Menu Close

c-a-0-a-1-1-f-x-2-dx-2-b-0-b-1-1-f-x-2-dx-2-c-L-1-2-L-2-2-




Question Number 206830 by Fabricista15 last updated on 27/Apr/24
c = (√((∫_a_0  ^a_1  (√(1+[f′(x)]^2 ))dx)^2 +(∫_b_0  ^b_1  (√(1+[f′(x)]^2 ))dx)^2 ))  c = (√(L_1 ^2 +L_2 ^2 ))
c=(a0a11+[f(x)]2dx)2+(b0b11+[f(x)]2dx)2c=L12+L22

Leave a Reply

Your email address will not be published. Required fields are marked *