6.2 Volumes/3: Difference between revisions

From Mr. V Wiki Math
Jump to navigation Jump to search
No edit summary
No edit summary
Line 10: Line 10:
&= \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}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex]
&= \pi\left[\left(-\frac{1}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right]\Bigg|_1^2\\[2ex]= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \pi\left[-\frac{1}{2}+1\right= \pi\left[\frac{1}{2}\right] \\[2ex]
&= \frac{\pi}{2} \\[2ex]
&= \frac{\pi}{2} \\[2ex]


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

Revision as of 02:28, 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}{\left(2\right)}\right)-\left(-\frac{1}{\left(1\right)}\right)\right] \\[2ex] &= \pi\left[-\frac{1}{2}+1\right= \pi\left[\frac{1}{2}\right] \\[2ex] &= \frac{\pi}{2} \\[2ex] \end{align} }