University calculus help

[Lavi]

-.- My prof did this question wrong, so if it ends up on the final, which is likely, I won't be able to do it properly:

http://www.wolframalpha.com/input/?i=integral+%28sinx%29%2Fx+from+pi%2F4+to+pi%2F2+%3C+sqrt2%2F2

Wolfram will not show the steps to proving it -.-

 

[DirgeNovak]

The answer is B. Put B.
Hope that absolutely hilarious joke keeps you going.

 

[Krunchybars]

I really hope you get the answer you are looking for but perhaps this really isnt the ideal place for calculus help. I could be wrong but I haven't got a clue.

 

[Actual]

Sorry I did physics, so we touched on this but I'm not familiar with si(x) which they derive:

http://www4b.wolframalpha.com/Calculate/MSP/MSP155519dd3af3698a45i9000027ciaei853ifghb7?MSPStoreType=image/gif&s=42&w=144&h=41

I do vaguely remember that if dealing with radeon (sp?) values it's easier to convert into regular numbers, make the calculations and then switch back to radeons. So pi=180, sin(pi/2)=1.

Of course that could be my memory lying to me.

I say find a class mate to ask for help.

 

[DuplicateValue]

Hmm well if you write it as [integral] (sin(x))(1/x), then I think you can have:

([integral] sin(x))([integral] 1/x)

The integral of sin(x) is cos(x) I think.
I'm not sure how to get the integral of 1/x without it being undefined. You could try substituting for U and working from there.

Once you do that, you have to find the new limits and all that.

I really don't know, I'm only in secondary school, this was probably no help/horribly wrong.

 

[Dindril]

... If I could remember the Derivative of sin(x), I could totally solve this... wow, I've been out of school for 6 months, and I've already forgot...

Edit: Figured it out! (not sure if this is the best method, but I'm remembering as I go.)

Derive: (Cos(x)/x)-(Sin(x)/x^2)
(working: This is taking a while...)
Input (pi/2): (Cos(pi/2)/(pi/2))-(Sin(pi/2)/(pi/2)^2)
(0/(pi/2))-(1/(pi/2)^2)
-1/(pi/2)^2
Input (pi/4): (Cos(pi/4)/(pi/4))-(Sin(pi/4)/(pi/4)^2)
((sqrt(2)/2)/(pi/4))-((sqrt(2)/2)/(pi/4)^2)

... Maybe I don't have it... regardless of what I try, it keep getting 0.1592<(sqrt(2)/2), which, though still true, doesn't match the -0.0945 on teh site...

 

[Lavi]

I really hope you get the answer you are looking for but perhaps this really isnt the ideal place for calculus help. I could be wrong but I haven't got a clue.

Rule for Uni: If the prof can't do it, why should I have to?

 

[DuplicateValue]

... If I could remember the Derivative of sin(x), I could totally solve this... wow, I've been out of school for 6 months, and I've already forgot...

cos(x) or -cos(x) I think.

 

[Fireshot25]

Look up "integration by parts" cause i think thats what you have to do. Let u=1/x and V'=sinx.

 

[Lavi]

What the prof tried:

pi/4 < x < pi/2
2/pi < 1/x < 4/pi
2/pi sinx < sinx/x < 4/pi sinx

Then integrate both sides, but you get 2sqrt2/pi -.-

 

[dstryfe]

You could try to integrate sin(x)/x through the power series of sin(x): sum from n=0 (-1)^n*x^(2n+1)/(2n+1)! . Just divide each term by x, then integrate. Not having paper or time handy...back to Guitar Hero!

Hope it helped.

 

[Dindril]

... As I read through this thread, I continue to relearn alot of stuff that I forgot in the past few months...

 

[Dindril]

... If I could remember the Derivative of sin(x), I could totally solve this... wow, I've been out of school for 6 months, and I've already forgot...

Edit: Figured it out! (not sure if this is the best method, but I'm remembering as I go.)

Derive: (Cos(x)/x)-(Sin(x)/x^2)
(working: This is taking a while...)
Input (pi/2): (Cos(pi/2)/(pi/2))-(Sin(pi/2)/(pi/2)^2)
(0/(pi/2))-(1/(pi/2)^2)
-1/(pi/2)^2
Input (pi/4): (Cos(pi/4)/(pi/4))-(Sin(pi/4)/(pi/4)^2)
((sqrt(2)/2)/(pi/4))-((sqrt(2)/2)/(pi/4)^2)

... Maybe I don't have it... regardless of what I try, it keep getting 0.1592<(sqrt(2)/2), which, though still true, doesn't match the -0.0945 on teh site...

... My work is very sloppy... Makes it hard to look back and see what I did wrong... (And, Yes, I'm quoting my own post)

 

[Rubashov]

-.- My prof did this question wrong, so if it ends up on the final, which is likely, I won't be able to do it properly:

http://www.wolframalpha.com/input/?i=integral+%28sinx%29%2Fx+from+pi%2F4+to+pi%2F2+%3C+sqrt2%2F2

Wolfram will not show the steps to proving it -.-

Because (sin (x))/x does not have an antiderivative that can be expressed in terms of an elementary function, the only way to come up with a formula for the integral is to express it as a power series, then find the integral of that power series.

As for how to show that the integral is less than sqrt(2)/2... No idea. Sorry.

 

[lacktheknack]

... If I could remember the Derivative of sin(x), I could totally solve this... wow, I've been out of school for 6 months, and I've already forgot...

cos(x) or -cos(x) I think.

cos(x). The derivative of cos(x) is -sin(x). Tricky bugger.

 

[Rubashov]

What the prof tried:

pi/4 < x < pi/2
2/pi < 1/x < 4/pi
2/pi sinx < sinx/x < 4/pi sinx

Then integrate both sides, but you get 2sqrt2/pi -.-

Oh, actually, this gives me an idea.

sin x is sqrt(2)/2 at pi/4 and 1 at pi/2, right? Well, when x is pi/4, 1/x is 4/pi and when x is pi/2, 1/x is 2/pi. ((sqrt(2)/2))(4/pi)= 2sqrt(2)/pi is greater than (2/pi), so it seems reasonable to say

(sin x)/x <= 2sqrt(2)/pi

on the interval [pi/4, pi/2].

Integrating both sides, we have:

integral(((sin x)/x)dx) <= 2xsqrt(2)/pi + C

Using this antiderivative, we calculate the definite integral:

integral(((sin x)/x)dx) <= (2sqrt(2)/pi)(pi/2) - (2sqrt(2)/pi)(pi/4) = (sqrt(2)/2).

I'm not sure how rigorous this proof is...but it seems to produce the correct result, at least.

 

[kaieth]

I'd do a couple of iterations of integration by parts, find a pattern, and then mumble something about due to the pattern, it'll stay under sqrt(2)/2 because of the 1/(n!) term.

 

[Lavi]

What the prof tried:

pi/4 < x < pi/2
2/pi < 1/x < 4/pi
2/pi sinx < sinx/x < 4/pi sinx

Then integrate both sides, but you get 2sqrt2/pi -.-

Oh, actually, this gives me an idea.

sin x is sqrt(2)/2 at pi/4 and 1 at pi/2, right? Well, when x is pi/4, 1/x is 4/pi and when x is pi/2, 1/x is 2/pi. ((sqrt(2)/2))(4/pi)= 2sqrt(2)/pi is greater than (2/pi), so it seems reasonable to say

(sin x)/x <= 2sqrt(2)/pi

on the interval [pi/4, pi/2].

Integrating both sides, we have:

integral(((sin x)/x)dx) <= 2xsqrt(2)/pi + C

Using this antiderivative, we calculate the definite integral:

integral(((sin x)/x)dx) <= (2sqrt(2)/pi)(pi/2) - (2sqrt(2)/pi)(pi/4) = (sqrt(2)/2).

I'm not sure how rigorous this proof is...but it seems to produce the correct result, at least.

xD Okay, appropriate result ftw. Thanks

 

[smithy_2045]

Hmm well if you write it as [integral] (sin(x))(1/x), then I think you can have:

([integral] sin(x))([integral] 1/x)

The integral of sin(x) is cos(x) I think.
I'm not sure how to get the integral of 1/x without it being undefined. You could try substituting for U and working from there.

Once you do that, you have to find the new limits and all that.

I really don't know, I'm only in secondary school, this was probably no help/horribly wrong.

int(1/x,dx) = log |x| + c if I remember correctly. Also, you can't split an integral like that.

 

[DuplicateValue]

Hmm well if you write it as [integral] (sin(x))(1/x), then I think you can have:

([integral] sin(x))([integral] 1/x)

The integral of sin(x) is cos(x) I think.
I'm not sure how to get the integral of 1/x without it being undefined. You could try substituting for U and working from there.

Once you do that, you have to find the new limits and all that.

I really don't know, I'm only in secondary school, this was probably no help/horribly wrong.

int(1/x,dx) = log |x| + c if I remember correctly. Also, you can't split an integral like that.

Yeah I figured I was probably wrong...