Inactive Tutor answered 04/19/14
Tutor
New to Wyzant
1) F(2)=∫02 sin(t2)dt
F(2)=0.5*[sin(0)+sin(0.52)]/2+0.5*[sin(0.52)+sin(1)]/2+0.5*[sin(1)+sin(1.52)]/2+0.5*[sin(1.52)+sin(22)]/2=0.5*[sin(0.52)+sin(1)+sin(1.52)+sin(4)/2]≈0.744
2) F(x) increases on the intervals where F'(x)>0. F'(x)=sin(x2). Since 0≤x≤3, F'(x)≥0 on the intervals [0,√π)]∪[√(2π);3] Therefore, those are the intervals where F(x) increases.
3) Average rate of change of F(x) is simply this: (1/3)∫03 sin(t2)dt. It is k, by the problem statement. So, the answer here is simply 3k.