1. ## Question about Area between curves, don't get..

Find the area of the region bounded by the graph of y= .25x^4 + x^3/3 - 2x^2 -4x and the line joining it's two relative minimum..

What do I do?

I have taken the derivative, set to to 0 and got x=2, what else do I do???

2. That's good. Now set it equal to zero.
Find these two mins, both x and y.
Then get the line that connects them, y=mx+b.
Finally integrate y2= .25x^4 + x^3/3 - 2x^2 -4x and y1=mx+b
with integrand y2-y1 and the bounds will be the two x's you obtained from the mins.

3. How do I find the tangent line? They don't give me a set of points in order to use for the point slope form.. I can find m, but what is y1 and x1??

Would I use one of the relative minima as my (x, y)??

4. They ask for the area between the graph and the line joining the two relative minima
So both the relative minima will be on the line, and you can use either of them to come up with an equation for the line
matheagle already posted how to do the rest - but in symbolic form
$
\int\limits_{left\_endpt}^{right\_endpt} {(y2(x) - y1(x))dx}
$

5. Originally Posted by billa
They ask for the area between the graph and the line joining the two relative minima
So both the relative minima will be on the line, and you can use either of them to come up with an equation for the line
matheagle already posted how to do the rest - but in symbolic form
$
\int\limits_{left\_endpt}^{right\_endpt} {(y2(x) - y1(x))dx}
$
y2-y1>0

6. Originally Posted by matheagle
y2-y1>0
I have no idea what you are talking about