I am trying to integrate my area function A(x) in my code, but it keeps telling me I need to put a "handle" on it. I thought putting bounds would be sufficient, I guess I was wrong. If you can help me, great, below is a a copy of my code. An image of what I am trying to integrate is in the attachments.
% Calculus way (True Values)
E = 30*10^6;
F = 300;
SLOPE = -0.25/6;
D = @(x) (SLOPE*x)+1; % This is the function of Diameter
A = @(x) (pi/4)*(D(x))^2; % This is the function of Area
True_Area = integral(F/(E*A),0,6)