Recent content by minimoocha

How did Archimedes discover the Quadrature of the parabola without the use of calculus? If someone could please explain, I would be eternally grateful.

Sorry I've been staring at maths all day. Thank you so much for the help! You are a lifesaver!

that makes much more sense- I can proceed :) \[ A=((-1/e)2^2+(2^2+4/e)2-2e^2/2)-((-1/e)0^2+(0^2+4/4)(0)-2e^0/2)) \] which simplifies to \[ A=((-1/e)(4))+((8/e)(2)-2e^1)-(2e^0) \] this is as far as i got :) \( A=((-1/e)(4))+((8/e)(2)-2e)-(2) \)

yes. $$$$ A=[-x^2e+ ((e^3+4x)/e^2))-1/2e^-0.5x] I think?

Actually, because I do not do very hard maths, it is just 0<x<2 That is wayyy to complicated for me to understand. Desmos deformed it because my graphing calculator says different. What I am doing is for my maths assignment: So I had to find the equation of the normal for the point (2,e) Which...

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The area of two lines that I need to find is 2.36, however i need this in exact form. The lines are y=-x/2e+1/e+e the other line is y=e^x/2 Since y=-x/2e+1/e+e is on top it is the first function. A=(the lower boundary is 0 and the top is 2) -x/2e+1/e+e-e^x/2 If you could please help!