How to integrate 1/2x-3 dx has bound for 4 and 2

Answer 1

To integrate the function 1/2x - 3 with bounds from 4 to 2, you would first integrate the function with respect to x, and then evaluate the integral at the upper and lower bounds.

The antiderivative of 1/2x - 3 with respect to x is (1/2)x^2/2 - 3x = 1/4x^2 - 3x.

To find the definite integral from 4 to 2, you would evaluate the antiderivative at the upper bound (4) and subtract the value of the antiderivative at the lower bound (2).

[1/4(4)^2 - 3(4)] - [1/4(2)^2 - 3(2)]
= [1/4(16) - 12] - [1/4(4) - 6]
= [4 - 12] - [1 - 6]
= -8 - (-5)
= -3

Therefore, the definite integral of 1/2x - 3 from 4 to 2 is -3.