Calculus 1 f(b)- f(a)/ b-a

I have a Calc test tomorrow and I don't know why this problem equals 6/pi:

h(t) = sin (3t) and [0, pi/6]

I get (and I don't know how to do this problem):
[3(sin pi/6) - 3(sin 0)]/ (pi/6) =

[3 (1/2) - 0] / (pi/6) = (3/2)(6/pi) = 9/pi

not 6/pi

What am I doing wrong here? Thanks for your help



I am not sure what I am doing wrong. The sin of pi/6 is 1/2

First you have to evaluate the function using f(b) - f(a)

a = sin(3*0) = 0
b = sin(3*(pi/6)) = sin(3pi/6) = 1
f(b) - f(a) = 1 - 0 = 1

b - a = pi/6 - 0 = pi/6

1/(pi/6) = 6/pi

I see your error. You pulled the 3 out of the sin(3t) part. You cannot do that as it is not a constant multiplier to sin. It is part of that function. You have to distribute it to the pi/6 in order to get the correct answer.