## Calculus (3rd Edition)

$A(x)=(x+2)^2$
Let $f(x) =2x+4$ The area function with a lower limit $a=-2$ is $A(x)$. $A(x)=\int_{-2}^x(2t+4)dt$ $A(x)=[t^2+4t]_{-2}^x$ $A(x)=(x^2+4x)-((-2)^2-4(2))$ $A(x)=x^2+4x+4$ $A(x)=(x+2)^2$