Expert: Paul Klarreich - 11/25/2007

Question
I'm not quite sure if this question is an application of the Mean Value Theorem, but I don't think so.  The question:

Let F(x)=the integral (from zero to x)of sin(t squared)dt for the interval (x greater than or equal to zero and less than or equal to three).

If the average rate of change of F on the closed interval [0,3] is given by "k", find the integral (from zero to 3) of sin(t squared)dt in terms of "k".

If using the Mean Value Theorem, would the answer just be 3sin(k squared)?  Since the problem states that the "average rate of change" of F is given by "k", doesn't this imply that one would have to find an expression for the derivative of F, which would give the slope, which you would then evaluate at the endpoints of the interval and divide by three?  I guess I'm confused by the wordage of the problem.

Answer
Questioner:   Mark
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  I'm not quite sure if this question is an application of the Mean Value Theorem, but I don't think so.  The question:

Let F(x)=the integral (from zero to x)of sin(t squared)dt for the interval (x greater than or equal to zero and less than or equal to three).

If the average rate of change of F on the closed interval [0,3] is given by "k", find the integral (from zero to 3) of sin(t squared)dt in terms of "k".

If using the Mean Value Theorem, would the answer just be 3sin(k squared)?  Since the problem states that the "average rate of change" of F is given by "k", doesn't this imply that one would have to find an expression for the derivative of F, which would give the slope, which you would then evaluate at the endpoints of the interval and divide by three?  I guess I'm confused by the wordage of the problem.
========================================
Hi, Mark,

All you need here is the definition of average rate of change.
Your F(x) =

{x
|   sin(t^2) dt  with x in [0,3]
}0


The Avg R-o-C of F(x) on [0,3] is

delta-F    F(3) - F(0)
-------- = ------------
delta-x      3 - 0

But  F(0) =

{0
|   sin(t^2) dt, which must be zero.
}0


So your R-O-C is  F(3)/3, which, you say, is "k".

So F(3) = 3k.

And, since F(3) =    
{3
|   sin(t^2) dt
}0

I think that's your answer. 3k, that is.  

I don't think you need any more.