What is the value of the definite integral ∫ from 1 to 2 of (3x^2) dx?

To find the value of the definite integral ∫ from 1 to 2 of (3x^2) dx, we start by calculating the indefinite integral of the function 3x². The integral of 3x² with respect to x is:

∫ 3x² dx = (3/3)x^3 + C = x^3 + C

Now we apply the Fundamental Theorem of Calculus to evaluate the definite integral from the limits 1 to 2:

∫ from 1 to 2 of (3x^2) dx = [x^3] from 1 to 2

= (2^3) - (1^3)

= 8 - 1

= 7

Thus, the value of the definite integral is 7. The computed result shows how the definite integral is a measure of the net area under the curve from x = 1 to x = 2. Given this procedure and outcome, the correct answer is indeed 7, consistent with the calculations.