Recent content by dfklajsdfald

i said my "attempt" i never claimed anything?? but anyway nevermind i'll figure it out on my own i just wanted some help and guidance not an argument or a forum of whose wrong whose right i'll delete this shortly

ohh okay that makes more sense. and i wasnt claiming that that's what the antiderivative was. i was saying that was how i was trying to solve the antiderviative but was getting the wrong answer so i needed guidance to go abouts finding the antiderivative correctly. you asked what my initial...

ya but the question is asking for antiderivative

what i meant was that it was my attempt at finding the antiderivative of xex. i know that the derivative of ex is ex therefor the anti derivative would be the same. so then for derivatives with base x you go xn-1 times the derivative of the power but since i need the anti derivative wouldn't i...

i don't know what ur asking, the question is find the antiderivative of the function x to the power of e to the power of x as stated at the start. I am not asking whether or not a solution is correct I am asking how abouts would i go to try and get the antiderivative because I've tried multiple...

actually scratch the last part. my attempt would be xe^x +1 becacuse normally for any function x u go x and add one to the power and divide by it

well i know that the derivative of ex is ex so the anti derivative of ex is ex and then my attempt would be exx but i don't know if that's right or even a good start

Homework Statement find the anti-derivative of xe ^x so its x to the power of e to the power of x Homework EquationsThe Attempt at a Solution i have 0 idea where to even start. this was a question on my quiz today

nope i just checked the question and what i said is exactly what's written. it says: the point (log a, log b) exists on the unit circle. find the value of a times b. round to the nearest thousandth

Homework Statement the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths. Homework Equations x2 + y2 = 1 The Attempt at a Solution x2+y2 = 1 loga2+logb2 =1 2loga+2logb = 1 2(loga+logb) = 1 loga + log b = 0.5 logb = 0.5−loga now i try...