Integration (1 Viewer)

[KeypadSDM]

Integrate this bigboy

This is mainly for those who have done the integration chapter, which a few of you should be doing now (maybe).

x = 1, y = 0:

dy/dx = y/[(x + 1)(x - 1)]

Find y as a function of x

Actually, that's insanely evil.

Hehehehe

 

[Affinity]

While you are on it find:

Integrate [0->1] [x(1-x)]^4/(x^2 + 1) dx

and hence show that pi is not 22/7

 

[George W. Bush]

Originally posted by KeypadSDM
Integrate this bigboy

This is mainly for those who have done the integration chapter, which a few of you should be doing now (maybe).

x = 1, y = 0:

dy/dx = y/[(x + 1)(x - 1)]

Find y as a function of x

Actually, that's insanely evil.

Hehehehe

Hrm, dy = ydx/[(x+1)(x-1)]

y = integral adx/(x+1) + integral bdx(x-1)

a(x-1) + b(x+1) = y
x=1, y=0 , therefore b=0

y = integral adx/(x+1)
y = aln|x+1| + c
c = -aln2
y = a(ln[|x+1|/2]

Hrm...

 

[Heinz]

Originally posted by Affinity
While you are on it find:

Integrate [0->1] [x(1-x)]^4/(x^2 + 1) dx

and hence show that pi is not 22/7

GRRRRRRR i spent 10mins typing it up on microsoft equation editor and something happened #$%@. I ended up with 22/7 - pi. But is there an easier method to do this other than expanding the numerator and then dividing out? I dont understand the second part. Because the integral doesnt equal zero, than 22/7 - pi doesnt equal zero? so then 22/7 doesnt equal to pi.

 

[George W. Bush]

Originally posted by Affinity
While you are on it find:

Integrate [0->1] [x(1-x)]^4/(x^2 + 1) dx

and hence show that pi is not 22/7

[0->1] [x(1-x)]^4/(x^2 + 1) dx = 4x - 4x^3/3 + x^5 - 2x^6/3 + x^7/7 - 4Arctanx [0->1]

= 3 + 1/7 - pi
= 22/7 - pi

If pi was 22/7, the area under the curve would be 0. As is it not , pi =! 22/7.

 

[George W. Bush]

Originally posted by Heinz
GRRRRRRR i spent 10mins typing it up on microsoft equation editor and something happened #$%@. I ended up with 22/7 - pi. But is there an easier method to do this other than expanding the numerator and then dividing out? I dont understand the second part. Because the integral doesnt equal zero, than 22/7 - pi doesnt equal zero? so then 22/7 doesnt equal to pi.

I used parts then expanded
4x^3(1-x)^3Arctanx - 8x^4(1-x)^3Arctanx

 

[Heinz]

Originally posted by George W. Bush
[0->1] [x(1-x)]^4/(x^2 + 1) dx = 4x - 4x^3/3 + x^5 - 2x^6/3 + x^7/7 - 4Arctanx [0->1]

= 3 + 1/7 - pi
= 22/7 - pi

If pi was 22/7, the area under the curve would be 0. As is it not , pi =! 22/7.

How come your workings so simple? Please tell me you have some secret to doing it quickly. It took me ages.

 

[George W. Bush]

Originally posted by Heinz
How come your workings so simple? Please tell me you have some secret to doing it quickly. It took me ages.

Grab a wad of paper and do that shit manual style

 

[Affinity]

Re: Re: Integration

Originally posted by George W. Bush
Hrm, dy = ydx/[(x+1)(x-1)]

y = integral adx/(x+1) + integral bdx(x-1)

a(x-1) + b(x+1) = y
x=1, y=0 , therefore b=0

y = integral adx/(x+1)
y = aln|x+1| + c
c = -aln2
y = a(ln[|x+1|/2]

Hrm...

Hmm incorrect

 

[maniacguy]

Answer : y = x^2 - 1
Method : I'll leave it as an exercise for the reader Suffice to say that the claim it's 'insanely evil' is a gross exaggeration. It depends on which method you're using, is all.

Edit:
As pointed out, this is wrong. The correct answer should be:

y = A*sqrt[ (1-x)/(1+x) ] for constant A

That'll teach me to do things by inspection. (Note: If this is wrong, then blame Maple as being incompetent, not me )

 

[Giant Lobster]

it cant be y = x^2 - 1 else y' = 1 for all x cept 1 and -1

lol i got a silly answer: x = -1
im most likely wrong, cos i didnt need the condition y = 0 when x = 1

 

[George W. Bush]

Originally posted by maniacguy
Answer : y = x^2 - 1
Method : I'll leave it as an exercise for the reader Suffice to say that the claim it's 'insanely evil' is a gross exaggeration. It depends on which method you're using, is all.

Edit:
As pointed out, this is wrong. The correct answer should be:

y = A*sqrt[ (1-x)/(1+x) ] for constant A

That'll teach me to do things by inspection. (Note: If this is wrong, then blame Maple as being incompetent, not me )

y=ax?

 

[KeypadSDM]

Fine, not INSANELY evil. But hard to say the least.

dy/dx = y/[(x + 1)(x - 1)]
1/y dy/dx = 1/[(x + 1)(x - 1)]
1/y dy/dx = 1/2[1/(x - 1) - 1/(x + 1)]
Ln[y] = 1/2Ln[x - 1] - 1/2Ln[x + 1] + c
Ln[y] = Ln[ Sqrt[(x - 1)/(x + 1)]] + c
y = e^c * Sqrt[(x - 1)/(x + 1)]

Thus, let e^c = A
y = ASqrt[(x - 1)/(x + 1)]

(When I made up this question it failed to occur to me that setting y = 0, x = 1 made no difference in the end. Stupid me.)