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

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Problem: 

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

Answer: 

$\int_1^2\frac{1}{x^3}\,dx=3/8$

Solution: 

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
\[
F(2)-F(1)=\frac{3}{8}.
\]