g(x) is increasing where f(x) is positive, since f(x) = g'(x)
for x in [-3,0)
g(x) = ∫[-2,x] 3 dt = 3t[-2,x] = 3x+6
for x in [0,6]
g(x) = ∫[-2,0] 3 dx + ∫[0,x] (-t+3) dt
= 3t[-2,0] + (-t^2/3+3t)[0,x]
= 6 + (-x^2/2 + 3x) - 0
= -x^2/2 + 3x + 6
f is a continuous function with a domain [−3, 9] such that
f(x)= 3 , -3 ≤ x < 0
-x+3 , 0 ≤ x ≤ 6
-3 , 6 < x ≤ 9
and let g(x)= ∫ f(t) dt where a=-2 b=x
On what interval is g increasing? Justify your answer.
For 0 ≤ x ≤ 6, express g(x) in terms of x.
2 answers
Just so people know, he is right on the second question and wrong on the first one.