Recent content by Shannabel

then i should hve 1-(e^x)/(1+e^x)? and use l'hospital's rule again and have -(e^x)/(e^x) =-1?

Homework Statement find the integral from (2 to infinity) of 1/(1+e^x) Homework Equations The Attempt at a Solution lim(t-->infinity) integral from (2 to t) of 1/(1+e^x) 1/1+e^x = (1+e^x-e^x)/(1+e^x) = 1-e^x(1+e6x) when you solve the integral: x-ln(1+e^x) between 2 and t...

thanks!

got it :) one other thing, at the beginning you started with f(f^(-1)(x))=x ... where did that come from?

that makes much more sense, thanks!

well i can't make it look right.. but it's 1/[sin^18(x)]

no, it should be \lim_{x \to0} 1/{sin^18(x)} \int_0^{sin^2(x)} tan^8(t)~dt

Homework Statement evaluate lim(x->0) (tan^8(t))dt(between 0 and sin^2x) Homework Equations The Attempt at a Solution [tan^8(sin^2(x))]/sin^18(x) my book says to use l'hospital's rule, so i continued with [8tan^7(sin^2x)*sec^2(sin^2(X))*2sinxcosx] but my book says i should...

so [f^(-1)(0)]' = 1/[f'(f^(-1)(0))] but where do i go from here? because i don't know what f^(-1)(0) is...

Homework Statement let f(x)=(4t^3+4t)dt(between 2 and x) if g(x) = f^(-1)(x), then g'(0)=? Homework Equations The Attempt at a Solution f'(x) = 4x^3+4x annd i already don't know where to go from here.. help?

lol... thankyou :)

my prof left the part of the course with function like sinh and cosh out of the course this year.

thankyou! :)

Homework Statement solve the integral [abs(x+1)(3+abs(x))]/(x+1) between -3 and 1 Homework Equations The Attempt at a Solution when x<-1 then [abs(x+1)(3+abs(x))]/(x+1) = [-(x+1)(3-x)]/(x+1) = -(3-x) when -1<x<0 then [abs(x+1)(3+abs(x))]/(x+1) = (x+1)(3-x)/(x+1) = 3-x when x>0...

by this same logic, shouldn't i have if x>2, abs(abs(x-2)-abs(x)) = -(abs(x-2-x)) = -(2) since abs(x-2) and abs(x) are positive but abs(abs(x-2)-abs(x)) is negative? i know it should be a positive 2 since that gets me to the right answer... but i don't see why?