Evaluate the following definite integral:𝐼=∫2−1(3𝑥2−𝑥+6)d𝑥

Sure, here are the steps to evaluate the definite integral:

Step 1: Find the antiderivative (also known as the indefinite integral) of the function.

The antiderivative of 3x^2 is x^3, of -x is -0.5x^2, and of 6 is 6x. So, the antiderivative of the function 3x^2 - x + 6 is x^3 - 0.5x^2 + 6x.

Step 2: Evaluate the antiderivative at the upper limit and the lower limit.

Substitute x = 2 into the antiderivative, we get 2^3 - 0.5*(2^2) + 6*2 = 8 - 2 + 12 = 18.

Substitute x = -1 into the antiderivative, we get (-1)^3 - 0.5*(-1)^2 + 6*(-1) = -1 - 0.5 - 6 = -7.5.

Step 3: Subtract the value of the antiderivative at the lower limit from the value at the upper limit.

So, the value of the definite integral is 18 - (-7.5) = 25.5.