Answer
$$\frac{{31}}{{160}}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{1}{{{x^6}}}} dx \cr
& {\text{negative exponent}} \cr
& = \int_1^2 {{x^{ - 6}}} dx \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \left( {\frac{{{x^{ - 5}}}}{{ - 5}}} \right)_1^2 \cr
& = - \frac{1}{5}\left( {\frac{1}{{{x^5}}}} \right)_1^2 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \frac{1}{5}\left( {\frac{1}{{{2^5}}} - \frac{1}{{{1^5}}}} \right) \cr
& {\text{simplify}} \cr
& = - \frac{1}{5}\left( { - \frac{{31}}{{32}}} \right) \cr
& = \frac{{31}}{{160}} \cr} $$