Question 126835
{{{int(4x^2-3x-5,dx)}}}


Remember

{{{int(f(x)+g(x),dx)=int(f(x),dx)+int(g(x),dx)}}}
{{{int(a*f(x),dx)=a*int(f(x),dx)}}}
and
{{{int(x^n,dx)=(x^(n+1)/(n+1))+c}}}


{{{int(4x^2-3x-5,dx)=4x^3/3-3x^2/2-5x+C}}}


{{{dC(x)/dx=3x^2-4x+7}}}  (the rendering system won't allow the C' notation)


{{{C(x)=int(dC(x)/dx,dx)+C}}} and the 500 fixed costs are the C.


So {{{C(x)=int(3x^2-4x+7,dx)+500=3x^3/3-4x^2/2+7+500=x^3-2x^2+507}}}