How do you evaluate: #int_{-2}^3 x^2 - 3 dx#?

Question

Isaiah Allen · Accepted Answer

#-10/3# STEP 1: Take the antiderivative of the function =#[1/3x^3 - 3x]_-2^3# STEP 2: Evaluate between the founds = #[(1/3) (3^3) - 3(3)] - [(1/3) (-2)^3 - 3(-2)]# STEP 3: Compute your final answer = #[9 - 9] - [-8/3 + 6]# = #8/3 - 18/3# = #-10/3#

Ezra Adams · Answer

undefined To evaluate the integral ∫_{-2}^3 (x^2 - 3) dx, follow these steps:

1. Integrate the function (x^2 - 3) with respect to x.
   ∫ (x^2 - 3) dx = (1/3)x^3 - 3x.

2. Evaluate the definite integral by substituting the upper limit (3) and the lower limit (-2) into the antiderivative and subtracting the result of the lower limit from the upper limit.
   ∫_{-2}^3 (x^2 - 3) dx = [(1/3)(3)^3 - 3(3)] - [(1/3)(-2)^3 - 3(-2)].

3. Calculate the values.
   = [(1/3)(27) - 9] - [(1/3)(-8) + 6]
   = [(9 - 9) - (-8/3 + 6)]
   = [0 - (-8/3 + 6)]
   = 8/3 - 6.

4. Simplify the result.
   = 8/3 - 18/3
   = -10/3.

Therefore, the value of the integral ∫_{-2}^3 (x^2 - 3) dx is -10/3.