5.3 The Fundamental Theorem of Calculus/25: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex] | |||
\ | &= \frac{3t^{-4+1}}{-4+1}\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\[2ex] | ||
&= \left[-(2)^{-3}\right]-\left[-(1)^{-3}\right] \\[2ex] | |||
&= \left[-\frac{1}{8}\right]+[1] = \frac{7}{8} | |||
\end{align} | |||
</math> | |||
Latest revision as of 21:01, 6 September 2022
![{\displaystyle {\begin{aligned}\int _{1}^{2}\left({\frac {3}{t^{4}}}\right)dt&=\int _{1}^{2}\left(3t^{-4}\right)\,dt\\[2ex]&={\frac {3t^{-4+1}}{-4+1}}{\bigg |}_{1}^{2}=-t^{-3}{\bigg |}_{1}^{2}\\[2ex]&=\left[-(2)^{-3}\right]-\left[-(1)^{-3}\right]\\[2ex]&=\left[-{\frac {1}{8}}\right]+[1]={\frac {7}{8}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a17338f8b3b287bbf56443dbf56ea5bf932081bd)