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I'm having probelms with another question, but it's a very simple one. The question is; Use simpson's rule to find an approximation to pi/8∫pi/4 tan x dx, correct to 2 decimal places. My only problem is that I'm not sure what interval to use. Could somone help me out here...? :o

One more time; Graph y = loge(x - x2). I'm not expecting anyone to be able to show me a completed diagram of the graph (there's one in the back of my book anyways), but could someone explain how I figure out things like the vertex and asymptotes? I can see that x =/= 0 and x =/= 1, but I'm not...

Thanks guys. Ugh... I don't know how I'm gonna make it though my extension 1 exam without you. ;)

So far the only work we've been assesed on in Extension 1 English has been out-of-class assesment tasks, however for the first time next week, we'll be doing an in-class assesment. I'm currently studying crime fiction, and I'm not sure exactly what to expect... The core text we've been focusing...

Sorry, I made a mistake when I first typed out the question, it should be Find ∫x2ex3 - 1 dx. It might mean ∫x2ex3 - 1 dx, I'm not sure. The textbook answer is 1/3 * ex3 - 1 + C, however it might also mean 1/3 * ex3 - 1...

Another question; Find ∫x2e3 - 1 dx. Don't forget the question in my last post though, I still need help with that. EDIT: ^ I'm really not sure. I don't suppose you could go through it...?

Ah... Right. Silly mistake again. :o EDIT: Hmm... How would I go about integrating (ln y)2? Would it be 1/2(ln y) or am I completely off the track? EDIT x 2: Oh crap, I'm completely forgetting my log laws again. (ln y)2 = 2ln y, right? So how do I go about integrating that?

Another one; Use simpson's rule with 5 function values to find the volume of the solid formed when the curve y = ex is rotated about the y-axis from y = 3 to y = 5, correct to 2 significant figures. My problem isn't so much in applying simpsons rule (that's really just substituting values in...

Whoops, silly mistake. I was thinking loge1 = 1. :rolleyes: Thanks.

Okay, new question; Find the exact gradient of the tangent to the curve y = ex + logex at the point where x = 1. This is my working; = y' = 1/x * ex + logex = m = 1/(1) * e(1) + loge(1) = m = 1 * e1 + 1 = m = e2 Only according to the book, that's wrong. According to it, the answer should be...

Another simple question that I'm just not quite sure how to approach; Find the equation of the locus of a point that is always 3 units away from the line 4x - 3y - 1 = 0

I think you've made a mistake there... y = x2/8, not x2/16, so the equation you eventually end up with is 2x - y - 8 = 0. Thanks for clearing that up though. EDIT: Beat'd

I'm having some difficulties with this question simply due to the way its worded. Could someone help me out? a) Find the coordinates of P on the parabola x = 4t, y = 2t2, where t = 2. This first part is easy, the answer is obviously (8, 8) This second part however... b) Find the equation of...

Here we go again; Two circles have radii r such that r + s = 25. Show that the sum of the areas of the circles is least when r = s.

What the heck? Those are exactly the same x-values I got the first I ran through the question, but I didn't get the same y-values... I must have substituted the numbers in to the derivative or something... Gah! I'm such an idiot. Thanks guys. :o

Ugh... MORE help needed; Find the greatest and least values of the function f(x) = 4x3 - 3x2 - 18x in the domain -2 ≤ x ≤ 3. Man I suck...

Hmm... The answer in the book is 16m2, ergo my working for that question is probably wrong. *sigh* This isn't good. I have a mathematics test on wednesday... :(

Ugh... I'm such an idiot. Thanks. Still stuck on the other questions though...

Thanks guys, but it looks like I've run in to more troubles; Find the maximum possible area if a straight 8m length of fencing is placed across a corner to enclose a triangular space. So, I figure = A = 1/2xy = x2 + y2 = 82 = y2 = 64 - x2 = y = (64 - x2)1/2 = A = 1/2x(64 - x2)1/2 = A' =...

It's weird, but yes, I did work out the same answer. Sorry about putting you through all this trouble. :(