Waht is the average (mean) value of 2t^3-3t^2+4 over the interval [-1,1]?

A) 0
B) 7/4
C) 3
D) 4
E) 6

I'm trying to do these test corrections and I can't figure out this answer. Please help! I know the answer isn't B, because I got it wrong.

average = 1/(max - min)∫f(x)dx on [max, min]

= 1/(1- (-1))∫(2t^3 - 3t^2 + 4)dt on [1, -1]

= 1/2*(1/2t^4 - t^3 + 4t) on [1, -1]

= 1/2(1/2(1^4) - (1)^3 + 4(1) - 1/2(-1)^4 + (-1)^3 - 4(-1))

= 1/2(-2 + 8)

= 6/2 = 3
ANS: C

1\2 (23t -3t2 +4)

Well, I would take the integral from -1 to 1 of that function first. This gives 6. Then, the total distance from -1 to 1 is 2, so take 6/2 = 3. The answer is C.

F(t) = int(f(t) * dt)

(F(b) - F(a)) / (b - a)

f(t) = 2t^3 - 3t^2 + 4
F(t) = (1/2) * t^4 - t^3 + 4t + C

(F(1) - F(-1)) / (1 - (-1)) =>
((1/2) * 1^4 - 1^3 + 4 * 1 - (1/2) * (-1)^4 + (-1)^3 - 4 * (-1)) / 2 =>
((1/2) - 1 + 4 - (1/2) - 1 + 4) / 2 =>
6/2 =>
3

dis motha fucka told me once dat i neva amount to nuthin. i told him he needa ****** shut his azz up or i will pop him. yo question is a ****** great question. i studied dis shyt rite here in motha ****** oxford with my ***** jay-c and big-mac.

for rizzle the integral of the function needa be taken, not da derivative. da formula for da avarage value of a function in da interval from a to b is

favg = 1/(b-a) ∫f(x) dx (definite integral from a to b)

F(T) = ∫(2t^3 - 3t + 4) dt = (1/2)t^4 - (3/2)t^2 + 4t

F(1) = (1/2)(1)^4 - (3/2)(1)^2 + 4(1) = 3
F(-1) = (1/2)(-1)^4 - (3/2)(-1)^2 + 4(-1) = -5
F(1) - F(-1) = 3 - (-5) = 8
1/(b-a) = 1/(1-(-1)) = 1/2
favg = (1/2) (8) = 4.
so da answer is D. dumb ****