Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Preliminary Questions - Page 257: 2

$3$ square units

Since $f$ is a negative function (under the x-axis) it follows that the area between $f$ and the $x$-axis with equation $y=0$ for $x \in[1,3]$ is: $$A=\int_{1}^{3}(0-f(x))~dx=\int_{1}^{3}-f(x)~dx=-\int_{1}^{3}f(x)~dx$$ Using the $FTC$ it follows: $$A=-\int_{1}^{3}f(x)~dx=-(F(3)-F(1))=F(1)-F(3)=7-4=3$$ So the area is $3$ square units.