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I think it's all making sense. Thanks a bunch to those who posted :) I'll post back if I fail later.

So I just wrote all that out on paper and it appears to make sense. y = lnx y' = 1/x And I now understand the function over the function thing, I was a bit confused before. So we've got: \frac{x}{x^{2}+ 2} using the formula, I'd say the answer is ln (x2+2) but there's a 1/2. I want to know...

So if I understand correctly, you need to differentiate it to then integrate it? 1/2 x 2x = x which is the original value, but it lets us integrate when its in that form? \int \frac{1}{2} \cdot 2xe^{x^2} dx = \frac{1}{2} \int 2xe^{x^2} dx = \frac{1}{2} \cdot e^{x^2} + C = \frac{1}{2}...

\int \frac{dx}{x} = \int \frac{1}{x} dx = log_{e}x + C \int \frac{f'(x)}{f(x)} dx = log_{e}f(x) + C I don't understand either of them or when to use either of them.

Cheers, first one anyone?

Another one. Find the integral of (x2 + 3/x) from 2 to 1. I ended up getting 2/1/3, which is half the answer. The whole answer is 2/1/3 + 3ln2. Where on earth does ln come into the equation? While we're at it, how would you do \int_{1}^{e^{3}}\frac{4}{x} dx

I don't understand the formulae given. Could someone explain it simply to me? I seem to have done the work once but don't understand it. An example or two would be nice. Thank you!

Kurt, for the last one, I'm not getting it, where does the 1/2 come from?

Wait so from -2pi to 0 is one full cycle of the tan graph? Cos in the textbook I see that the whole thing from -2pi to 2pi is one full cycle but it only goes from 0 to 2pi. Blahh.

I can't figure out how to do this. I did this a month ago and now I'm revising and just can't get it. It's really killing me. 5. e) Differentiate: ln (7x - 2) 5. g) differentiate: log2 x 6. d) Find the integral of: (e2x + 5) limits being 1 to 0. 6. e) '' xex2 (this one is indefinite)...

If you're wanting to become a programmer, which course would you be looking more at..? That post blew everything I had in mind.

Doesn't the period mean the distance it takes to complete a "cycle" of the graph? So if the period of 2pi, don't you get tan graph from -2pi to 0 and then another tan graph from 0 to 2pi? I must be missing something. smp, for your #1 answer was 0, so you were right. Cheers for all the help so far

it was cosx(degrees) that looks right, thanks. Thanks. 2 more: 1. Integrate from pi to 0 cos(pi + x) dx. 2. Sketch y = tan x/2 for -2pi <= x <= 2pi With 2. I sketched the tan graph twice, but it shows the tan graph once from -2pi to 2pi. The period is pi/b, and I got 2pi for the period...

Thanks for 1, yeah 3 is a weird question. The answers show it as a degrees sign. The final one ended up being 180/pi sin x(deg) + C. In regards to 2, can you expand on it a bit more? Just not getting it atm, pretty much from line 2 to line 3.

Find \int cos x^{\circ} dx (needed latex)

And

When do I use absolute values when I'm doing these questions. EG: Find the area bounded by the curve y=9-x and the line y=5. I think my teacher roughly mentioned it, but I've forgotten already, is there any golden rule? Or anything to know.

You mean "under" the y-axis? As in the negative side. If so I think I understand what you mean. Thanks. Got another one that I failed at. My limits were 3 and 2, my answer was 24.8333 and the correct answer was 1/6.

Thankfully natural selection exists and you'll be eradicated from the gene pool for being such a fucking ignorant stain on society. Also, I don't see why the stereotype "emo" is given such a hard time. I've been called emo from all the dicktarded wogalicious kids from my middle school, but I...

I just wasn't thinking straight, heh. Also I'm not understanding where the abs. value signs go? Because I keep getting a negative. I did: 9 - 9 -9 - | -1/3 - 1 +3 | Which gives - 10/2/3. From that I'd say it actually goes around the 9s. Though how are you supposed to know where to put the abs...