5.3 The Fundamental Theorem of Calculus/25: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 2: | Line 2: | ||
\begin{align} | \begin{align} | ||
\int_{1}^{2}\left(\frac{3}{t^4}\right)dt &= \int_{1}^{2}\left(3t^{-4}\right)\,dt \\[2ex] | \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} \\ | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 20:06, 25 August 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}{-3}}{\bigg |}_{1}^{2}=-t^{-}3{\bigg |}_{1}^{2}\\\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}