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) | \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] | &= \left[-(2)^-3\right]-\left[-(1)^-3\right] \\[2ex] | ||
Revision as of 19:57, 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) &= \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} }