7.1 Integration By Parts/25: Difference between revisions
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(Created page with "<math> \int_{0}^{1}\frac{y}{e^{2y}}dy u=y, -du=dy, dv=\frac{dy}{e^{2}y} &=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} </math> <math> \begin{align} \int_{0}^{1}\frac{y}{e^{2y}}dy &=\left[\frac{-1}{2}ye^{-2y}\right]\bigg|_{0}^{1} + \frac{1}{2}\int_{0}{1}e^{-2y}dy \\[2ex] &= \left(\frac{-1}{2}e^{-2}+0\right)-\frac{1}{4}\left[e^{-2y}\right]\bigg|_{0}^(1) \\[2ex] &= \frac{-1}{2}e^{-2}-\frac{1}{4}e^{-2}+\frac{1}{4} \\[2ex] &= \frac{1}{4}-\frac{3}{4}e^{-2} \\[2ex] \end{align} </math>") |
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\int_{0}^{1}\frac{y}{e^{2y}}dy | \int_{0}^{1}\frac{y}{e^{2y}}dy | ||
u=y, -du=dy, | u=y, -du=dy, | ||
dv=\frac{dy}{e^{2}y} | dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} | ||
</math> | </math> | ||
Revision as of 17:55, 13 December 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {y}{e^{2y}}}dy&=\left[{\frac {-1}{2}}ye^{-2y}\right]{\bigg |}_{0}^{1}+{\frac {1}{2}}\int _{0}{1}e^{-2y}dy\\[2ex]&=\left({\frac {-1}{2}}e^{-2}+0\right)-{\frac {1}{4}}\left[e^{-2y}\right]{\bigg |}_{0}^{(}1)\\[2ex]&={\frac {-1}{2}}e^{-2}-{\frac {1}{4}}e^{-2}+{\frac {1}{4}}\\[2ex]&={\frac {1}{4}}-{\frac {3}{4}}e^{-2}\\[2ex]\end{aligned}}}