Hi. I was wondering if anybody could help me out with this problem.
F(x) equals the Integration from x^2 to x^3 of (lnt)dt. We have to find all critical points of f(x).
What I (think) I know.
I think you have to use the fundamental theroem of calculus (1 and 2?), but I'm unsure of how to apply it. A friend helped me a bit and so far we have gotten this:
ln(x^3)3x - ln(x^2)2x = 0
from there though, I'm not sure what you do. I'm not sure (if that is the correct procedure) how you go about solving for those points.
Any help is greatly appreciated.
If I understand what you mean....
F(x) = INT.(x^2 --->x^3)[ln(t)]dt
So,
F(x) = [1/t]|(x^2 ---> x^3)
F(x) = 1/(x^3) -1/(x^2) --------------(i)
Now to the critical points of F(x), we get the 1st derivative of F(x) and equate that to zero.
F(x) = x^(-3) -x^(-2) ---------rewriting (i) in another form.
F'(x) = (-3)[x^(-4)] -(-2)[x^(-3)]
F'(x) = -3/(x^4) +2/(x^3)
Set that ot zero,
0 = -3/(x^4) +2/(x^3)
Clear the fractions, multiply both sides by (x^4)(x^3),
0 = -3x^3 +2x^4
0 = (-3 +2x)x^3
x = 0 or 3/2
Hence, the critical points of F(x) are:
When x = 0,
F(0) = 1/0 -1/0 = indeterminate.
Meaning, zero is not a root of F(x). So discard x=0.
When x = 3/2,
F(3/2) = 1/[(3/2)^3] -1/[(3/2)^2]
F(3/2) = 8/27 -4/9
F(3/2) = (8 -12)/27 = -4/27
So point (3/2,-4/27) --------------------the only critical point.
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:-), Heck, I made a honest mistake. INT.[ln x] dx = 1/x. Yeah, right. Because INT.[1/u]du = ln(u) +C. :-)
I will not edit this wrong solution, nor will i delete it. Let me be wrong once in a while.
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