http://i161.photobucket.com/albums/t...0/calc3112.jpg

I really don't know where to start with this since theere is no f(t) equation. Please help!

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- March 13th 2010, 11:21 AMrawkstarIntegration on an Interval
http://i161.photobucket.com/albums/t...0/calc3112.jpg

I really don't know where to start with this since theere is no f(t) equation. Please help! - March 13th 2010, 12:05 PMione
- March 13th 2010, 12:37 PMHallsofIvy
I suspect it was not the difference between "f(x)" and "f(t)" that was rawkstar's problem. I suspect he simply means that he is not given a formula for f(x).

And that's the problem with memorizing formulas rather than learning concepts. You don't**need**to know the formula if you really understand what is being asked- you read the answers off the graph.

For example, the "concavity" tells how the slope of the graph is changing. A function is "concave upward" as long as the first derivative is increasing.

a) By the "fundamental theorem of calculus", the first derivative of is, simply, f(x). Looking at the graph, f(x) is increasing between -2 and -1, decreasing for x between -1 and 3/2, then increasing again for x between 3/2 and 3.

b) If the equation of the tangent line to y= G(x) at x= 0 is y= mx+ b, then m is the derivative of G(x) at x= 0 which means it is the value of f(x) at x= 0 which can be read off the graph.

Of course, the "b" is G(0) and that is the area under the graph of f(x) from -2 to 0. And that is the area of a rectangle of height 2 and base 2 plus the area of a semi- circle of radius 1.

c) The "average value" of a function f(x) on [a, b] is .

You are told that, on 0 to 3, the average value of f(x) is 0- that tells you that so that . And .

And you have already found G(0) in (b). - March 13th 2010, 02:23 PMione