Answer:
(B) 1
Step-by-step explanation:
g(x) = ∫₂ˣ f(t) dt
Use second fundamental theorem of calculus to find g'(x).
g'(x) = f(x) d(x)/dx − f(2) d(2)/dx
g'(x) = f(x)
g(x) is a maximum when g'(x) = 0.
0 = f(x)
x = 3
So the maximum value of g(x) is:
g(3) = ∫₂³ f(t) dt
g(3) = ½ (2) (1)
g(3) = 1
0
zero divided by anything that is integral is zero and 3 in a integer
well that took alot of guesses...
