6.2 Volumes/3: Difference between revisions

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&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex]
&= \pi\left[left(-\frac{1}{2}\right)-\left(-\frac{1}{1}\right)\right]\Bigg|_1^2\\[2ex]


\end{align}
\end{align}
</math>
</math>

Revision as of 02:08, 29 November 2022

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \pi\int_1^2\left[\left(\frac{1}{x}\right)^2\right]dx & = \pi\int_1^2\left[\left(\frac{1}{x^2}\right)\right]dx \\[2ex] &= \pi\left[-\frac{1}{x}\right]\Bigg|_1^2\\[2ex] &= \pi\left[left(-\frac{1}{2}\right)-\left(-\frac{1}{1}\right)\right]\Bigg|_1^2\\[2ex] \end{align} }