Determine $$\int_1^2\frac{1}{x^3}\,dx.$$

The Fundamental Theorem of Calculus states that
$$\int_1^2\frac{1}{x^3}\,dx=F(2)-F(1)$$
where $F(x)$ is the antiderivative of $f(x)=1/x^3$. We determined $F(x)$ here and found that
$$F(x)=-\frac{1}{2x^2}+C.$$

So now
$$F(2)=-\frac{1}{8}+C$$
and
$$F(1)=-\frac{1}{2}C$$
so that
$$F(2)-F(1)=\frac{3}{8}.$$