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By product rule, \frac{d}{dx}(xe^x) = xe^x + e^x Rearranging this, \frac{d}{dx}(xe^x) - e^x = xe^x Integrating both sides with respect to x with limits x = 0 to 1 \int ^1_0 xe^x dx = \int^1_0 \frac{d}{dx}(xe^x) - e^xdx = [xe^x - e^x]^1_0 = e - e - (0 - 1) = 1

oh, i thought you said today? nevermind then

LOL yes it does. So if you want to use it, get on MSN within the next 2 hrs and 23 minutes.

For future use.

Thanks. :) Was that bear on your pen a few weeks ago?

Haha don't worry.. The method was very confusing at first for me as well. No textbook sufficiently explained it for me and showed proper working out for a question.

Yeah i've realised.. Been sick and making more mistakes lately :/

$We know that $ sin2x = 2sinxcosx$ sin^2xcos^2x = sinxcosx \times sinxcosx = \frac{1}{2}sin2x \times \frac{1}{2}sin2x = \frac{1}{4}sin^2 2x \therefore \int sin^2xcos^2x dx = \frac{1}{4} \int sin^22xdx = \frac{1}{8} \int (1 - cos4x) dx $And go from there$

You should be preparing for your exams. :P

Thought i screwed up my 2U preliminary final. Felt like absolute crap afterwards finding i got different answers to everyone. Few weeks later, exams came back, 83/84. Win. On the note of actually screwing up an assessment, my 3U preliminary half yearly.. Horrible marks. Being too cocky and...

Ahh yeah that makes sense. Should be more careful..

<3

When in doubt, let z = x + iy Let z = x + iy Im(x + iy) = |x + iy| y = \sqrt{x^2 + y^2} y^2 = x^2 + y^2 x^2 = 0 x = 0 However |x^2 + y^2| \geq 0, so the solution is the positive imaginary axis, i.e. y \geq 0 \cap x = 0

Same answer as him lol

1. What i think are difficult/vital concepts Prelim physics: Using the inverse square law troubles some people. Using the correct formulae in moving about and electricity in the home Fully understanding vectors Understanding potential difference Prelim chemistry: Fully understanding and knowing...

Let the other root be B Product of roots: -\alpha \beta = -1 \beta = \frac{1}{\alpha} C

Yep.

Acids: Weak: ethanoic acid, citric acid, oxalic acid, methanoic acid Strong: hydrochloric acid, sulfuric acid, nitric acid, perchloric acid Bases: Weak: sodium carbonate, calcium carbonate, sodium bicarbonate Strong: sodium hydroxide, calcium hydroxide

Talk about electrolysis.

There are no electrons in the solution. When a copper compound dissolutes on water, it forms its copper ion and the anion; no electrons are released. However, when zinc solid is placed in to the solution, the copper ions will grab electrons from the solid zinc as it has a stronger pull for...