7.1 Integration By Parts/25: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{0}^{1}\frac{y}{e^{2y}}dy, | \int_{0}^{1}\frac{y}{e^{2y}}dy, \\[2ex] | ||
& u=y, -du=dy, | & u=y, -du=dy, \\[2ex] | ||
& dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} | & dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} | ||
\end{align} | \end{align} | ||
Revision as of 17:59, 13 December 2022
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{0}^{1}\frac{y}{e^{2y}}dy, \\[2ex] & u=y, -du=dy, \\[2ex] & dv=\frac{dy}{e^{2}y}=e^{-2y}dy, v=\frac{-1}{2}e^{-2y} \end{align} }
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}}}