What is the total change of f(x), if f'(x) = 2x - 3x^2, over the interval [0, 3]?
if f ' (x) = 2x- 3x^2
f(x) = x^2 - x^3 + c
f(0) = 0-0+c
f(3) = 9 + 27 + c = 36+c
change from f(0) to f(3)
= 36+c - c
= 36
My only options are:
-12
6
-18
-6
Oops. Flipped a minus sign
f(3) = 9-27+c = -18+c
total change is thus -18
To find the total change of f(x) over the interval [0, 3], we can use the Fundamental Theorem of Calculus. The fundamental theorem states that if F(x) is an antiderivative of f(x), then the total change of f(x) over an interval [a, b] is equal to F(b) - F(a).
First, let's find the antiderivative of f'(x) = 2x - 3x^2. To do this, we need to find a function F(x) whose derivative is equal to f'(x).
The antiderivative of 2x is x^2, and the antiderivative of -3x^2 is -x^3. Therefore, the antiderivative of f'(x) = 2x - 3x^2 is F(x) = x^2 - x^3 + C, where C is the constant of integration.
Now, we can find the total change of f(x) over the interval [0, 3]. Plug in the upper and lower bounds into the antiderivative F(x):
F(3) - F(0) = (3^2 - 3^3) - (0^2 - 0^3)
= (9 - 27) - (0 - 0)
= -18
Therefore, the total change of f(x) over the interval [0, 3] is -18.