determine the area o the curve (x) x^2+2x-8 between x=0 and x=3
To find the area of the curve between x = 0 and x = 3, we need to evaluate the definite integral of the curve.
The integral of the curve f(x) = x^2 + 2x - 8 with respect to x is given by:
∫[0,3] (x^2 + 2x - 8) dx
To evaluate this integral, we can use the power rule for integration.
First, we integrate each term of the function separately:
∫(x^2) dx = (1/3) * x^3 + C
∫(2x) dx = x^2 + C
∫(-8) dx = -8x + C
Adding these integrals together, we get:
∫[0,3] (x^2 + 2x - 8) dx = (1/3) * x^3 + x^2 - 8x + C
To find the definite integral between x = 0 and x = 3, we subtract the value of the integral at the lower limit from the value of the integral at the upper limit:
∫[0,3] (x^2 + 2x - 8) dx = [(1/3) * 3^3 + 3^2 - 8 * 3] - [(1/3) * 0^3 + 0^2 - 8 * 0]
Simplifying this expression, we get:
∫[0,3] (x^2 + 2x - 8) dx = [(1/3) * 27 + 9 - 24] - (0)
= [9 + 9 - 24] - (0)
= -6
Therefore, the area of the curve between x = 0 and x = 3 is -6 units squared.