1. ## calculus question (integrals)

this is an extra credit question on my take home test..

i think i attached it to the post here but it would read...

the integral from -1 to 1 of [ f(x) - f(-x) ] / 2

he wants us to explain why it is = 0 for any f(x)

I'm pretty sure it has to do with the numerator being 0 because you are subtracting the opposite of f(x) from f(x) ... ? I think

2. edit again
it does cancel out as the exponents that get added will allow the signs to flip and let you cancel it out. use an example with X and N (for exponent) if necessary to show that.

3. also, dont forget to show C in your work to get full credit!

4. substitute t for -x as a dummy variable of the f(-x) integral.

5. 0004

6. Originally Posted by Doobs
substitute t for -x as a dummy variable of the f(-x) integral.
X and -X would be better...

7. Originally Posted by Reko
X and -X would be better...
I don't get what you're saying

8. Originally Posted by Doobs
I don't get what you're saying
ITs easier to keep it all in terms of X dx rather than have a Xdx and a t dt in the same problem.

9. Oh ok. If you do the substitution of t=-x and -dt=dx, then you end up with identical integrals except with different dummy variables, and you can just cancel out both integrals without having to integrate anything.

10. are you sure that works? because then you have F(x) - F(t), where t = -x, as opposed to plain x.

I am def. rusty on this stuff, especially the u sub methods. I'll double check my older notes tonight if i remember.

Here is what I was thinking:
Can you leave the -x as -x and then integrate the parts separately (breaking it up into sparate integrals of f(x)/2 - f(-x)/2 first), which if you use X as x you would have (x^[n+1])/(2*[n+1]) - ([-x]^(n+1)/(2*[n+1])?
If n+1 is positive, then you would have a positive - positive = 0
If n+1 is odd, then since you are multiplying negative, you would have a negative - negative = 0 as well.

I'm just not sure if you can keep that -x in the integral without having to pull it out a -1...

any ideas?

11. Neeerds!

This:
 Video

12. Originally Posted by Reko
are you sure that works? because then you have F(x) - F(t), where t = -x, as opposed to plain x.

I am def. rusty on this stuff, especially the u sub methods. I'll double check my older notes tonight if i remember.

Here is what I was thinking:
Can you leave the -x as -x and then integrate the parts separately (breaking it up into sparate integrals of f(x)/2 - f(-x)/2 first), which if you use X as x you would have (x^[n+1])/(2*[n+1]) - ([-x]^(n+1)/(2*[n+1])?
If n+1 is positive, then you would have a positive - positive = 0
If n+1 is odd, then since you are multiplying negative, you would have a negative - negative = 0 as well.

I'm just not sure if you can keep that -x in the integral without having to pull it out a -1...

any ideas?
I think this is what i was trying for.. so if this IS possible. I thnk its it

13. Well, its extra credit right? I can't figure anything out on it other than that as an idea, or maaaaaaybe finding a way to get
involved, but I'm not sure that applies here. Never know though....

Good luck

14. Originally Posted by beaverfootball
this is an extra credit question on my take home test..

i think i attached it to the post here but it would read...

the integral from -1 to 1 of [ f(x) - f(-x) ] / 2

he wants us to explain why it is = 0 for any f(x)

I'm pretty sure it has to do with the numerator being 0 because you are subtracting the opposite of f(x) from f(x) ... ? I think
You have two definite integrals, Int[-1,1, f(x) dx] = F(1) - F(-1)

The other, Int[-1,1, f(-x) dx], after doing a few manipulation rules...

Let u = -x; du = -dx; x = -1 => u = 1; x = 1 => u = -1

So it can be re-written as Int[1,-1, -f(u) du] = Int[-1,1, f(u) du]

So the 2nd integral evaluates to F(1) - F(-1) as well.

So you end up subtracting (F(1) - F(-1)) - (F(1) - F(-1)) = 0. The 1/2 is irrelevant as far as I can see.

15. Originally Posted by eps
You have two definite integrals, Int[-1,1, f(x) dx] = F(1) - F(-1)

The other, Int[-1,1, f(-x) dx], after doing a few manipulation rules...

Let u = -x; du = -dx; x = -1 => u = 1; x = 1 => u = -1

So it can be re-written as Int[1,-1, -f(u) du] = Int[-1,1, f(u) du]

So the 2nd integral evaluates to F(1) - F(-1) as well.

So you end up subtracting (F(1) - F(-1)) - (F(1) - F(-1)) = 0. The 1/2 is irrelevant as far as I can see.

By golly!

I assume that would work for anything that is X though? Since it calls for f(x)?

16. eps was right

17. Wow guys, this was awsome seeing a lifting community with brains to go with their strength. We've got it all here.

18. Originally Posted by Reko
WHAT SORT OF WIZARDRY IS THIS???

WHAT SORT OF WIZARDRY IS THIS???
Definite integrals.

20. Originally Posted by Reko
Definite integrals.
math is ****ing ghey.

21. It has been 13 years since I had that class and couldn't even begin to tell what any of this means - translation = calculus is worthless once you leave college

22. lol I'm gonna have to relearn most of it actually.

23. I've gone through calc III...and retained just about none of it. BTW, calc III was one year ago.

**** calculus!

24. Originally Posted by borracho
I've gone through calc III...and retained just about none of it. BTW, calc III was one year ago.

**** calculus!
Awesome. I thought I was bad for taking two semesters in college and forgetting, from 13 years ago.

25. i took college alegbra 1 twice.

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