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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- Mar 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! - Mar 13th 2010, 12:05 PMione
- Mar 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 $\displaystyle G(x)= \int_2^x f(t)dt$ 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 $\displaystyle \frac{\int_a^b f(x) dx}{b-a}$.

You are told that, on 0 to 3, the average value of f(x) is 0- that tells you that $\displaystyle \frac{\int_0^3 f(x)dx}{3}= 0$ so that $\displaystyle \int_0^3 f(x)dx= 0$. And $\displaystyle \int_0^3 f(x)dx= G(3)- G(0)$.

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