5.3 The Fundamental Theorem of Calculus/25

${\begin{aligned}\int _{1}^{2}\left({\frac {3}{t^{4}}}\right)dt&=\int _{1}^{2}\left(3t^{-4}\right)\,dt\\[2ex]&={\frac {3t^{-4}}{-3}}{\bigg |}_{1}^{2}=-t^{-3}{\bigg |}_{1}^{2}\\[2ex]&=\left[-(2)^{-3}\right]-\left[-(1)^{-3}\right]\\[2ex]&=1-{\frac {1}{8}}={\frac {7}{8}}\end{aligned}}$