Recent content by rzaidan

Hi icystrike ; M(e)=1 since e is defined to be the real number such that the area under the function f(x)=1/x and x=1 ,x=e and the x-axis is equal to 1. Best Wishes Riad Zaidan

Hi Mark44 , The problem was in ((removing squares)) from both sides by taking square roots and not in finding another way to solve the problem. In general, thanks a lot for you... Best Regards Riad Zaidan Al-Quds Open University

Hi 1/2" When you take the square root for both sides ,you are to take the absolute values when removing squares; that is t-2=6 or t-2=-6 so you will have t=8 , t=-4 as you get. Best regards Riad Zaidan

Hi g_edgar THankyou for your hint , and my work was in the real numbers. Best Regards Riad Zaidan

Hi vikcool812 : Note that ln(|x|)=ln(x) when x>0 and =ln(-x) when x<0 So d/dx (ln(|x|))=d/dx(ln(x)) when x>0 and d/dx (ln(|x|))=d/dx(ln(-x)) when x<0 therefore d/dx(ln(x)) =1/x when x>0 by definition of ln(x) and d/dx(ln(-x))=-1/-x = 1/x when x<0...

Hi tiny-tim thanks for the detailes Riad Zaidan

Hi Legendre Since you work is in the first quardrant, then θ takes the values from 0 to π/2 and r takes the values from 0 _which corresponds to θ=0 i.e r= 2 sin (2 *0)=0 _ to r= 2 sin (2 θ) since r is a variable and the integral is of the form: π/2 2sin(2θ) π/2...

Hi soopo you can integrata as follows: sec x + tan x ∫ (1/cos x) dx=∫ sec x dx=∫ sec x ______________ dx sec x + tan x (sec x)^2 + sec x tan x...

Hi Mol_Bolom when you find the derivative by definition (d/dx)(f(x)=lim(f(x+h)-f(x))/h you can find the limit of d/dh(f(x+h)-f(x)) h__ 0 ---------------...

Dear arildno the problem was to prove that "If (a+b)/2= sqrt(ab), then a= b".So this is given Best Regards

Hello Thread This is the first time that I participate this site and I have the following proof for your problem: (a-b)^2=a^2-2ab+b^2=a^2+2ab+b^2-4ab=(a+b)^2-4ab =4[((a+b)/2)^2-ab] =4([(a+b)/2)-sqrt{ab})([(a+b)[(a+b)/2)+sqrt{ab})...