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17, HSC year, male no, no, EDIT[maybe... finding an example] 3. Just search around in the non-school forum for agro threads. Check out a thread I created called the Anti-Club Club which was rather ill received :p. I had predicted that there would be a negative reaction, the unfortunate...

See survey board. http://www.boredofstudies.org/community/forumdisplay.php?f=489

I still have mine too. The guys who created gumby were on crack. Go here and check out the videos titled "clay peacock" http://www.gumbyworld.com/gumbysgalaxy/video/galasia.html . I dunno if they're satanic or conformist or what... but they scare me.

Simpsons rule when you have a lot of strips. There is always something that gets miscalculated. Integration makes the job so much easier...

Thanks to Descartes we have the cartesian plane and the cartesian self.

If it's an even function then f(x) = f(-x) f(x) = 1/(x²+4) f(-x) = 1/((-x)²+4) = 1/( x²+4) hence f(x) = f(-x) and f(x) = 1/(x²+4) is an even function

haha, yeah that's what I figured.

I think you probably should have done it the other way round. How did you combine the (3\2∫ x²e^3x dx) term with the (∫xe^3xdx) term? Anyhow, using the other way ∫ xe^3xdx let u=x and dv/dx = e^3x du/dx=1 and v =...

Fort Street High 4U Math 4U English Physics Chemistry

no problem, I think the gist of 46 is back there somewhere as well. This thread has become quite chaotic.

yeah man, since you're using int. by parts: ∫ u'v = uv - ∫ uv' where u' = cosx and v=cos^n-1x you have to integrate cosx which = sinx and differentiate cos^n-1x which = (n-1)cos^n-2x(-sinx) combine these and you get [- ∫ uv'] = (n-1) &int...

46. use integration by parts: ∫ u'v = uv - ∫ uv' where u' = sinxcos²x and v=sin^n-1x When integrating sinxcos²x use the substitution u= cosx See if you can approach the form I_n = (n-1)/3.I_n-2 - (n-1)/3.I_n

I think it should be (positive) (n-1) ∫ cos^n-2xsin²xdx

For 48. Integration by parts: ∫ u'v = uv - ∫ uv' where u' = cosx and v=cos^n-1x Remember the identity sin²x = 1 - cos²x and don't hesitate to rearange I_n and I_n-2 terms

49. Have a go at substitutions like (u=π/2 - x) or (u = x - π/2) and see how it works out. Keep in mind that [-a --->a]∫f(x)dx = - [a --->-a] ∫f(x)dx and vice versa

47. I_n = [0-->π/2] ∫ xⁿsinxdx do the ∫u'v = uv -∫ uv' where u'=sinx and v=xⁿ I_n = [-xⁿcosx]{between 0-->π/2} + n∫x^n-1cosx dx = [0-->π/2] n∫x^n-1cosx dx then use the same process...

Haha interesting timing, I did all of those yesterday. I'll write up what I can be bothered to do, for the most part I'll try and just give you hints because it's best if you can try and do them yourself. Give me 20 mins or so.

18. If I_n = [1-->e] ∫ (lnx)ⁿdx, show that I_n = e - I_n-1 I_n = ∫ (lnx)ⁿdx = ∫ 1.(lnx)ⁿdx so use integration by parts where: ∫u'v = uv - ∫uv' u' = 1 -------> u = x v = (lnx)ⁿ=...

I've only used mathematica

Yup, then geustimate, bisect intervals, use newton's method, whatever you need to do to get you estimation to the rigthnumber of d.p's.