To evaluate the integral ∫(3x^2 + 2x 5) dx from x = 0 to x = 2, we'll follow these steps:

Step 1: Identify the integral to be solved
The given integral is ∫(3x^2 + 2x 5) dx.

2: Integrate the function term by term
To integrate the function 3x^2 + 2x 5, we'll integrate each term separately:
The integral of 3x^2 is x^3.
The integral of 2x is x^2.
The integral of 5 is 5x.

3: Write the indefinite integral
Combining the results from Step 2, the indefinite integral is:
∫(3x^2 + 2x 5) dx = x^3 + x^2 5x + C.

4: Apply the Fundamental Theorem of Calculus
To find the definite integral from x = 0 to x = 2, we evaluate the antiderivative at the bounds and subtract:
[x^3 + x^2 5x] from 0 to 2 = (2^3 + 2^2 5*2) (0^3 + 0^2 5*0).

5: Perform the calculations
Calculate the value at x = 2:
(2^3 + 2^2 5*2) = 8 + 4 10 = 2.

6: Calculate the value at the lower bound
At x = 0, the value is 0.

7: Subtract the values to find the definite integral
Subtract the value at the lower bound from the value at the upper bound:
2 0 = 2.

The final answer is: $\boxed{2}$