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Integration watch

1. Does anyone know how to do the 2nd one in particular? The first one is done in cylindrical polars but as for the 2nd I'm not sure.
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2. hmm i worked it out to be 1/2. Can anyone confirm this?

*worked it out wrong *
3. (int over R) x^2 e^(-x^2) dx dy
= (int from 0 to 1) x^3 e^(-x^2) dx + (int from 1 to 2) x e^(-x^2) dx
= [1/2 - e^-1] + [1/2 - (1/2)e^-1]
= 1 - (3/2) e^-1.
4. (Original post by Lagz)
Does anyone know how to do the 2nd one in particular? The first one is done in cylindrical polars but as for the 2nd I'm not sure.
Do you mean part b) or the second integral on the page? For INT{0,inf} x^2 e^(-x^2) dx i get sqrt(pi)/4.
5. right I have worked through it now.

i: Sqrt(Pi/4)
ii: Sqrt(Pi/4)
iii,iv: easy.

erm. One of the first two looks a bit suspect? The areas under them are *as I worked them out* the same???? Anyone know which if any are wrong

cheers for help so far

Olly
6. ok i fixed it now second one should be half as much
7. As I said before, I got sqrt(pi)/4 rather than sqrt(pi/4) for (ii).
8. (Original post by mikesgt2)
Do you mean part b) or the second integral on the page? For INT{0,inf} x^2 e^(-x^2) dx i get sqrt(pi)/4.
you got working for this? I worked it out ok in the end in polars but needed to use some craaazy trig.

Infact dont worry I now see the easy way to do it (parts x and xe^(-x^2))
9. (Original post by Lagz)
you got working for this? I worked it out ok in the end in polars but needed to use some craaazy trig.
I worked out a reduction formula for I(n) = INT{0,inf} x^n e^(-x^2) dx. Apply parts with u=e^(-x^2) and dv/dx=x^n. Then, du/dx=-2xe^(-x^2) and v=x^(n+1)/(n+1). Therefore,

I(n) = INT{0,inf} x^n e^(-x^2) dx = [ e^(-x^2) x^(n+1)/(n+1) ] + 2/(n+1) INT x^(n+1) x e^(-x^2) dx

The bit in the square brackets in meant to have the limits on the right bracket; the square bracket bit evaluates to zero. We then have

I(n) = 2/(n+1) INT x^(n+1) x e^(-x^2) dx = 2/(n+1) INT x^(n+2) e^(-x^2) dx

That integral is equal to I(n+2), so we obtain the relation

I(n) = 2/(n+1) I(n+2)

We need INT{0,inf} x^2 e^(-x^2) = I(2) so substitute n=0 into the above formula to get

I(0) = 2I(2) => I(2) = I(0)/2.

And we evaluated I(0) in the first part of the question as sqrt(pi/4) so:

I(2) = sqrt(pi/4)/2 = sqrt(pi)/4

Well, that was my method. However, a reduction formula is not really necessary, you can just apply integration by parts to INT{0,inf} e^(-x^2) dx with u=e^(-x^2) and dv=dx.
10. Exercise 4G p138, qu.23, Edexcel P3

Ok the question is:

23) A lump of radioactive substance is disintegrating. At time t days after it was first observed to have mass 10 grams, its mass is m grams and

dm/dt = -km

where k is a positive constant

Find the time, in days, for the substance to reduce to 1 gram in mass, given that its half-life is 8 days.
(The half-life is the time in which half of any mass of the substance will decay).

---------------
Thanks
11. solve the differential equation to get m in terms of t,and use the fact that the halflife is eight days to work aout the value of the constants.
Once u have done that u know that m=1 and find the corresponding t value by rearranging
12. That first one is a NST 1A paper question if I'm not much mistaken!

Ben
13. (Original post by Jonny W)
(int over R) x^2 e^(-x^2) dx dy
= (int from 0 to 1) x^3 e^(-x^2) dx + (int from 1 to 2) x e^(-x^2) dx
= [1/2 - e^-1] + [1/2 - (1/2)e^-1]
= 1 - (3/2) e^-1.
Why have you evaluated from 0 to 1 for the first bit?

I get the answer as (3/2)e^(-1) - e^(-3). But I could well be, and probably am, wrong.

Ben
14. I HATE E'S AND LOGS
xox

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