5.3 The Fundamental Theorem of Calculus/25

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

\int_{1}^{2}\left(\frac{3}{t^4}\right)dt = \int_{1}^{2}\left(\{3t^-4}\right) \\

&= \frac{3t^-4}{-3}\bigg|_{1}^{2} = -t^-3 \bigg|_{1}^{2} \\

&= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex]

&= 1-\frac{1}{8} = \frac{7}{8}