Function given by integral

[matwinky]

Hi guys ,i have a question regarding this problem.
Consider the function F integral of 1/(t(t-1)) from x to x+4.
a) whats is the domain this one i had already done.
b)determinate where F is positive ,null or negative.Do not do the integral.This is the problematic question
c)determinate the point of maximun minimun relatives ,and do a grafic of the function F.This one i had problems too.
Thank you.

 

[Ishuda]

Hi guys ,i have a question regarding this problem.
Consider the function F integral of 1/(t(t-1)) from x to x+4.
a) whats is the domain this one i had already done.
b)determinate where F is positive ,null or negative.Do not do the integral.This is the problematic question
c)determinate the point of maximun minimun relatives ,and do a grafic of the function F.This one i had problems too.
Thank you.

For (b) what are you allowed? For example
f(x) = \(\displaystyle \frac{1}{t (t-1)} = \frac{1}{t-1} - \frac{1}{t}\)
and we know that, since the power of t in the denominator is 1, "the singularity will not be removed by integration" [use the u = t-1 substitution if you need to on the first fraction]. Thus t and t-1 should both be less than zero or both be greater than zero so that the interval of integration does not span the singularities. This translates to x<-3 or x>0 [if I haven't messed up again]. If that is the case F is positive everywhere in that domain.

For (c) are you allowed to do the integral? Anyway, if F is the integral then F' is given, i.e. f(x+4) - f(x).

EDIT: Fix the confusion I mixed up about t limits and x limits

 

[Steven G]

Ishuda is generally right (if not always right) so I am a bit nervous going against what was posted but here goes. if x<-3, then x+4<1. So for example t can go from -3- to 1-. The problem is that on this interval t and t-1 do not always have the same sign as you wanted (just consider t = 1/2). I also do not think that F>0 only when the integrand is always positive over the limits.
I will wait to hear about my fate for taking on a giant like Ishuda.

 

[Ishuda]

Ishuda is generally right (if not always right) so I am a bit nervous going against what was posted but here goes. if x<-3, then x+4<1. So for example t can go from -3- to 1-. The problem is that on this interval t and t-1 do not always have the same sign as you wanted (just consider t = 1/2). I also do not think that F>0 only when the integrand is always positive over the limits.
I will wait to hear about my fate for taking on a giant like Ishuda.

First "t can go from -3- to 1-": You are talking about limits of integration I presume. That's what messed me up to start with and the reason for the edit and parenthetical statement.
Let's look at it again and look at the function. It asymptotes at 0+ at -\(\displaystyle \infty\) and +\(\displaystyle \infty\). It goes to +\(\displaystyle \infty\) at 0- and 1+. It goes to -\(\displaystyle \infty\) at 0+ and 1-. It has a single relative maximum of -4 at x=1/2. Because the function does not have a "removable singularity by integration" [such as t^-1/2 would have, for example], any integration interval which crosses those points of singularity will be undefined. For the limits of integration that means that (a)both x and x+4 must be less than zero or (b)both x and x+4 must be between 0+ and 1- or (c) both x and x+4 must be greater than 1. Case (b) is clearly impossible so
(a) x and x+4 less than zero is x< 0 and x+4<0 so that x<-4 satisfies both [I did mess up again - that is (a) is for t < 0 and t-1 less than zero so t<0 satisfies both NOT t-1 satisfies both]
(b) x and x+4 greater than 1 is x> 1 and x+4 > 1 so that x>1 satisfies both [this is for t and t-1 both positive and t>1 satisfies both].

About the "I also do not think that F>0 only when": I didn't think I said or implied that but if I did, I was wrong. What I meant by "that being the case" is that if the limits of integration were restricted to either only the (a) case or to the (b) case then the integrand is positive and thus F is positive.

 

[Steven G]

First "t can go from -3- to 1-": You are talking about limits of integration I presume. That's what messed me up to start with and the reason for the edit and parenthetical statement.
Let's look at it again and look at the function. It asymptotes at 0+ at -\(\displaystyle \infty\) and +\(\displaystyle \infty\). It goes to +\(\displaystyle \infty\) at 0- and 1+. It goes to -\(\displaystyle \infty\) at 0+ and 1-. It has a single relative maximum of -4 at x=1/2. Because the function does not have a "removable singularity by integration" [such as t^-1/2 would have, for example], any integration interval which crosses those points of singularity will be undefined. For the limits of integration that means that (a)both x and x+4 must be less than zero or (b)both x and x+4 must be between 0+ and 1- or (c) both x and x+4 must be greater than 1. Case (b) is clearly impossible so
(a) x and x+4 less than zero is x< 0 and x+4<0 so that x<-4 satisfies both [I did mess up again - that is (a) is for t < 0 and t-1 less than zero so t<0 satisfies both NOT t-1 satisfies both]
(b) x and x+4 greater than 1 is x> 1 and x+4 > 1 so that x>1 satisfies both [this is for t and t-1 both positive and t>1 satisfies both].

About the "I also do not think that F>0 only when": I didn't think I said or implied that but if I did, I was wrong. What I meant by "that being the case" is that if the limits of integration were restricted to either only the (a) case or to the (b) case then the integrand is positive and thus F is positive.

Initially i thought what you wrote was very confusing and possibly wrong. However after reading your comment three times I realized that you were not only correct but explained everything perfectly. Thanks for taking the time to give me a detailed explanation.