## Calculus (3rd Edition)

$A(x)=x^2+4x$
Let $f(x)=2x+4$ The area function with the lower limit $a+0$ is A(x). $A(x)=\int_0^x(2t+4)dt$ $A(x)=[t^2+4t]_0^x$ $A(x)=x^2+4x$