5.3 The Fundamental Theorem of Calculus/25: Difference between revisions
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\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} \\ | &= \frac{3t^{-4}{-3\bigg|_{1}^{2} = -t^{-3} \bigg|_{1}^{2} \\ | ||
&= \left[-(2)^ | &= \left[-(2)^-3\right]-\left[-(1)^{-3}\right] \\[2ex] | ||
&= 1+\frac{-1}{8} = \frac{7}{8} | &= 1+\frac{-1}{8} = \frac{7}{8} | ||
\end{align} | \end{align} | ||
Revision as of 20:05, 25 August 2022
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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{align} }
\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}