x \left( e^3 - 1 \right) \right] \, dx$

Split into two integrals:

$\int_0^2 \frac{e^{3x}}{2} \left( e^6 - 1 \right) \, dx - \int_0^2 e^x \left( e^3 - 1 \right) \, dx$

First term:

$\int_0^2 \frac{e^{3x}}{2} \left( e^6 - 1 \right) \, dx = \frac{\left( e^6 - 1 \right)}{2} \int_0^2 e^{3x} \, dx$

$= \frac{\left( e^6 - 1 \right)}{2} \cdot \frac{1}{3} e^{3x} \Big|_0^2$

$= \frac{\left( e^6 - 1 \right)}{6} \left( e^6 - 1 \right)$

Second term:

$\int_0^2 e^x \left( e^3 - 1 \right) \, dx = \left( e^3 - 1 \right) \int_0^2 e^x \, dx$

$= \left( e^3 - 1 \right) \cdot e^x \Big|_0^2$

$= \left( e^3 - 1 \right) \left( e^2 - 1 \right)$

Final Result:

Combine both terms:

$\int_0^2 \int_0^3 \int_0^{2x+y} e^{x+y+z} \, dz \, dy \, dx = \frac{\left( e^6 - 1 \right)^2}{6} - \left( e^3 - 1 \right) \left( e^2 - 1 \right)$

Would you like a detailed simplification of this result?